Transformation of random variables and joint distributions
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Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation
$y_i$ ~ NormalDistribution[0,$σ_y$ ]
I want to obtain with Mathematica:
The distribution of:
$x = bar{y} = frac {sum_{i=1}^ny_i}{n}$The joint distribution of $ (x,y_i )$
probability-or-statistics distributions
New contributor
$endgroup$
add a comment |
$begingroup$
Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation
$y_i$ ~ NormalDistribution[0,$σ_y$ ]
I want to obtain with Mathematica:
The distribution of:
$x = bar{y} = frac {sum_{i=1}^ny_i}{n}$The joint distribution of $ (x,y_i )$
probability-or-statistics distributions
New contributor
$endgroup$
5
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What have you tried? For example, have you seen the documentation onTransformedDistribution
andProbabilityDistribution
?
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– JimB
2 days ago
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@JimB . I tried thisTransformedDistribution[Sum[y, {i, n}]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result isNormalDistribution[0, [Sigma]y]
. However, the correct result should beNormalDistribution[0, [Sigma]y / Sqrt[n]]
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– Andrea2810
2 days ago
1
$begingroup$
You need to "index" the variabley
or else Mathematica thinks it is a single variable.
$endgroup$
– JimB
2 days ago
add a comment |
$begingroup$
Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation
$y_i$ ~ NormalDistribution[0,$σ_y$ ]
I want to obtain with Mathematica:
The distribution of:
$x = bar{y} = frac {sum_{i=1}^ny_i}{n}$The joint distribution of $ (x,y_i )$
probability-or-statistics distributions
New contributor
$endgroup$
Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation
$y_i$ ~ NormalDistribution[0,$σ_y$ ]
I want to obtain with Mathematica:
The distribution of:
$x = bar{y} = frac {sum_{i=1}^ny_i}{n}$The joint distribution of $ (x,y_i )$
probability-or-statistics distributions
probability-or-statistics distributions
New contributor
New contributor
edited yesterday
J. M. is slightly pensive♦
98.5k10309466
98.5k10309466
New contributor
asked 2 days ago
Andrea2810Andrea2810
334
334
New contributor
New contributor
5
$begingroup$
What have you tried? For example, have you seen the documentation onTransformedDistribution
andProbabilityDistribution
?
$endgroup$
– JimB
2 days ago
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@JimB . I tried thisTransformedDistribution[Sum[y, {i, n}]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result isNormalDistribution[0, [Sigma]y]
. However, the correct result should beNormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
2 days ago
1
$begingroup$
You need to "index" the variabley
or else Mathematica thinks it is a single variable.
$endgroup$
– JimB
2 days ago
add a comment |
5
$begingroup$
What have you tried? For example, have you seen the documentation onTransformedDistribution
andProbabilityDistribution
?
$endgroup$
– JimB
2 days ago
$begingroup$
@JimB . I tried thisTransformedDistribution[Sum[y, {i, n}]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result isNormalDistribution[0, [Sigma]y]
. However, the correct result should beNormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
2 days ago
1
$begingroup$
You need to "index" the variabley
or else Mathematica thinks it is a single variable.
$endgroup$
– JimB
2 days ago
5
5
$begingroup$
What have you tried? For example, have you seen the documentation on
TransformedDistribution
and ProbabilityDistribution
?$endgroup$
– JimB
2 days ago
$begingroup$
What have you tried? For example, have you seen the documentation on
TransformedDistribution
and ProbabilityDistribution
?$endgroup$
– JimB
2 days ago
$begingroup$
@JimB . I tried this
TransformedDistribution[Sum[y, {i, n}]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result is NormalDistribution[0, [Sigma]y]
. However, the correct result should be NormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
2 days ago
$begingroup$
@JimB . I tried this
TransformedDistribution[Sum[y, {i, n}]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result is NormalDistribution[0, [Sigma]y]
. However, the correct result should be NormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
2 days ago
1
1
$begingroup$
You need to "index" the variable
y
or else Mathematica thinks it is a single variable.$endgroup$
– JimB
2 days ago
$begingroup$
You need to "index" the variable
y
or else Mathematica thinks it is a single variable.$endgroup$
– JimB
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
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I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.
First the distribution of the mean:
marginalDistribution = TransformedDistribution[Sum[y[i], {i, n}]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], {i, n}],
Assumptions -> [Sigma] > 0]
{#, marginalDistribution/.n->#} &/@Range[2,10]
$$
begin{array}{cc}
2 & text{NormalDistribution}left[0,frac{sigma }{sqrt{2}}right] \
3 & text{NormalDistribution}left[0,frac{sigma }{sqrt{3}}right] \
4 & text{NormalDistribution}left[0,frac{sigma }{2}right] \
5 & text{NormalDistribution}left[0,frac{sigma }{sqrt{5}}right] \
6 & text{NormalDistribution}left[0,frac{sigma }{sqrt{6}}right] \
7 & text{NormalDistribution}left[0,frac{sigma }{sqrt{7}}right] \
8 & text{NormalDistribution}left[0,frac{sigma }{2 sqrt{2}}right] \
9 & text{NormalDistribution}left[0,frac{sigma }{3}right] \
10 & text{NormalDistribution}left[0,frac{sigma }{sqrt{10}}right] \
end{array}
$$
So we see that the marginal distribution of $bar{y}$ is
NormalDistribution[0, σ/Sqrt[n]]
The joint distribution of $bar{y}$ and, say, $y_1$ is given by
jointDistribution = TransformedDistribution[{y[1], Sum[y[i], {i, n}]/n},
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], {i, n}]]
{#, jointDistribution /. n -> #} & /@ Range[2, 10] // TableForm
$$
begin{array}{cc}
2 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{2} \
frac{sigma ^2}{2} & frac{sigma ^2}{2} \
end{array}
right)right] \
3 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{3} \
frac{sigma ^2}{3} & frac{sigma ^2}{3} \
end{array}
right)right] \
4 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{4} \
frac{sigma ^2}{4} & frac{sigma ^2}{4} \
end{array}
right)right] \
5 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{5} \
frac{sigma ^2}{5} & frac{sigma ^2}{5} \
end{array}
right)right] \
6 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{6} \
frac{sigma ^2}{6} & frac{sigma ^2}{6} \
end{array}
right)right] \
7 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{7} \
frac{sigma ^2}{7} & frac{sigma ^2}{7} \
end{array}
right)right] \
8 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{8} \
frac{sigma ^2}{8} & frac{sigma ^2}{8} \
end{array}
right)right] \
9 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{9} \
frac{sigma ^2}{9} & frac{sigma ^2}{9} \
end{array}
right)right] \
10 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{10} \
frac{sigma ^2}{10} & frac{sigma ^2}{10} \
end{array}
right)right] \
end{array}
$$
So the general distribution is a multivariate normal
MultinormalDistribution[{0, 0}, {{σ^2, σ^2/n}, {σ^2/n, σ^2/n}}]
The general form of the joint density function can then be found with
FullSimplify[PDF[MultinormalDistribution[{0, 0}, {{σ^2, σ^2/n}, {σ^2/n, σ^2/n}}], {y, ybar}],
Assumptions -> {σ > 0, n > 1}]
$$frac{n e^{-frac{n left(n text{ybar}^2+y^2-2 y text{ybar}right)}{2 (n-1) sigma ^2}}}{2 pi sqrt{n-1} sigma ^2}$$
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Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
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– mjw
2 days ago
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@mjw Good. Answers should always be scrutinized and challenged if desired.
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– JimB
2 days ago
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Nice! In addition to trying to understand the technique, I checked the marginal integrals. Looks great!
$endgroup$
– mjw
2 days ago
add a comment |
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Here is the distribution of $x=overline{y}$ (Part I of your question):
a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
p[n_] := TransformedDistribution[Sum[y[k]/n, {k, n}], a[n]];
Now
x [Distributed] p[5] (* n=5, for example *)
The result is
x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
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I am not sure, but shouldn't be n instead of 5 hereTransformedDistribution[Sum[y[k]/n, {k, 5}], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw
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– Andrea2810
2 days ago
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Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
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– mjw
2 days ago
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Let's go with five because it is clearer. The result isNormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica putsAbs
around $sigma$. Obviously, $sigma>0$.
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– mjw
2 days ago
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Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
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– Andrea2810
2 days ago
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a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
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– mjw
2 days ago
|
show 5 more comments
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just modified @mjw's answer,
n = 100;(*for example*)ClearAll[y];
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
meanDist = TransformedDistribution[Sum[y[k]/100, {k, 100}], a]
JointDistribution can be composed by ProductDistribution,
if these random variables are independent.
if not,you have to use Copula
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, {100000}];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, {100000}];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]
=>
{{1., -0.00256777}, {-0.00256777, 1.}}
I'm not sure about correlation,but it's okay.
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I believe that the distributions are not independent. Since $overline{x}$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
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– mjw
2 days ago
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I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
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– mjw
2 days ago
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Exactly, the two variables are not independent unfortunately
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– Andrea2810
2 days ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.
First the distribution of the mean:
marginalDistribution = TransformedDistribution[Sum[y[i], {i, n}]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], {i, n}],
Assumptions -> [Sigma] > 0]
{#, marginalDistribution/.n->#} &/@Range[2,10]
$$
begin{array}{cc}
2 & text{NormalDistribution}left[0,frac{sigma }{sqrt{2}}right] \
3 & text{NormalDistribution}left[0,frac{sigma }{sqrt{3}}right] \
4 & text{NormalDistribution}left[0,frac{sigma }{2}right] \
5 & text{NormalDistribution}left[0,frac{sigma }{sqrt{5}}right] \
6 & text{NormalDistribution}left[0,frac{sigma }{sqrt{6}}right] \
7 & text{NormalDistribution}left[0,frac{sigma }{sqrt{7}}right] \
8 & text{NormalDistribution}left[0,frac{sigma }{2 sqrt{2}}right] \
9 & text{NormalDistribution}left[0,frac{sigma }{3}right] \
10 & text{NormalDistribution}left[0,frac{sigma }{sqrt{10}}right] \
end{array}
$$
So we see that the marginal distribution of $bar{y}$ is
NormalDistribution[0, σ/Sqrt[n]]
The joint distribution of $bar{y}$ and, say, $y_1$ is given by
jointDistribution = TransformedDistribution[{y[1], Sum[y[i], {i, n}]/n},
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], {i, n}]]
{#, jointDistribution /. n -> #} & /@ Range[2, 10] // TableForm
$$
begin{array}{cc}
2 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{2} \
frac{sigma ^2}{2} & frac{sigma ^2}{2} \
end{array}
right)right] \
3 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{3} \
frac{sigma ^2}{3} & frac{sigma ^2}{3} \
end{array}
right)right] \
4 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{4} \
frac{sigma ^2}{4} & frac{sigma ^2}{4} \
end{array}
right)right] \
5 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{5} \
frac{sigma ^2}{5} & frac{sigma ^2}{5} \
end{array}
right)right] \
6 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{6} \
frac{sigma ^2}{6} & frac{sigma ^2}{6} \
end{array}
right)right] \
7 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{7} \
frac{sigma ^2}{7} & frac{sigma ^2}{7} \
end{array}
right)right] \
8 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{8} \
frac{sigma ^2}{8} & frac{sigma ^2}{8} \
end{array}
right)right] \
9 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{9} \
frac{sigma ^2}{9} & frac{sigma ^2}{9} \
end{array}
right)right] \
10 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{10} \
frac{sigma ^2}{10} & frac{sigma ^2}{10} \
end{array}
right)right] \
end{array}
$$
So the general distribution is a multivariate normal
MultinormalDistribution[{0, 0}, {{σ^2, σ^2/n}, {σ^2/n, σ^2/n}}]
The general form of the joint density function can then be found with
FullSimplify[PDF[MultinormalDistribution[{0, 0}, {{σ^2, σ^2/n}, {σ^2/n, σ^2/n}}], {y, ybar}],
Assumptions -> {σ > 0, n > 1}]
$$frac{n e^{-frac{n left(n text{ybar}^2+y^2-2 y text{ybar}right)}{2 (n-1) sigma ^2}}}{2 pi sqrt{n-1} sigma ^2}$$
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Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
2 days ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
2 days ago
$begingroup$
Nice! In addition to trying to understand the technique, I checked the marginal integrals. Looks great!
$endgroup$
– mjw
2 days ago
add a comment |
$begingroup$
I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.
First the distribution of the mean:
marginalDistribution = TransformedDistribution[Sum[y[i], {i, n}]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], {i, n}],
Assumptions -> [Sigma] > 0]
{#, marginalDistribution/.n->#} &/@Range[2,10]
$$
begin{array}{cc}
2 & text{NormalDistribution}left[0,frac{sigma }{sqrt{2}}right] \
3 & text{NormalDistribution}left[0,frac{sigma }{sqrt{3}}right] \
4 & text{NormalDistribution}left[0,frac{sigma }{2}right] \
5 & text{NormalDistribution}left[0,frac{sigma }{sqrt{5}}right] \
6 & text{NormalDistribution}left[0,frac{sigma }{sqrt{6}}right] \
7 & text{NormalDistribution}left[0,frac{sigma }{sqrt{7}}right] \
8 & text{NormalDistribution}left[0,frac{sigma }{2 sqrt{2}}right] \
9 & text{NormalDistribution}left[0,frac{sigma }{3}right] \
10 & text{NormalDistribution}left[0,frac{sigma }{sqrt{10}}right] \
end{array}
$$
So we see that the marginal distribution of $bar{y}$ is
NormalDistribution[0, σ/Sqrt[n]]
The joint distribution of $bar{y}$ and, say, $y_1$ is given by
jointDistribution = TransformedDistribution[{y[1], Sum[y[i], {i, n}]/n},
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], {i, n}]]
{#, jointDistribution /. n -> #} & /@ Range[2, 10] // TableForm
$$
begin{array}{cc}
2 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{2} \
frac{sigma ^2}{2} & frac{sigma ^2}{2} \
end{array}
right)right] \
3 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{3} \
frac{sigma ^2}{3} & frac{sigma ^2}{3} \
end{array}
right)right] \
4 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{4} \
frac{sigma ^2}{4} & frac{sigma ^2}{4} \
end{array}
right)right] \
5 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{5} \
frac{sigma ^2}{5} & frac{sigma ^2}{5} \
end{array}
right)right] \
6 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{6} \
frac{sigma ^2}{6} & frac{sigma ^2}{6} \
end{array}
right)right] \
7 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{7} \
frac{sigma ^2}{7} & frac{sigma ^2}{7} \
end{array}
right)right] \
8 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{8} \
frac{sigma ^2}{8} & frac{sigma ^2}{8} \
end{array}
right)right] \
9 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{9} \
frac{sigma ^2}{9} & frac{sigma ^2}{9} \
end{array}
right)right] \
10 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{10} \
frac{sigma ^2}{10} & frac{sigma ^2}{10} \
end{array}
right)right] \
end{array}
$$
So the general distribution is a multivariate normal
MultinormalDistribution[{0, 0}, {{σ^2, σ^2/n}, {σ^2/n, σ^2/n}}]
The general form of the joint density function can then be found with
FullSimplify[PDF[MultinormalDistribution[{0, 0}, {{σ^2, σ^2/n}, {σ^2/n, σ^2/n}}], {y, ybar}],
Assumptions -> {σ > 0, n > 1}]
$$frac{n e^{-frac{n left(n text{ybar}^2+y^2-2 y text{ybar}right)}{2 (n-1) sigma ^2}}}{2 pi sqrt{n-1} sigma ^2}$$
$endgroup$
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
2 days ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
2 days ago
$begingroup$
Nice! In addition to trying to understand the technique, I checked the marginal integrals. Looks great!
$endgroup$
– mjw
2 days ago
add a comment |
$begingroup$
I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.
First the distribution of the mean:
marginalDistribution = TransformedDistribution[Sum[y[i], {i, n}]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], {i, n}],
Assumptions -> [Sigma] > 0]
{#, marginalDistribution/.n->#} &/@Range[2,10]
$$
begin{array}{cc}
2 & text{NormalDistribution}left[0,frac{sigma }{sqrt{2}}right] \
3 & text{NormalDistribution}left[0,frac{sigma }{sqrt{3}}right] \
4 & text{NormalDistribution}left[0,frac{sigma }{2}right] \
5 & text{NormalDistribution}left[0,frac{sigma }{sqrt{5}}right] \
6 & text{NormalDistribution}left[0,frac{sigma }{sqrt{6}}right] \
7 & text{NormalDistribution}left[0,frac{sigma }{sqrt{7}}right] \
8 & text{NormalDistribution}left[0,frac{sigma }{2 sqrt{2}}right] \
9 & text{NormalDistribution}left[0,frac{sigma }{3}right] \
10 & text{NormalDistribution}left[0,frac{sigma }{sqrt{10}}right] \
end{array}
$$
So we see that the marginal distribution of $bar{y}$ is
NormalDistribution[0, σ/Sqrt[n]]
The joint distribution of $bar{y}$ and, say, $y_1$ is given by
jointDistribution = TransformedDistribution[{y[1], Sum[y[i], {i, n}]/n},
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], {i, n}]]
{#, jointDistribution /. n -> #} & /@ Range[2, 10] // TableForm
$$
begin{array}{cc}
2 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{2} \
frac{sigma ^2}{2} & frac{sigma ^2}{2} \
end{array}
right)right] \
3 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{3} \
frac{sigma ^2}{3} & frac{sigma ^2}{3} \
end{array}
right)right] \
4 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{4} \
frac{sigma ^2}{4} & frac{sigma ^2}{4} \
end{array}
right)right] \
5 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{5} \
frac{sigma ^2}{5} & frac{sigma ^2}{5} \
end{array}
right)right] \
6 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{6} \
frac{sigma ^2}{6} & frac{sigma ^2}{6} \
end{array}
right)right] \
7 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{7} \
frac{sigma ^2}{7} & frac{sigma ^2}{7} \
end{array}
right)right] \
8 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{8} \
frac{sigma ^2}{8} & frac{sigma ^2}{8} \
end{array}
right)right] \
9 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{9} \
frac{sigma ^2}{9} & frac{sigma ^2}{9} \
end{array}
right)right] \
10 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{10} \
frac{sigma ^2}{10} & frac{sigma ^2}{10} \
end{array}
right)right] \
end{array}
$$
So the general distribution is a multivariate normal
MultinormalDistribution[{0, 0}, {{σ^2, σ^2/n}, {σ^2/n, σ^2/n}}]
The general form of the joint density function can then be found with
FullSimplify[PDF[MultinormalDistribution[{0, 0}, {{σ^2, σ^2/n}, {σ^2/n, σ^2/n}}], {y, ybar}],
Assumptions -> {σ > 0, n > 1}]
$$frac{n e^{-frac{n left(n text{ybar}^2+y^2-2 y text{ybar}right)}{2 (n-1) sigma ^2}}}{2 pi sqrt{n-1} sigma ^2}$$
$endgroup$
I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.
First the distribution of the mean:
marginalDistribution = TransformedDistribution[Sum[y[i], {i, n}]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], {i, n}],
Assumptions -> [Sigma] > 0]
{#, marginalDistribution/.n->#} &/@Range[2,10]
$$
begin{array}{cc}
2 & text{NormalDistribution}left[0,frac{sigma }{sqrt{2}}right] \
3 & text{NormalDistribution}left[0,frac{sigma }{sqrt{3}}right] \
4 & text{NormalDistribution}left[0,frac{sigma }{2}right] \
5 & text{NormalDistribution}left[0,frac{sigma }{sqrt{5}}right] \
6 & text{NormalDistribution}left[0,frac{sigma }{sqrt{6}}right] \
7 & text{NormalDistribution}left[0,frac{sigma }{sqrt{7}}right] \
8 & text{NormalDistribution}left[0,frac{sigma }{2 sqrt{2}}right] \
9 & text{NormalDistribution}left[0,frac{sigma }{3}right] \
10 & text{NormalDistribution}left[0,frac{sigma }{sqrt{10}}right] \
end{array}
$$
So we see that the marginal distribution of $bar{y}$ is
NormalDistribution[0, σ/Sqrt[n]]
The joint distribution of $bar{y}$ and, say, $y_1$ is given by
jointDistribution = TransformedDistribution[{y[1], Sum[y[i], {i, n}]/n},
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], {i, n}]]
{#, jointDistribution /. n -> #} & /@ Range[2, 10] // TableForm
$$
begin{array}{cc}
2 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{2} \
frac{sigma ^2}{2} & frac{sigma ^2}{2} \
end{array}
right)right] \
3 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{3} \
frac{sigma ^2}{3} & frac{sigma ^2}{3} \
end{array}
right)right] \
4 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{4} \
frac{sigma ^2}{4} & frac{sigma ^2}{4} \
end{array}
right)right] \
5 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{5} \
frac{sigma ^2}{5} & frac{sigma ^2}{5} \
end{array}
right)right] \
6 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{6} \
frac{sigma ^2}{6} & frac{sigma ^2}{6} \
end{array}
right)right] \
7 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{7} \
frac{sigma ^2}{7} & frac{sigma ^2}{7} \
end{array}
right)right] \
8 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{8} \
frac{sigma ^2}{8} & frac{sigma ^2}{8} \
end{array}
right)right] \
9 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{9} \
frac{sigma ^2}{9} & frac{sigma ^2}{9} \
end{array}
right)right] \
10 & text{MultinormalDistribution}left[{0,0},left(
begin{array}{cc}
sigma ^2 & frac{sigma ^2}{10} \
frac{sigma ^2}{10} & frac{sigma ^2}{10} \
end{array}
right)right] \
end{array}
$$
So the general distribution is a multivariate normal
MultinormalDistribution[{0, 0}, {{σ^2, σ^2/n}, {σ^2/n, σ^2/n}}]
The general form of the joint density function can then be found with
FullSimplify[PDF[MultinormalDistribution[{0, 0}, {{σ^2, σ^2/n}, {σ^2/n, σ^2/n}}], {y, ybar}],
Assumptions -> {σ > 0, n > 1}]
$$frac{n e^{-frac{n left(n text{ybar}^2+y^2-2 y text{ybar}right)}{2 (n-1) sigma ^2}}}{2 pi sqrt{n-1} sigma ^2}$$
edited 2 days ago
answered 2 days ago
JimBJimB
18.2k12863
18.2k12863
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
2 days ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
2 days ago
$begingroup$
Nice! In addition to trying to understand the technique, I checked the marginal integrals. Looks great!
$endgroup$
– mjw
2 days ago
add a comment |
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
2 days ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
2 days ago
$begingroup$
Nice! In addition to trying to understand the technique, I checked the marginal integrals. Looks great!
$endgroup$
– mjw
2 days ago
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
2 days ago
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
2 days ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
2 days ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
2 days ago
$begingroup$
Nice! In addition to trying to understand the technique, I checked the marginal integrals. Looks great!
$endgroup$
– mjw
2 days ago
$begingroup$
Nice! In addition to trying to understand the technique, I checked the marginal integrals. Looks great!
$endgroup$
– mjw
2 days ago
add a comment |
$begingroup$
Here is the distribution of $x=overline{y}$ (Part I of your question):
a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
p[n_] := TransformedDistribution[Sum[y[k]/n, {k, n}], a[n]];
Now
x [Distributed] p[5] (* n=5, for example *)
The result is
x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
$endgroup$
$begingroup$
I am not sure, but shouldn't be n instead of 5 hereTransformedDistribution[Sum[y[k]/n, {k, 5}], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 days ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 days ago
$begingroup$
Let's go with five because it is clearer. The result isNormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica putsAbs
around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 days ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
2 days ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
$endgroup$
– mjw
2 days ago
|
show 5 more comments
$begingroup$
Here is the distribution of $x=overline{y}$ (Part I of your question):
a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
p[n_] := TransformedDistribution[Sum[y[k]/n, {k, n}], a[n]];
Now
x [Distributed] p[5] (* n=5, for example *)
The result is
x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
$endgroup$
$begingroup$
I am not sure, but shouldn't be n instead of 5 hereTransformedDistribution[Sum[y[k]/n, {k, 5}], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 days ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 days ago
$begingroup$
Let's go with five because it is clearer. The result isNormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica putsAbs
around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 days ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
2 days ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
$endgroup$
– mjw
2 days ago
|
show 5 more comments
$begingroup$
Here is the distribution of $x=overline{y}$ (Part I of your question):
a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
p[n_] := TransformedDistribution[Sum[y[k]/n, {k, n}], a[n]];
Now
x [Distributed] p[5] (* n=5, for example *)
The result is
x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
$endgroup$
Here is the distribution of $x=overline{y}$ (Part I of your question):
a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
p[n_] := TransformedDistribution[Sum[y[k]/n, {k, n}], a[n]];
Now
x [Distributed] p[5] (* n=5, for example *)
The result is
x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
edited 2 days ago
answered 2 days ago
mjwmjw
1,03710
1,03710
$begingroup$
I am not sure, but shouldn't be n instead of 5 hereTransformedDistribution[Sum[y[k]/n, {k, 5}], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 days ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 days ago
$begingroup$
Let's go with five because it is clearer. The result isNormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica putsAbs
around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 days ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
2 days ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
$endgroup$
– mjw
2 days ago
|
show 5 more comments
$begingroup$
I am not sure, but shouldn't be n instead of 5 hereTransformedDistribution[Sum[y[k]/n, {k, 5}], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 days ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 days ago
$begingroup$
Let's go with five because it is clearer. The result isNormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica putsAbs
around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 days ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
2 days ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
$endgroup$
– mjw
2 days ago
$begingroup$
I am not sure, but shouldn't be n instead of 5 here
TransformedDistribution[Sum[y[k]/n, {k, 5}], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw$endgroup$
– Andrea2810
2 days ago
$begingroup$
I am not sure, but shouldn't be n instead of 5 here
TransformedDistribution[Sum[y[k]/n, {k, 5}], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw$endgroup$
– Andrea2810
2 days ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 days ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 days ago
$begingroup$
Let's go with five because it is clearer. The result is
NormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica puts Abs
around $sigma$. Obviously, $sigma>0$.$endgroup$
– mjw
2 days ago
$begingroup$
Let's go with five because it is clearer. The result is
NormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica puts Abs
around $sigma$. Obviously, $sigma>0$.$endgroup$
– mjw
2 days ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
2 days ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
2 days ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
$endgroup$
– mjw
2 days ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
$endgroup$
– mjw
2 days ago
|
show 5 more comments
$begingroup$
just modified @mjw's answer,
n = 100;(*for example*)ClearAll[y];
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
meanDist = TransformedDistribution[Sum[y[k]/100, {k, 100}], a]
JointDistribution can be composed by ProductDistribution,
if these random variables are independent.
if not,you have to use Copula
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, {100000}];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, {100000}];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]
=>
{{1., -0.00256777}, {-0.00256777, 1.}}
I'm not sure about correlation,but it's okay.
$endgroup$
$begingroup$
I believe that the distributions are not independent. Since $overline{x}$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 days ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 days ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 days ago
add a comment |
$begingroup$
just modified @mjw's answer,
n = 100;(*for example*)ClearAll[y];
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
meanDist = TransformedDistribution[Sum[y[k]/100, {k, 100}], a]
JointDistribution can be composed by ProductDistribution,
if these random variables are independent.
if not,you have to use Copula
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, {100000}];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, {100000}];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]
=>
{{1., -0.00256777}, {-0.00256777, 1.}}
I'm not sure about correlation,but it's okay.
$endgroup$
$begingroup$
I believe that the distributions are not independent. Since $overline{x}$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 days ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 days ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 days ago
add a comment |
$begingroup$
just modified @mjw's answer,
n = 100;(*for example*)ClearAll[y];
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
meanDist = TransformedDistribution[Sum[y[k]/100, {k, 100}], a]
JointDistribution can be composed by ProductDistribution,
if these random variables are independent.
if not,you have to use Copula
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, {100000}];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, {100000}];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]
=>
{{1., -0.00256777}, {-0.00256777, 1.}}
I'm not sure about correlation,but it's okay.
$endgroup$
just modified @mjw's answer,
n = 100;(*for example*)ClearAll[y];
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], {k, 1, n}];
meanDist = TransformedDistribution[Sum[y[k]/100, {k, 100}], a]
JointDistribution can be composed by ProductDistribution,
if these random variables are independent.
if not,you have to use Copula
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, {100000}];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, {100000}];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]
=>
{{1., -0.00256777}, {-0.00256777, 1.}}
I'm not sure about correlation,but it's okay.
edited 2 days ago
answered 2 days ago
XminerXminer
33818
33818
$begingroup$
I believe that the distributions are not independent. Since $overline{x}$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 days ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 days ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 days ago
add a comment |
$begingroup$
I believe that the distributions are not independent. Since $overline{x}$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 days ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 days ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 days ago
$begingroup$
I believe that the distributions are not independent. Since $overline{x}$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 days ago
$begingroup$
I believe that the distributions are not independent. Since $overline{x}$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 days ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 days ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 days ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 days ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 days ago
add a comment |
Andrea2810 is a new contributor. Be nice, and check out our Code of Conduct.
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5
$begingroup$
What have you tried? For example, have you seen the documentation on
TransformedDistribution
andProbabilityDistribution
?$endgroup$
– JimB
2 days ago
$begingroup$
@JimB . I tried this
TransformedDistribution[Sum[y, {i, n}]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result isNormalDistribution[0, [Sigma]y]
. However, the correct result should beNormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
2 days ago
1
$begingroup$
You need to "index" the variable
y
or else Mathematica thinks it is a single variable.$endgroup$
– JimB
2 days ago