How to plot logistic regression decision boundary?












4












$begingroup$


I am running logistic regression on a small dataset which looks like this:



enter image description here



After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.



Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly.



enter image description here



Extracting data



clear all; close all; clc;

alpha = 0.01;
num_iters = 1000;

%% Plotting data
x1 = linspace(0,3,50);
mqtrue = 5;
cqtrue = 30;
dat1 = mqtrue*x1+5*randn(1,50);

x2 = linspace(7,10,50);
dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

x = [x1 x2]'; % X

subplot(2,2,1);
dat = [dat1 dat2]'; % Y

scatter(x1, dat1); hold on;
scatter(x2, dat2, '*'); hold on;
classdata = (dat>40);


Computing Cost, Gradient and plotting



%  Setup the data matrix appropriately, and add ones for the intercept term
[m, n] = size(x);

% Add intercept term to x and X_test
x = [ones(m, 1) x];

% Initialize fitting parameters
theta = zeros(n + 1, 1);
%initial_theta = [0.2; 0.2];

J_history = zeros(num_iters, 1);

plot_x = [min(x(:,2))-2, max(x(:,2))+2]

for iter = 1:num_iters
% Compute and display initial cost and gradient
[cost, grad] = logistic_costFunction(theta, x, classdata);
theta = theta - alpha * grad;
J_history(iter) = cost;

fprintf('Iteration #%d - Cost = %d... rn',iter, cost);


subplot(2,2,2);
hold on; grid on;
plot(iter, J_history(iter), '.r'); title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));
xlabel('Iterations')
ylabel('MSE')
drawnow

subplot(2,2,3);
grid on;
plot3(theta(1), theta(2), J_history(iter),'o')
title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
xlabel('Tita0')
ylabel('Tita1')
zlabel('Cost')
hold on;
drawnow

subplot(2,2,1);
grid on;
% Calculate the decision boundary line
plot_y = theta(2).*plot_x + theta(1); % <--- Boundary line
% Plot, and adjust axes for better viewing
plot(plot_x, plot_y)
hold on;
drawnow

end

fprintf('Cost at initial theta (zeros): %fn', cost);
fprintf('Gradient at initial theta (zeros): n');
fprintf(' %f n', grad);


The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.










share|improve this question









$endgroup$

















    4












    $begingroup$


    I am running logistic regression on a small dataset which looks like this:



    enter image description here



    After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.



    Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly.



    enter image description here



    Extracting data



    clear all; close all; clc;

    alpha = 0.01;
    num_iters = 1000;

    %% Plotting data
    x1 = linspace(0,3,50);
    mqtrue = 5;
    cqtrue = 30;
    dat1 = mqtrue*x1+5*randn(1,50);

    x2 = linspace(7,10,50);
    dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

    x = [x1 x2]'; % X

    subplot(2,2,1);
    dat = [dat1 dat2]'; % Y

    scatter(x1, dat1); hold on;
    scatter(x2, dat2, '*'); hold on;
    classdata = (dat>40);


    Computing Cost, Gradient and plotting



    %  Setup the data matrix appropriately, and add ones for the intercept term
    [m, n] = size(x);

    % Add intercept term to x and X_test
    x = [ones(m, 1) x];

    % Initialize fitting parameters
    theta = zeros(n + 1, 1);
    %initial_theta = [0.2; 0.2];

    J_history = zeros(num_iters, 1);

    plot_x = [min(x(:,2))-2, max(x(:,2))+2]

    for iter = 1:num_iters
    % Compute and display initial cost and gradient
    [cost, grad] = logistic_costFunction(theta, x, classdata);
    theta = theta - alpha * grad;
    J_history(iter) = cost;

    fprintf('Iteration #%d - Cost = %d... rn',iter, cost);


    subplot(2,2,2);
    hold on; grid on;
    plot(iter, J_history(iter), '.r'); title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));
    xlabel('Iterations')
    ylabel('MSE')
    drawnow

    subplot(2,2,3);
    grid on;
    plot3(theta(1), theta(2), J_history(iter),'o')
    title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
    xlabel('Tita0')
    ylabel('Tita1')
    zlabel('Cost')
    hold on;
    drawnow

    subplot(2,2,1);
    grid on;
    % Calculate the decision boundary line
    plot_y = theta(2).*plot_x + theta(1); % <--- Boundary line
    % Plot, and adjust axes for better viewing
    plot(plot_x, plot_y)
    hold on;
    drawnow

    end

    fprintf('Cost at initial theta (zeros): %fn', cost);
    fprintf('Gradient at initial theta (zeros): n');
    fprintf(' %f n', grad);


    The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.










    share|improve this question









    $endgroup$















      4












      4








      4


      1



      $begingroup$


      I am running logistic regression on a small dataset which looks like this:



      enter image description here



      After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.



      Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly.



      enter image description here



      Extracting data



      clear all; close all; clc;

      alpha = 0.01;
      num_iters = 1000;

      %% Plotting data
      x1 = linspace(0,3,50);
      mqtrue = 5;
      cqtrue = 30;
      dat1 = mqtrue*x1+5*randn(1,50);

      x2 = linspace(7,10,50);
      dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

      x = [x1 x2]'; % X

      subplot(2,2,1);
      dat = [dat1 dat2]'; % Y

      scatter(x1, dat1); hold on;
      scatter(x2, dat2, '*'); hold on;
      classdata = (dat>40);


      Computing Cost, Gradient and plotting



      %  Setup the data matrix appropriately, and add ones for the intercept term
      [m, n] = size(x);

      % Add intercept term to x and X_test
      x = [ones(m, 1) x];

      % Initialize fitting parameters
      theta = zeros(n + 1, 1);
      %initial_theta = [0.2; 0.2];

      J_history = zeros(num_iters, 1);

      plot_x = [min(x(:,2))-2, max(x(:,2))+2]

      for iter = 1:num_iters
      % Compute and display initial cost and gradient
      [cost, grad] = logistic_costFunction(theta, x, classdata);
      theta = theta - alpha * grad;
      J_history(iter) = cost;

      fprintf('Iteration #%d - Cost = %d... rn',iter, cost);


      subplot(2,2,2);
      hold on; grid on;
      plot(iter, J_history(iter), '.r'); title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));
      xlabel('Iterations')
      ylabel('MSE')
      drawnow

      subplot(2,2,3);
      grid on;
      plot3(theta(1), theta(2), J_history(iter),'o')
      title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
      xlabel('Tita0')
      ylabel('Tita1')
      zlabel('Cost')
      hold on;
      drawnow

      subplot(2,2,1);
      grid on;
      % Calculate the decision boundary line
      plot_y = theta(2).*plot_x + theta(1); % <--- Boundary line
      % Plot, and adjust axes for better viewing
      plot(plot_x, plot_y)
      hold on;
      drawnow

      end

      fprintf('Cost at initial theta (zeros): %fn', cost);
      fprintf('Gradient at initial theta (zeros): n');
      fprintf(' %f n', grad);


      The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.










      share|improve this question









      $endgroup$




      I am running logistic regression on a small dataset which looks like this:



      enter image description here



      After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.



      Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly.



      enter image description here



      Extracting data



      clear all; close all; clc;

      alpha = 0.01;
      num_iters = 1000;

      %% Plotting data
      x1 = linspace(0,3,50);
      mqtrue = 5;
      cqtrue = 30;
      dat1 = mqtrue*x1+5*randn(1,50);

      x2 = linspace(7,10,50);
      dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

      x = [x1 x2]'; % X

      subplot(2,2,1);
      dat = [dat1 dat2]'; % Y

      scatter(x1, dat1); hold on;
      scatter(x2, dat2, '*'); hold on;
      classdata = (dat>40);


      Computing Cost, Gradient and plotting



      %  Setup the data matrix appropriately, and add ones for the intercept term
      [m, n] = size(x);

      % Add intercept term to x and X_test
      x = [ones(m, 1) x];

      % Initialize fitting parameters
      theta = zeros(n + 1, 1);
      %initial_theta = [0.2; 0.2];

      J_history = zeros(num_iters, 1);

      plot_x = [min(x(:,2))-2, max(x(:,2))+2]

      for iter = 1:num_iters
      % Compute and display initial cost and gradient
      [cost, grad] = logistic_costFunction(theta, x, classdata);
      theta = theta - alpha * grad;
      J_history(iter) = cost;

      fprintf('Iteration #%d - Cost = %d... rn',iter, cost);


      subplot(2,2,2);
      hold on; grid on;
      plot(iter, J_history(iter), '.r'); title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));
      xlabel('Iterations')
      ylabel('MSE')
      drawnow

      subplot(2,2,3);
      grid on;
      plot3(theta(1), theta(2), J_history(iter),'o')
      title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
      xlabel('Tita0')
      ylabel('Tita1')
      zlabel('Cost')
      hold on;
      drawnow

      subplot(2,2,1);
      grid on;
      % Calculate the decision boundary line
      plot_y = theta(2).*plot_x + theta(1); % <--- Boundary line
      % Plot, and adjust axes for better viewing
      plot(plot_x, plot_y)
      hold on;
      drawnow

      end

      fprintf('Cost at initial theta (zeros): %fn', cost);
      fprintf('Gradient at initial theta (zeros): n');
      fprintf(' %f n', grad);


      The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.







      machine-learning logistic-regression






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 21 hours ago









      Rrz0Rrz0

      1738




      1738






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          Your decision boundary is a surface in 3D as your points are in 2D.



          With Wolfram Language



          Create the data sets.



          mqtrue = 5;
          cqtrue = 30;
          With[{x = Subdivide[0, 3, 50]},
          dat1 = Transpose@{x, mqtrue x + 5 RandomReal[1, Length@x]};
          ];
          With[{x = Subdivide[7, 10, 50]},
          dat2 = Transpose@{x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x]};
          ];


          View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.



          ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]



          Mathematica graphics




          I Append the response variable to the data.



          datPlot =
          ListPointPlot3D[
          MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}
          ]



          enter image description here




          Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.



          With[{dat = Join[dat1, dat2]},
          model =
          LogitModelFit[
          MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
          {x, y}, {x, y}]
          ]



          Mathematica graphics




          From the FittedModel "Properties" we need "Function".



          model["Properties"]



          {AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries,
          AIC, DevianceTableEntries, ParameterConfidenceRegion,
          AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors,
          BasisFunctions, DevianceTableResidualDeviances, ParameterPValues,
          BestFit, EfronPseudoRSquared, ParameterTable,
          BestFitParameters, EstimatedDispersion, ParameterTableEntries,
          BIC, FitResiduals, ParameterZStatistics,
          CookDistances, Function, PearsonChiSquare,
          CorrelationMatrix, HatDiagonal, PearsonResiduals,
          CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse,
          CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties,
          CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance,
          Data, LinearPredictor, ResidualDegreesOfFreedom,
          DesignMatrix, LogLikelihood, Response,
          DevianceResiduals, NullDeviance, StandardizedDevianceResiduals,
          Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals,
          DevianceTable, ParameterConfidenceIntervals, WorkingResiduals,
          DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable}




          model["Function"]



          Mathematica graphics




          Use this for prediction



          model["Function"][8, 54]



          0.0196842



          and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D



          modelPlot =
          Show[
          datPlot,
          Plot3D[
          model["Function"][x, y],
          Evaluate[
          Sequence @@
          MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
          Mesh -> None,
          PlotStyle -> Opacity[.25, Green],
          PlotPoints -> 30
          ]
          ]


          enter image description here



          With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.



          Manipulate[
          Show[
          modelPlot,
          ParametricPlot3D[
          {x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
          ParametricPlot3D[
          {u, y, model["Function"][u, y]}, {u, 0, 10}, PlotStyle -> Purple],
          PlotLabel ->
          StringTemplate["model[`1`, `2`] = `3`"] @@ {x, y, model["Function"][x, y]}
          ],
          {{x, 6, Style["x", Orange, Bold]}, 0, 10, Appearance -> "Labeled"},
          {{y, 40, Style["y", Purple, Bold]}, 0, 80, Appearance -> "Labeled"}
          ]


          enter image description here



          You can also project contours of this surface into 2D (Plot). For example, keeping y fixed and letting x vary.



          yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
          Manipulate[
          Show[
          ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers",
          PlotTheme -> "Detailed"],
          Plot[yMax model["Function"][x, y], {x, 0, 10}, PlotStyle -> Purple,
          Exclusions -> None]
          ],
          {{y, 40}, 0, yMax, Appearance -> "Labeled"}
          ]


          enter image description here



          Update



          You can also plot contours of the probability in 2D.



          plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

          Manipulate[
          db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
          Show[
          plot,
          Plot[db, {x, 0, 10}, PlotStyle -> Red]
          ],
          {{p, .5}, 0, 1, Appearance -> "Labeled"}
          ]


          enter image description here



          Hope this helps.






          share|improve this answer











          $endgroup$













          • $begingroup$
            Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbol{x},P(y=1|boldsymbol{x}))$, as you have beautifully illustrated, needs $d+1$ for visualization.
            $endgroup$
            – Esmailian
            10 hours ago






          • 1




            $begingroup$
            @Esmailian See update.
            $endgroup$
            – Edmund
            7 hours ago



















          1












          $begingroup$

          Regarding the code



          You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.



          Decision boundary



          Assuming that input is $boldsymbol{x}=(x_1, x_2)$ ((x, dat) or (x, y) in the code), and parameter is $boldsymbol{theta}=(theta_0, theta_1,theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary:
          $$x_2 = -frac{theta_1}{theta_2} x_1 - frac{theta_0}{theta_2}$$
          which can be drawn as a segment by connecting two points $(0, - frac{theta_0}{theta_2})$ and $(- frac{theta_0}{theta_1}, 0)$.
          However, if $theta_2=0$, the line would be $x_1=-frac{theta_0}{theta_1}$.



          Where this comes from?



          Decision boundary of Logistic regression is the set of all points $boldsymbol{x}$ that satisfy
          $${Bbb P}(y=1|boldsymbol{x})={Bbb P}(y=0|boldsymbol{x}) = frac{1}{2}.$$
          Given
          $${Bbb P}(y=1|boldsymbol{x})=frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}}$$
          where $boldsymbol{theta}=(theta_0, theta_1,cdots,theta_d)$, and $boldsymbol{x}$ is extended to $boldsymbol{x_+}=(1, x_1, cdots, x_d)$ for the sake of readability to have$$boldsymbol{theta}^tboldsymbol{x_+}=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
          decision boundary can be derived as follows
          $$begin{align*}
          &frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{2} \
          &Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = 0\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
          end{align*}$$

          For two dimensional input $boldsymbol{x}=(x_1, x_2)$ we have
          $$begin{align*}
          & theta_0 + theta_1 x_1+theta_2 x_2 = 0 \
          & Rightarrow x_2 = -frac{theta_1}{theta_2} x_1 - frac{theta_0}{theta_2}
          end{align*}$$

          which is the separation line that should be drawn in $(x_1, x_2)$ plane.



          Weighted decision boundary



          If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary:
          $$w{Bbb P}(y=1|boldsymbol{x}) = {Bbb P}(y=0|boldsymbol{x}) = frac{w}{w+1}$$



          For example, $w=2$ means point $boldsymbol{x}$ will be assigned to positive class if ${Bbb P}(y=1|boldsymbol{x}) > 0.33$ (or equivalently if ${Bbb P}(y=0|boldsymbol{x}) < 0.66$), which implies favoring the positive class (increasing the true positive rate).



          Here is the line for this general case:
          $$begin{align*}
          &frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{w+1} \
          &Rightarrow e^{-boldsymbol{theta}^tboldsymbol{x_+}} = w\
          &Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = -text{ln}w\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -text{ln}w
          end{align*}$$






          share|improve this answer











          $endgroup$














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            2 Answers
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            2 Answers
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            4












            $begingroup$

            Your decision boundary is a surface in 3D as your points are in 2D.



            With Wolfram Language



            Create the data sets.



            mqtrue = 5;
            cqtrue = 30;
            With[{x = Subdivide[0, 3, 50]},
            dat1 = Transpose@{x, mqtrue x + 5 RandomReal[1, Length@x]};
            ];
            With[{x = Subdivide[7, 10, 50]},
            dat2 = Transpose@{x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x]};
            ];


            View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.



            ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]



            Mathematica graphics




            I Append the response variable to the data.



            datPlot =
            ListPointPlot3D[
            MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}
            ]



            enter image description here




            Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.



            With[{dat = Join[dat1, dat2]},
            model =
            LogitModelFit[
            MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
            {x, y}, {x, y}]
            ]



            Mathematica graphics




            From the FittedModel "Properties" we need "Function".



            model["Properties"]



            {AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries,
            AIC, DevianceTableEntries, ParameterConfidenceRegion,
            AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors,
            BasisFunctions, DevianceTableResidualDeviances, ParameterPValues,
            BestFit, EfronPseudoRSquared, ParameterTable,
            BestFitParameters, EstimatedDispersion, ParameterTableEntries,
            BIC, FitResiduals, ParameterZStatistics,
            CookDistances, Function, PearsonChiSquare,
            CorrelationMatrix, HatDiagonal, PearsonResiduals,
            CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse,
            CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties,
            CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance,
            Data, LinearPredictor, ResidualDegreesOfFreedom,
            DesignMatrix, LogLikelihood, Response,
            DevianceResiduals, NullDeviance, StandardizedDevianceResiduals,
            Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals,
            DevianceTable, ParameterConfidenceIntervals, WorkingResiduals,
            DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable}




            model["Function"]



            Mathematica graphics




            Use this for prediction



            model["Function"][8, 54]



            0.0196842



            and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D



            modelPlot =
            Show[
            datPlot,
            Plot3D[
            model["Function"][x, y],
            Evaluate[
            Sequence @@
            MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
            Mesh -> None,
            PlotStyle -> Opacity[.25, Green],
            PlotPoints -> 30
            ]
            ]


            enter image description here



            With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.



            Manipulate[
            Show[
            modelPlot,
            ParametricPlot3D[
            {x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
            ParametricPlot3D[
            {u, y, model["Function"][u, y]}, {u, 0, 10}, PlotStyle -> Purple],
            PlotLabel ->
            StringTemplate["model[`1`, `2`] = `3`"] @@ {x, y, model["Function"][x, y]}
            ],
            {{x, 6, Style["x", Orange, Bold]}, 0, 10, Appearance -> "Labeled"},
            {{y, 40, Style["y", Purple, Bold]}, 0, 80, Appearance -> "Labeled"}
            ]


            enter image description here



            You can also project contours of this surface into 2D (Plot). For example, keeping y fixed and letting x vary.



            yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
            Manipulate[
            Show[
            ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers",
            PlotTheme -> "Detailed"],
            Plot[yMax model["Function"][x, y], {x, 0, 10}, PlotStyle -> Purple,
            Exclusions -> None]
            ],
            {{y, 40}, 0, yMax, Appearance -> "Labeled"}
            ]


            enter image description here



            Update



            You can also plot contours of the probability in 2D.



            plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

            Manipulate[
            db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
            Show[
            plot,
            Plot[db, {x, 0, 10}, PlotStyle -> Red]
            ],
            {{p, .5}, 0, 1, Appearance -> "Labeled"}
            ]


            enter image description here



            Hope this helps.






            share|improve this answer











            $endgroup$













            • $begingroup$
              Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbol{x},P(y=1|boldsymbol{x}))$, as you have beautifully illustrated, needs $d+1$ for visualization.
              $endgroup$
              – Esmailian
              10 hours ago






            • 1




              $begingroup$
              @Esmailian See update.
              $endgroup$
              – Edmund
              7 hours ago
















            4












            $begingroup$

            Your decision boundary is a surface in 3D as your points are in 2D.



            With Wolfram Language



            Create the data sets.



            mqtrue = 5;
            cqtrue = 30;
            With[{x = Subdivide[0, 3, 50]},
            dat1 = Transpose@{x, mqtrue x + 5 RandomReal[1, Length@x]};
            ];
            With[{x = Subdivide[7, 10, 50]},
            dat2 = Transpose@{x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x]};
            ];


            View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.



            ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]



            Mathematica graphics




            I Append the response variable to the data.



            datPlot =
            ListPointPlot3D[
            MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}
            ]



            enter image description here




            Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.



            With[{dat = Join[dat1, dat2]},
            model =
            LogitModelFit[
            MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
            {x, y}, {x, y}]
            ]



            Mathematica graphics




            From the FittedModel "Properties" we need "Function".



            model["Properties"]



            {AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries,
            AIC, DevianceTableEntries, ParameterConfidenceRegion,
            AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors,
            BasisFunctions, DevianceTableResidualDeviances, ParameterPValues,
            BestFit, EfronPseudoRSquared, ParameterTable,
            BestFitParameters, EstimatedDispersion, ParameterTableEntries,
            BIC, FitResiduals, ParameterZStatistics,
            CookDistances, Function, PearsonChiSquare,
            CorrelationMatrix, HatDiagonal, PearsonResiduals,
            CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse,
            CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties,
            CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance,
            Data, LinearPredictor, ResidualDegreesOfFreedom,
            DesignMatrix, LogLikelihood, Response,
            DevianceResiduals, NullDeviance, StandardizedDevianceResiduals,
            Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals,
            DevianceTable, ParameterConfidenceIntervals, WorkingResiduals,
            DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable}




            model["Function"]



            Mathematica graphics




            Use this for prediction



            model["Function"][8, 54]



            0.0196842



            and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D



            modelPlot =
            Show[
            datPlot,
            Plot3D[
            model["Function"][x, y],
            Evaluate[
            Sequence @@
            MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
            Mesh -> None,
            PlotStyle -> Opacity[.25, Green],
            PlotPoints -> 30
            ]
            ]


            enter image description here



            With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.



            Manipulate[
            Show[
            modelPlot,
            ParametricPlot3D[
            {x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
            ParametricPlot3D[
            {u, y, model["Function"][u, y]}, {u, 0, 10}, PlotStyle -> Purple],
            PlotLabel ->
            StringTemplate["model[`1`, `2`] = `3`"] @@ {x, y, model["Function"][x, y]}
            ],
            {{x, 6, Style["x", Orange, Bold]}, 0, 10, Appearance -> "Labeled"},
            {{y, 40, Style["y", Purple, Bold]}, 0, 80, Appearance -> "Labeled"}
            ]


            enter image description here



            You can also project contours of this surface into 2D (Plot). For example, keeping y fixed and letting x vary.



            yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
            Manipulate[
            Show[
            ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers",
            PlotTheme -> "Detailed"],
            Plot[yMax model["Function"][x, y], {x, 0, 10}, PlotStyle -> Purple,
            Exclusions -> None]
            ],
            {{y, 40}, 0, yMax, Appearance -> "Labeled"}
            ]


            enter image description here



            Update



            You can also plot contours of the probability in 2D.



            plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

            Manipulate[
            db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
            Show[
            plot,
            Plot[db, {x, 0, 10}, PlotStyle -> Red]
            ],
            {{p, .5}, 0, 1, Appearance -> "Labeled"}
            ]


            enter image description here



            Hope this helps.






            share|improve this answer











            $endgroup$













            • $begingroup$
              Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbol{x},P(y=1|boldsymbol{x}))$, as you have beautifully illustrated, needs $d+1$ for visualization.
              $endgroup$
              – Esmailian
              10 hours ago






            • 1




              $begingroup$
              @Esmailian See update.
              $endgroup$
              – Edmund
              7 hours ago














            4












            4








            4





            $begingroup$

            Your decision boundary is a surface in 3D as your points are in 2D.



            With Wolfram Language



            Create the data sets.



            mqtrue = 5;
            cqtrue = 30;
            With[{x = Subdivide[0, 3, 50]},
            dat1 = Transpose@{x, mqtrue x + 5 RandomReal[1, Length@x]};
            ];
            With[{x = Subdivide[7, 10, 50]},
            dat2 = Transpose@{x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x]};
            ];


            View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.



            ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]



            Mathematica graphics




            I Append the response variable to the data.



            datPlot =
            ListPointPlot3D[
            MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}
            ]



            enter image description here




            Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.



            With[{dat = Join[dat1, dat2]},
            model =
            LogitModelFit[
            MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
            {x, y}, {x, y}]
            ]



            Mathematica graphics




            From the FittedModel "Properties" we need "Function".



            model["Properties"]



            {AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries,
            AIC, DevianceTableEntries, ParameterConfidenceRegion,
            AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors,
            BasisFunctions, DevianceTableResidualDeviances, ParameterPValues,
            BestFit, EfronPseudoRSquared, ParameterTable,
            BestFitParameters, EstimatedDispersion, ParameterTableEntries,
            BIC, FitResiduals, ParameterZStatistics,
            CookDistances, Function, PearsonChiSquare,
            CorrelationMatrix, HatDiagonal, PearsonResiduals,
            CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse,
            CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties,
            CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance,
            Data, LinearPredictor, ResidualDegreesOfFreedom,
            DesignMatrix, LogLikelihood, Response,
            DevianceResiduals, NullDeviance, StandardizedDevianceResiduals,
            Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals,
            DevianceTable, ParameterConfidenceIntervals, WorkingResiduals,
            DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable}




            model["Function"]



            Mathematica graphics




            Use this for prediction



            model["Function"][8, 54]



            0.0196842



            and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D



            modelPlot =
            Show[
            datPlot,
            Plot3D[
            model["Function"][x, y],
            Evaluate[
            Sequence @@
            MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
            Mesh -> None,
            PlotStyle -> Opacity[.25, Green],
            PlotPoints -> 30
            ]
            ]


            enter image description here



            With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.



            Manipulate[
            Show[
            modelPlot,
            ParametricPlot3D[
            {x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
            ParametricPlot3D[
            {u, y, model["Function"][u, y]}, {u, 0, 10}, PlotStyle -> Purple],
            PlotLabel ->
            StringTemplate["model[`1`, `2`] = `3`"] @@ {x, y, model["Function"][x, y]}
            ],
            {{x, 6, Style["x", Orange, Bold]}, 0, 10, Appearance -> "Labeled"},
            {{y, 40, Style["y", Purple, Bold]}, 0, 80, Appearance -> "Labeled"}
            ]


            enter image description here



            You can also project contours of this surface into 2D (Plot). For example, keeping y fixed and letting x vary.



            yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
            Manipulate[
            Show[
            ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers",
            PlotTheme -> "Detailed"],
            Plot[yMax model["Function"][x, y], {x, 0, 10}, PlotStyle -> Purple,
            Exclusions -> None]
            ],
            {{y, 40}, 0, yMax, Appearance -> "Labeled"}
            ]


            enter image description here



            Update



            You can also plot contours of the probability in 2D.



            plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

            Manipulate[
            db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
            Show[
            plot,
            Plot[db, {x, 0, 10}, PlotStyle -> Red]
            ],
            {{p, .5}, 0, 1, Appearance -> "Labeled"}
            ]


            enter image description here



            Hope this helps.






            share|improve this answer











            $endgroup$



            Your decision boundary is a surface in 3D as your points are in 2D.



            With Wolfram Language



            Create the data sets.



            mqtrue = 5;
            cqtrue = 30;
            With[{x = Subdivide[0, 3, 50]},
            dat1 = Transpose@{x, mqtrue x + 5 RandomReal[1, Length@x]};
            ];
            With[{x = Subdivide[7, 10, 50]},
            dat2 = Transpose@{x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x]};
            ];


            View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.



            ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]



            Mathematica graphics




            I Append the response variable to the data.



            datPlot =
            ListPointPlot3D[
            MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}
            ]



            enter image description here




            Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.



            With[{dat = Join[dat1, dat2]},
            model =
            LogitModelFit[
            MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
            {x, y}, {x, y}]
            ]



            Mathematica graphics




            From the FittedModel "Properties" we need "Function".



            model["Properties"]



            {AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries,
            AIC, DevianceTableEntries, ParameterConfidenceRegion,
            AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors,
            BasisFunctions, DevianceTableResidualDeviances, ParameterPValues,
            BestFit, EfronPseudoRSquared, ParameterTable,
            BestFitParameters, EstimatedDispersion, ParameterTableEntries,
            BIC, FitResiduals, ParameterZStatistics,
            CookDistances, Function, PearsonChiSquare,
            CorrelationMatrix, HatDiagonal, PearsonResiduals,
            CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse,
            CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties,
            CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance,
            Data, LinearPredictor, ResidualDegreesOfFreedom,
            DesignMatrix, LogLikelihood, Response,
            DevianceResiduals, NullDeviance, StandardizedDevianceResiduals,
            Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals,
            DevianceTable, ParameterConfidenceIntervals, WorkingResiduals,
            DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable}




            model["Function"]



            Mathematica graphics




            Use this for prediction



            model["Function"][8, 54]



            0.0196842



            and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D



            modelPlot =
            Show[
            datPlot,
            Plot3D[
            model["Function"][x, y],
            Evaluate[
            Sequence @@
            MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
            Mesh -> None,
            PlotStyle -> Opacity[.25, Green],
            PlotPoints -> 30
            ]
            ]


            enter image description here



            With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.



            Manipulate[
            Show[
            modelPlot,
            ParametricPlot3D[
            {x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
            ParametricPlot3D[
            {u, y, model["Function"][u, y]}, {u, 0, 10}, PlotStyle -> Purple],
            PlotLabel ->
            StringTemplate["model[`1`, `2`] = `3`"] @@ {x, y, model["Function"][x, y]}
            ],
            {{x, 6, Style["x", Orange, Bold]}, 0, 10, Appearance -> "Labeled"},
            {{y, 40, Style["y", Purple, Bold]}, 0, 80, Appearance -> "Labeled"}
            ]


            enter image description here



            You can also project contours of this surface into 2D (Plot). For example, keeping y fixed and letting x vary.



            yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
            Manipulate[
            Show[
            ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers",
            PlotTheme -> "Detailed"],
            Plot[yMax model["Function"][x, y], {x, 0, 10}, PlotStyle -> Purple,
            Exclusions -> None]
            ],
            {{y, 40}, 0, yMax, Appearance -> "Labeled"}
            ]


            enter image description here



            Update



            You can also plot contours of the probability in 2D.



            plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

            Manipulate[
            db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
            Show[
            plot,
            Plot[db, {x, 0, 10}, PlotStyle -> Red]
            ],
            {{p, .5}, 0, 1, Appearance -> "Labeled"}
            ]


            enter image description here



            Hope this helps.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 7 hours ago

























            answered 16 hours ago









            EdmundEdmund

            250311




            250311












            • $begingroup$
              Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbol{x},P(y=1|boldsymbol{x}))$, as you have beautifully illustrated, needs $d+1$ for visualization.
              $endgroup$
              – Esmailian
              10 hours ago






            • 1




              $begingroup$
              @Esmailian See update.
              $endgroup$
              – Edmund
              7 hours ago


















            • $begingroup$
              Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbol{x},P(y=1|boldsymbol{x}))$, as you have beautifully illustrated, needs $d+1$ for visualization.
              $endgroup$
              – Esmailian
              10 hours ago






            • 1




              $begingroup$
              @Esmailian See update.
              $endgroup$
              – Edmund
              7 hours ago
















            $begingroup$
            Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbol{x},P(y=1|boldsymbol{x}))$, as you have beautifully illustrated, needs $d+1$ for visualization.
            $endgroup$
            – Esmailian
            10 hours ago




            $begingroup$
            Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbol{x},P(y=1|boldsymbol{x}))$, as you have beautifully illustrated, needs $d+1$ for visualization.
            $endgroup$
            – Esmailian
            10 hours ago




            1




            1




            $begingroup$
            @Esmailian See update.
            $endgroup$
            – Edmund
            7 hours ago




            $begingroup$
            @Esmailian See update.
            $endgroup$
            – Edmund
            7 hours ago











            1












            $begingroup$

            Regarding the code



            You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.



            Decision boundary



            Assuming that input is $boldsymbol{x}=(x_1, x_2)$ ((x, dat) or (x, y) in the code), and parameter is $boldsymbol{theta}=(theta_0, theta_1,theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary:
            $$x_2 = -frac{theta_1}{theta_2} x_1 - frac{theta_0}{theta_2}$$
            which can be drawn as a segment by connecting two points $(0, - frac{theta_0}{theta_2})$ and $(- frac{theta_0}{theta_1}, 0)$.
            However, if $theta_2=0$, the line would be $x_1=-frac{theta_0}{theta_1}$.



            Where this comes from?



            Decision boundary of Logistic regression is the set of all points $boldsymbol{x}$ that satisfy
            $${Bbb P}(y=1|boldsymbol{x})={Bbb P}(y=0|boldsymbol{x}) = frac{1}{2}.$$
            Given
            $${Bbb P}(y=1|boldsymbol{x})=frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}}$$
            where $boldsymbol{theta}=(theta_0, theta_1,cdots,theta_d)$, and $boldsymbol{x}$ is extended to $boldsymbol{x_+}=(1, x_1, cdots, x_d)$ for the sake of readability to have$$boldsymbol{theta}^tboldsymbol{x_+}=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
            decision boundary can be derived as follows
            $$begin{align*}
            &frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{2} \
            &Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = 0\
            &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
            end{align*}$$

            For two dimensional input $boldsymbol{x}=(x_1, x_2)$ we have
            $$begin{align*}
            & theta_0 + theta_1 x_1+theta_2 x_2 = 0 \
            & Rightarrow x_2 = -frac{theta_1}{theta_2} x_1 - frac{theta_0}{theta_2}
            end{align*}$$

            which is the separation line that should be drawn in $(x_1, x_2)$ plane.



            Weighted decision boundary



            If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary:
            $$w{Bbb P}(y=1|boldsymbol{x}) = {Bbb P}(y=0|boldsymbol{x}) = frac{w}{w+1}$$



            For example, $w=2$ means point $boldsymbol{x}$ will be assigned to positive class if ${Bbb P}(y=1|boldsymbol{x}) > 0.33$ (or equivalently if ${Bbb P}(y=0|boldsymbol{x}) < 0.66$), which implies favoring the positive class (increasing the true positive rate).



            Here is the line for this general case:
            $$begin{align*}
            &frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{w+1} \
            &Rightarrow e^{-boldsymbol{theta}^tboldsymbol{x_+}} = w\
            &Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = -text{ln}w\
            &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -text{ln}w
            end{align*}$$






            share|improve this answer











            $endgroup$


















              1












              $begingroup$

              Regarding the code



              You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.



              Decision boundary



              Assuming that input is $boldsymbol{x}=(x_1, x_2)$ ((x, dat) or (x, y) in the code), and parameter is $boldsymbol{theta}=(theta_0, theta_1,theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary:
              $$x_2 = -frac{theta_1}{theta_2} x_1 - frac{theta_0}{theta_2}$$
              which can be drawn as a segment by connecting two points $(0, - frac{theta_0}{theta_2})$ and $(- frac{theta_0}{theta_1}, 0)$.
              However, if $theta_2=0$, the line would be $x_1=-frac{theta_0}{theta_1}$.



              Where this comes from?



              Decision boundary of Logistic regression is the set of all points $boldsymbol{x}$ that satisfy
              $${Bbb P}(y=1|boldsymbol{x})={Bbb P}(y=0|boldsymbol{x}) = frac{1}{2}.$$
              Given
              $${Bbb P}(y=1|boldsymbol{x})=frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}}$$
              where $boldsymbol{theta}=(theta_0, theta_1,cdots,theta_d)$, and $boldsymbol{x}$ is extended to $boldsymbol{x_+}=(1, x_1, cdots, x_d)$ for the sake of readability to have$$boldsymbol{theta}^tboldsymbol{x_+}=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
              decision boundary can be derived as follows
              $$begin{align*}
              &frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{2} \
              &Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = 0\
              &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
              end{align*}$$

              For two dimensional input $boldsymbol{x}=(x_1, x_2)$ we have
              $$begin{align*}
              & theta_0 + theta_1 x_1+theta_2 x_2 = 0 \
              & Rightarrow x_2 = -frac{theta_1}{theta_2} x_1 - frac{theta_0}{theta_2}
              end{align*}$$

              which is the separation line that should be drawn in $(x_1, x_2)$ plane.



              Weighted decision boundary



              If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary:
              $$w{Bbb P}(y=1|boldsymbol{x}) = {Bbb P}(y=0|boldsymbol{x}) = frac{w}{w+1}$$



              For example, $w=2$ means point $boldsymbol{x}$ will be assigned to positive class if ${Bbb P}(y=1|boldsymbol{x}) > 0.33$ (or equivalently if ${Bbb P}(y=0|boldsymbol{x}) < 0.66$), which implies favoring the positive class (increasing the true positive rate).



              Here is the line for this general case:
              $$begin{align*}
              &frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{w+1} \
              &Rightarrow e^{-boldsymbol{theta}^tboldsymbol{x_+}} = w\
              &Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = -text{ln}w\
              &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -text{ln}w
              end{align*}$$






              share|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Regarding the code



                You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.



                Decision boundary



                Assuming that input is $boldsymbol{x}=(x_1, x_2)$ ((x, dat) or (x, y) in the code), and parameter is $boldsymbol{theta}=(theta_0, theta_1,theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary:
                $$x_2 = -frac{theta_1}{theta_2} x_1 - frac{theta_0}{theta_2}$$
                which can be drawn as a segment by connecting two points $(0, - frac{theta_0}{theta_2})$ and $(- frac{theta_0}{theta_1}, 0)$.
                However, if $theta_2=0$, the line would be $x_1=-frac{theta_0}{theta_1}$.



                Where this comes from?



                Decision boundary of Logistic regression is the set of all points $boldsymbol{x}$ that satisfy
                $${Bbb P}(y=1|boldsymbol{x})={Bbb P}(y=0|boldsymbol{x}) = frac{1}{2}.$$
                Given
                $${Bbb P}(y=1|boldsymbol{x})=frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}}$$
                where $boldsymbol{theta}=(theta_0, theta_1,cdots,theta_d)$, and $boldsymbol{x}$ is extended to $boldsymbol{x_+}=(1, x_1, cdots, x_d)$ for the sake of readability to have$$boldsymbol{theta}^tboldsymbol{x_+}=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
                decision boundary can be derived as follows
                $$begin{align*}
                &frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{2} \
                &Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = 0\
                &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
                end{align*}$$

                For two dimensional input $boldsymbol{x}=(x_1, x_2)$ we have
                $$begin{align*}
                & theta_0 + theta_1 x_1+theta_2 x_2 = 0 \
                & Rightarrow x_2 = -frac{theta_1}{theta_2} x_1 - frac{theta_0}{theta_2}
                end{align*}$$

                which is the separation line that should be drawn in $(x_1, x_2)$ plane.



                Weighted decision boundary



                If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary:
                $$w{Bbb P}(y=1|boldsymbol{x}) = {Bbb P}(y=0|boldsymbol{x}) = frac{w}{w+1}$$



                For example, $w=2$ means point $boldsymbol{x}$ will be assigned to positive class if ${Bbb P}(y=1|boldsymbol{x}) > 0.33$ (or equivalently if ${Bbb P}(y=0|boldsymbol{x}) < 0.66$), which implies favoring the positive class (increasing the true positive rate).



                Here is the line for this general case:
                $$begin{align*}
                &frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{w+1} \
                &Rightarrow e^{-boldsymbol{theta}^tboldsymbol{x_+}} = w\
                &Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = -text{ln}w\
                &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -text{ln}w
                end{align*}$$






                share|improve this answer











                $endgroup$



                Regarding the code



                You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.



                Decision boundary



                Assuming that input is $boldsymbol{x}=(x_1, x_2)$ ((x, dat) or (x, y) in the code), and parameter is $boldsymbol{theta}=(theta_0, theta_1,theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary:
                $$x_2 = -frac{theta_1}{theta_2} x_1 - frac{theta_0}{theta_2}$$
                which can be drawn as a segment by connecting two points $(0, - frac{theta_0}{theta_2})$ and $(- frac{theta_0}{theta_1}, 0)$.
                However, if $theta_2=0$, the line would be $x_1=-frac{theta_0}{theta_1}$.



                Where this comes from?



                Decision boundary of Logistic regression is the set of all points $boldsymbol{x}$ that satisfy
                $${Bbb P}(y=1|boldsymbol{x})={Bbb P}(y=0|boldsymbol{x}) = frac{1}{2}.$$
                Given
                $${Bbb P}(y=1|boldsymbol{x})=frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}}$$
                where $boldsymbol{theta}=(theta_0, theta_1,cdots,theta_d)$, and $boldsymbol{x}$ is extended to $boldsymbol{x_+}=(1, x_1, cdots, x_d)$ for the sake of readability to have$$boldsymbol{theta}^tboldsymbol{x_+}=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
                decision boundary can be derived as follows
                $$begin{align*}
                &frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{2} \
                &Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = 0\
                &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
                end{align*}$$

                For two dimensional input $boldsymbol{x}=(x_1, x_2)$ we have
                $$begin{align*}
                & theta_0 + theta_1 x_1+theta_2 x_2 = 0 \
                & Rightarrow x_2 = -frac{theta_1}{theta_2} x_1 - frac{theta_0}{theta_2}
                end{align*}$$

                which is the separation line that should be drawn in $(x_1, x_2)$ plane.



                Weighted decision boundary



                If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary:
                $$w{Bbb P}(y=1|boldsymbol{x}) = {Bbb P}(y=0|boldsymbol{x}) = frac{w}{w+1}$$



                For example, $w=2$ means point $boldsymbol{x}$ will be assigned to positive class if ${Bbb P}(y=1|boldsymbol{x}) > 0.33$ (or equivalently if ${Bbb P}(y=0|boldsymbol{x}) < 0.66$), which implies favoring the positive class (increasing the true positive rate).



                Here is the line for this general case:
                $$begin{align*}
                &frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{w+1} \
                &Rightarrow e^{-boldsymbol{theta}^tboldsymbol{x_+}} = w\
                &Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = -text{ln}w\
                &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -text{ln}w
                end{align*}$$







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 12 hours ago

























                answered 19 hours ago









                EsmailianEsmailian

                3,486420




                3,486420






























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