MLE of the unknown radius
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$begingroup$
Consider this question,
Suppose that $(X_1, Y_1),(X_2, Y_2), . . . ,(X_n, Y_n)$ are the
coordinates of $n$ points chosen independently and uniformly at random
within a circle with center $(0, 0)$ and unknown radius $r$. Obtain
the MLE $hat{r}_n$ of $r$.
My attempt:
I have thought about the question but I am not able to put it formally. This is my reasoning.
$X_i$ and $Y_i$ are both $uniformly$ $distributed$ independent random variables on a circle of radius $r$ and center at $(0,0)$. Since $(X_i, Y_i)$ are the coordinates of the $i^text{th}$ point, transforming these coordinates to polar coordinates we get for the $i^text{th}$ point $(theta_i,a_i)$ (say) where $theta_i$ follows $Uniform(0,2pi)$ and $a_i$ follows $Uniform(0,r)$ independently. Then $a_{(n)} = max(a_i)$ is the MLE of $r$.
Is this reasoning correct? How do I put it formally? And if not, how should this problem be solved?
self-study estimation maximum-likelihood inference
$endgroup$
add a comment |
$begingroup$
Consider this question,
Suppose that $(X_1, Y_1),(X_2, Y_2), . . . ,(X_n, Y_n)$ are the
coordinates of $n$ points chosen independently and uniformly at random
within a circle with center $(0, 0)$ and unknown radius $r$. Obtain
the MLE $hat{r}_n$ of $r$.
My attempt:
I have thought about the question but I am not able to put it formally. This is my reasoning.
$X_i$ and $Y_i$ are both $uniformly$ $distributed$ independent random variables on a circle of radius $r$ and center at $(0,0)$. Since $(X_i, Y_i)$ are the coordinates of the $i^text{th}$ point, transforming these coordinates to polar coordinates we get for the $i^text{th}$ point $(theta_i,a_i)$ (say) where $theta_i$ follows $Uniform(0,2pi)$ and $a_i$ follows $Uniform(0,r)$ independently. Then $a_{(n)} = max(a_i)$ is the MLE of $r$.
Is this reasoning correct? How do I put it formally? And if not, how should this problem be solved?
self-study estimation maximum-likelihood inference
$endgroup$
add a comment |
$begingroup$
Consider this question,
Suppose that $(X_1, Y_1),(X_2, Y_2), . . . ,(X_n, Y_n)$ are the
coordinates of $n$ points chosen independently and uniformly at random
within a circle with center $(0, 0)$ and unknown radius $r$. Obtain
the MLE $hat{r}_n$ of $r$.
My attempt:
I have thought about the question but I am not able to put it formally. This is my reasoning.
$X_i$ and $Y_i$ are both $uniformly$ $distributed$ independent random variables on a circle of radius $r$ and center at $(0,0)$. Since $(X_i, Y_i)$ are the coordinates of the $i^text{th}$ point, transforming these coordinates to polar coordinates we get for the $i^text{th}$ point $(theta_i,a_i)$ (say) where $theta_i$ follows $Uniform(0,2pi)$ and $a_i$ follows $Uniform(0,r)$ independently. Then $a_{(n)} = max(a_i)$ is the MLE of $r$.
Is this reasoning correct? How do I put it formally? And if not, how should this problem be solved?
self-study estimation maximum-likelihood inference
$endgroup$
Consider this question,
Suppose that $(X_1, Y_1),(X_2, Y_2), . . . ,(X_n, Y_n)$ are the
coordinates of $n$ points chosen independently and uniformly at random
within a circle with center $(0, 0)$ and unknown radius $r$. Obtain
the MLE $hat{r}_n$ of $r$.
My attempt:
I have thought about the question but I am not able to put it formally. This is my reasoning.
$X_i$ and $Y_i$ are both $uniformly$ $distributed$ independent random variables on a circle of radius $r$ and center at $(0,0)$. Since $(X_i, Y_i)$ are the coordinates of the $i^text{th}$ point, transforming these coordinates to polar coordinates we get for the $i^text{th}$ point $(theta_i,a_i)$ (say) where $theta_i$ follows $Uniform(0,2pi)$ and $a_i$ follows $Uniform(0,r)$ independently. Then $a_{(n)} = max(a_i)$ is the MLE of $r$.
Is this reasoning correct? How do I put it formally? And if not, how should this problem be solved?
self-study estimation maximum-likelihood inference
self-study estimation maximum-likelihood inference
edited 19 hours ago
Sanket Agrawal
asked 20 hours ago
Sanket AgrawalSanket Agrawal
646
646
add a comment |
add a comment |
1 Answer
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active
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$begingroup$
Uniform sampling over geometric shapes: With these kinds of problems, where your sample points are uniformly distributed over some fixed geometric shape, it is almost always easiest to proceed by first deriving the distribution function for the relevant quantity of interest. This is quite simple because the probability of falling within any subset of the overall sampling space is equal to the proportion that this subset takes up in the total area/volume of the total sampling space.
This question is substantially simplified if we recognise that the points give information about the radius $r$ only through their distance from the centre (i.e., the zero point). Let $R_1,R_2,R_3,...$ be these distance values corresponding to each of the points and let $A(r) = pi cdot r^2$ be the area of a circle with radius $r$. Since each sample point is uniform in the circle we must have:
$$mathbb{P}(R_i leq t) = frac{A(t)}{A(r)} = frac{pi cdot t^2}{pi cdot r^2} = Big( frac{t}{r} Big)^2
quad quad quad text{for all } 0 leq t leq r.$$
This gives the corresponding sample density $p_r(t) = 2t/r^2$ over the support $0 leq t leq r$. As you can see, this sampling density is not uniform over the support. This is actually unsurprising --- the probability density for the distance $R_i$ is proportional to the circumference of a circle with radius equal to that distance.
The maximum likelihood estimator: The likelihood function for $n$ observed data points is:
$$L_mathbf{r}(r) = prod_{i=1}^n p_r(r_i) propto frac{1}{r^{2n}} cdot mathbb{I}(r geq r_{(n)}).$$
We can see that the likelihood function is strictly decreasing in $r$, so the maximum likelihood estimator (MLE) occurs at the boundary point $hat{r}_n = r_{(n)}$. That is, the MLE for the true radius is the maximum of the lengths from the zero point to the sample points. Since the true parameter $r$ is at least as large as the MLE, the MLE is biased, and will tend to underestimate the true radius. Specifically, we have:
$$begin{equation} begin{aligned}
mathbb{E}(hat{R}_n) = mathbb{E}(R_{(n)})
&= int limits_0^r mathbb{P}(R_{(n)} > t) dt \[6pt]
&= int limits_0^r (1 - mathbb{P}(R_{(n)} leq t)) dt \[6pt]
&= int limits_0^r Big( 1 - frac{t^{2n}}{r^{2n}} Big) dt \[6pt]
&= Bigg[ t - frac{1}{2n+1} cdot frac{t^{2n+1}}{r^{2n}} Bigg]_{t=0}^{t=r} \[6pt]
&= r Big( 1 - frac{1}{2n+1} Big) \[6pt]
&= frac{2n}{2n+1} cdot r. \[6pt]
end{aligned} end{equation}$$
Thus, a bias-corrected scaled version of the MLE is:
$$tilde{r}_n = frac{2n+1}{2n} cdot r_{(n)}.$$
Incidentally, this process can easily be extended to a hypersphere in higher dimensions. In each case the MLE is the maximum distance from the zero point. For a hypershere in $k$ dimensions, the bias-corrected scaled MLE is:
$$tilde{r}_n = frac{kn+1}{kn} cdot r_{(n)}.$$
$endgroup$
$begingroup$
@Silverfish - Sorry, it's me that was having the mind-blank. I have edited to correct. Thanks for spotting that.
$endgroup$
– Ben
2 hours ago
$begingroup$
No problem, thanks for the edit! (Will come back and delete this comment tomorrow.)
$endgroup$
– Silverfish
1 hour ago
add a comment |
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$begingroup$
Uniform sampling over geometric shapes: With these kinds of problems, where your sample points are uniformly distributed over some fixed geometric shape, it is almost always easiest to proceed by first deriving the distribution function for the relevant quantity of interest. This is quite simple because the probability of falling within any subset of the overall sampling space is equal to the proportion that this subset takes up in the total area/volume of the total sampling space.
This question is substantially simplified if we recognise that the points give information about the radius $r$ only through their distance from the centre (i.e., the zero point). Let $R_1,R_2,R_3,...$ be these distance values corresponding to each of the points and let $A(r) = pi cdot r^2$ be the area of a circle with radius $r$. Since each sample point is uniform in the circle we must have:
$$mathbb{P}(R_i leq t) = frac{A(t)}{A(r)} = frac{pi cdot t^2}{pi cdot r^2} = Big( frac{t}{r} Big)^2
quad quad quad text{for all } 0 leq t leq r.$$
This gives the corresponding sample density $p_r(t) = 2t/r^2$ over the support $0 leq t leq r$. As you can see, this sampling density is not uniform over the support. This is actually unsurprising --- the probability density for the distance $R_i$ is proportional to the circumference of a circle with radius equal to that distance.
The maximum likelihood estimator: The likelihood function for $n$ observed data points is:
$$L_mathbf{r}(r) = prod_{i=1}^n p_r(r_i) propto frac{1}{r^{2n}} cdot mathbb{I}(r geq r_{(n)}).$$
We can see that the likelihood function is strictly decreasing in $r$, so the maximum likelihood estimator (MLE) occurs at the boundary point $hat{r}_n = r_{(n)}$. That is, the MLE for the true radius is the maximum of the lengths from the zero point to the sample points. Since the true parameter $r$ is at least as large as the MLE, the MLE is biased, and will tend to underestimate the true radius. Specifically, we have:
$$begin{equation} begin{aligned}
mathbb{E}(hat{R}_n) = mathbb{E}(R_{(n)})
&= int limits_0^r mathbb{P}(R_{(n)} > t) dt \[6pt]
&= int limits_0^r (1 - mathbb{P}(R_{(n)} leq t)) dt \[6pt]
&= int limits_0^r Big( 1 - frac{t^{2n}}{r^{2n}} Big) dt \[6pt]
&= Bigg[ t - frac{1}{2n+1} cdot frac{t^{2n+1}}{r^{2n}} Bigg]_{t=0}^{t=r} \[6pt]
&= r Big( 1 - frac{1}{2n+1} Big) \[6pt]
&= frac{2n}{2n+1} cdot r. \[6pt]
end{aligned} end{equation}$$
Thus, a bias-corrected scaled version of the MLE is:
$$tilde{r}_n = frac{2n+1}{2n} cdot r_{(n)}.$$
Incidentally, this process can easily be extended to a hypersphere in higher dimensions. In each case the MLE is the maximum distance from the zero point. For a hypershere in $k$ dimensions, the bias-corrected scaled MLE is:
$$tilde{r}_n = frac{kn+1}{kn} cdot r_{(n)}.$$
$endgroup$
$begingroup$
@Silverfish - Sorry, it's me that was having the mind-blank. I have edited to correct. Thanks for spotting that.
$endgroup$
– Ben
2 hours ago
$begingroup$
No problem, thanks for the edit! (Will come back and delete this comment tomorrow.)
$endgroup$
– Silverfish
1 hour ago
add a comment |
$begingroup$
Uniform sampling over geometric shapes: With these kinds of problems, where your sample points are uniformly distributed over some fixed geometric shape, it is almost always easiest to proceed by first deriving the distribution function for the relevant quantity of interest. This is quite simple because the probability of falling within any subset of the overall sampling space is equal to the proportion that this subset takes up in the total area/volume of the total sampling space.
This question is substantially simplified if we recognise that the points give information about the radius $r$ only through their distance from the centre (i.e., the zero point). Let $R_1,R_2,R_3,...$ be these distance values corresponding to each of the points and let $A(r) = pi cdot r^2$ be the area of a circle with radius $r$. Since each sample point is uniform in the circle we must have:
$$mathbb{P}(R_i leq t) = frac{A(t)}{A(r)} = frac{pi cdot t^2}{pi cdot r^2} = Big( frac{t}{r} Big)^2
quad quad quad text{for all } 0 leq t leq r.$$
This gives the corresponding sample density $p_r(t) = 2t/r^2$ over the support $0 leq t leq r$. As you can see, this sampling density is not uniform over the support. This is actually unsurprising --- the probability density for the distance $R_i$ is proportional to the circumference of a circle with radius equal to that distance.
The maximum likelihood estimator: The likelihood function for $n$ observed data points is:
$$L_mathbf{r}(r) = prod_{i=1}^n p_r(r_i) propto frac{1}{r^{2n}} cdot mathbb{I}(r geq r_{(n)}).$$
We can see that the likelihood function is strictly decreasing in $r$, so the maximum likelihood estimator (MLE) occurs at the boundary point $hat{r}_n = r_{(n)}$. That is, the MLE for the true radius is the maximum of the lengths from the zero point to the sample points. Since the true parameter $r$ is at least as large as the MLE, the MLE is biased, and will tend to underestimate the true radius. Specifically, we have:
$$begin{equation} begin{aligned}
mathbb{E}(hat{R}_n) = mathbb{E}(R_{(n)})
&= int limits_0^r mathbb{P}(R_{(n)} > t) dt \[6pt]
&= int limits_0^r (1 - mathbb{P}(R_{(n)} leq t)) dt \[6pt]
&= int limits_0^r Big( 1 - frac{t^{2n}}{r^{2n}} Big) dt \[6pt]
&= Bigg[ t - frac{1}{2n+1} cdot frac{t^{2n+1}}{r^{2n}} Bigg]_{t=0}^{t=r} \[6pt]
&= r Big( 1 - frac{1}{2n+1} Big) \[6pt]
&= frac{2n}{2n+1} cdot r. \[6pt]
end{aligned} end{equation}$$
Thus, a bias-corrected scaled version of the MLE is:
$$tilde{r}_n = frac{2n+1}{2n} cdot r_{(n)}.$$
Incidentally, this process can easily be extended to a hypersphere in higher dimensions. In each case the MLE is the maximum distance from the zero point. For a hypershere in $k$ dimensions, the bias-corrected scaled MLE is:
$$tilde{r}_n = frac{kn+1}{kn} cdot r_{(n)}.$$
$endgroup$
$begingroup$
@Silverfish - Sorry, it's me that was having the mind-blank. I have edited to correct. Thanks for spotting that.
$endgroup$
– Ben
2 hours ago
$begingroup$
No problem, thanks for the edit! (Will come back and delete this comment tomorrow.)
$endgroup$
– Silverfish
1 hour ago
add a comment |
$begingroup$
Uniform sampling over geometric shapes: With these kinds of problems, where your sample points are uniformly distributed over some fixed geometric shape, it is almost always easiest to proceed by first deriving the distribution function for the relevant quantity of interest. This is quite simple because the probability of falling within any subset of the overall sampling space is equal to the proportion that this subset takes up in the total area/volume of the total sampling space.
This question is substantially simplified if we recognise that the points give information about the radius $r$ only through their distance from the centre (i.e., the zero point). Let $R_1,R_2,R_3,...$ be these distance values corresponding to each of the points and let $A(r) = pi cdot r^2$ be the area of a circle with radius $r$. Since each sample point is uniform in the circle we must have:
$$mathbb{P}(R_i leq t) = frac{A(t)}{A(r)} = frac{pi cdot t^2}{pi cdot r^2} = Big( frac{t}{r} Big)^2
quad quad quad text{for all } 0 leq t leq r.$$
This gives the corresponding sample density $p_r(t) = 2t/r^2$ over the support $0 leq t leq r$. As you can see, this sampling density is not uniform over the support. This is actually unsurprising --- the probability density for the distance $R_i$ is proportional to the circumference of a circle with radius equal to that distance.
The maximum likelihood estimator: The likelihood function for $n$ observed data points is:
$$L_mathbf{r}(r) = prod_{i=1}^n p_r(r_i) propto frac{1}{r^{2n}} cdot mathbb{I}(r geq r_{(n)}).$$
We can see that the likelihood function is strictly decreasing in $r$, so the maximum likelihood estimator (MLE) occurs at the boundary point $hat{r}_n = r_{(n)}$. That is, the MLE for the true radius is the maximum of the lengths from the zero point to the sample points. Since the true parameter $r$ is at least as large as the MLE, the MLE is biased, and will tend to underestimate the true radius. Specifically, we have:
$$begin{equation} begin{aligned}
mathbb{E}(hat{R}_n) = mathbb{E}(R_{(n)})
&= int limits_0^r mathbb{P}(R_{(n)} > t) dt \[6pt]
&= int limits_0^r (1 - mathbb{P}(R_{(n)} leq t)) dt \[6pt]
&= int limits_0^r Big( 1 - frac{t^{2n}}{r^{2n}} Big) dt \[6pt]
&= Bigg[ t - frac{1}{2n+1} cdot frac{t^{2n+1}}{r^{2n}} Bigg]_{t=0}^{t=r} \[6pt]
&= r Big( 1 - frac{1}{2n+1} Big) \[6pt]
&= frac{2n}{2n+1} cdot r. \[6pt]
end{aligned} end{equation}$$
Thus, a bias-corrected scaled version of the MLE is:
$$tilde{r}_n = frac{2n+1}{2n} cdot r_{(n)}.$$
Incidentally, this process can easily be extended to a hypersphere in higher dimensions. In each case the MLE is the maximum distance from the zero point. For a hypershere in $k$ dimensions, the bias-corrected scaled MLE is:
$$tilde{r}_n = frac{kn+1}{kn} cdot r_{(n)}.$$
$endgroup$
Uniform sampling over geometric shapes: With these kinds of problems, where your sample points are uniformly distributed over some fixed geometric shape, it is almost always easiest to proceed by first deriving the distribution function for the relevant quantity of interest. This is quite simple because the probability of falling within any subset of the overall sampling space is equal to the proportion that this subset takes up in the total area/volume of the total sampling space.
This question is substantially simplified if we recognise that the points give information about the radius $r$ only through their distance from the centre (i.e., the zero point). Let $R_1,R_2,R_3,...$ be these distance values corresponding to each of the points and let $A(r) = pi cdot r^2$ be the area of a circle with radius $r$. Since each sample point is uniform in the circle we must have:
$$mathbb{P}(R_i leq t) = frac{A(t)}{A(r)} = frac{pi cdot t^2}{pi cdot r^2} = Big( frac{t}{r} Big)^2
quad quad quad text{for all } 0 leq t leq r.$$
This gives the corresponding sample density $p_r(t) = 2t/r^2$ over the support $0 leq t leq r$. As you can see, this sampling density is not uniform over the support. This is actually unsurprising --- the probability density for the distance $R_i$ is proportional to the circumference of a circle with radius equal to that distance.
The maximum likelihood estimator: The likelihood function for $n$ observed data points is:
$$L_mathbf{r}(r) = prod_{i=1}^n p_r(r_i) propto frac{1}{r^{2n}} cdot mathbb{I}(r geq r_{(n)}).$$
We can see that the likelihood function is strictly decreasing in $r$, so the maximum likelihood estimator (MLE) occurs at the boundary point $hat{r}_n = r_{(n)}$. That is, the MLE for the true radius is the maximum of the lengths from the zero point to the sample points. Since the true parameter $r$ is at least as large as the MLE, the MLE is biased, and will tend to underestimate the true radius. Specifically, we have:
$$begin{equation} begin{aligned}
mathbb{E}(hat{R}_n) = mathbb{E}(R_{(n)})
&= int limits_0^r mathbb{P}(R_{(n)} > t) dt \[6pt]
&= int limits_0^r (1 - mathbb{P}(R_{(n)} leq t)) dt \[6pt]
&= int limits_0^r Big( 1 - frac{t^{2n}}{r^{2n}} Big) dt \[6pt]
&= Bigg[ t - frac{1}{2n+1} cdot frac{t^{2n+1}}{r^{2n}} Bigg]_{t=0}^{t=r} \[6pt]
&= r Big( 1 - frac{1}{2n+1} Big) \[6pt]
&= frac{2n}{2n+1} cdot r. \[6pt]
end{aligned} end{equation}$$
Thus, a bias-corrected scaled version of the MLE is:
$$tilde{r}_n = frac{2n+1}{2n} cdot r_{(n)}.$$
Incidentally, this process can easily be extended to a hypersphere in higher dimensions. In each case the MLE is the maximum distance from the zero point. For a hypershere in $k$ dimensions, the bias-corrected scaled MLE is:
$$tilde{r}_n = frac{kn+1}{kn} cdot r_{(n)}.$$
edited 2 hours ago
answered 17 hours ago
BenBen
28.8k233129
28.8k233129
$begingroup$
@Silverfish - Sorry, it's me that was having the mind-blank. I have edited to correct. Thanks for spotting that.
$endgroup$
– Ben
2 hours ago
$begingroup$
No problem, thanks for the edit! (Will come back and delete this comment tomorrow.)
$endgroup$
– Silverfish
1 hour ago
add a comment |
$begingroup$
@Silverfish - Sorry, it's me that was having the mind-blank. I have edited to correct. Thanks for spotting that.
$endgroup$
– Ben
2 hours ago
$begingroup$
No problem, thanks for the edit! (Will come back and delete this comment tomorrow.)
$endgroup$
– Silverfish
1 hour ago
$begingroup$
@Silverfish - Sorry, it's me that was having the mind-blank. I have edited to correct. Thanks for spotting that.
$endgroup$
– Ben
2 hours ago
$begingroup$
@Silverfish - Sorry, it's me that was having the mind-blank. I have edited to correct. Thanks for spotting that.
$endgroup$
– Ben
2 hours ago
$begingroup$
No problem, thanks for the edit! (Will come back and delete this comment tomorrow.)
$endgroup$
– Silverfish
1 hour ago
$begingroup$
No problem, thanks for the edit! (Will come back and delete this comment tomorrow.)
$endgroup$
– Silverfish
1 hour ago
add a comment |
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