Nehtdx

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 $hatr_n$ 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 $i^textth$ point, transforming these coordinates to polar coordinates we get for the $i^textth$ 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?

Your reasoning is not quite correct: if $X$ and $Y$ were both independent uniform on $(-r, r)$, then there is positive probability that, for example, the point $(X, Y)$ is close to the point $(r, r)$, which does not lie on the circle of radius $r$ centered at the origin.

Edit.
I mistakenly stopped reading after the second sentence of your attempt, where you claim that each $X_i$ and $Y_i$ are independently distributed on $(-r, r)$.
While that claim is incorrect, the logic following that claim leads to the correct solution (and is frankly more elegant than my solution).
You may think of my solution below as a formal way to approach this problem.

What you need to do instead is figure out the joint distribution of each pair $(X, Y)$.
You are told that $(X, Y)$ is distributed uniformly on the circle of radius $r$ centered at $0$.

First, let's recall a general definition of what it means to be uniformly distributed on a subset of the plane:
Let $A$ be a subset of the plane with a defined positive area (formally, a Borel subset of $mathbbR^2$ of positive Lebesgue measure, but don't worry about that if those words aren't familiar).
Then we say that a random vector $(X, Y)$ is distributed uniformly on $A$ if the joint density of $(X, Y)$ is given by
$$
f_X, Y(x, y)
= begincases
frac1operatornamearea(A) & textif $(x, y) in A$ \
0 & textotherwise
endcases
$$
This should remind you of the definition of a random variable uniformly distributed on an interval in $mathbbR$:
We say that a random variable $U$ is distributed uniformly on an interval $[a, b]$ if its density is
$$
f_U(u)
= begincases
frac1b - a & textif $a le u le b$ \
0 & textotherwise
endcases
$$
which can be rewritten as
$$
f_U(u)
= begincases
frac1operatornamelength([a, b]) & textif $u in [a, b]$ \
0 & textotherwise
endcases
$$
To go from this univariate case to the bivariate case, you replace intervals by subsets of the plane, and length by area (hopefully you can see that this can be done in any number of dimensions by replacing length or area by higher dimensional volume).

Back to the question at hand.
What you need to do instead is figure out the joint distribution of each pair $(X, Y)$.
You are told that $(X, Y)$ is distributed uniformly on the circle of radius $r$ centered at $0$.
Let's call this circle $C_r$; it has area $pi r^2$.
By what we said above, if $(X, Y)$ is distributed uniformly on $C_r$, then the joint density of $(X, Y)$ is
$$
beginaligned
f_X, Y(x, y mid r)
&= begincases
frac1operatornamearea(C_r) & textif $(x, y) in C_r$ \
0 & textotherwise
endcases \
&= begincases
frac1pi r^2 & textif $x^2 + y^2 le r^2$ \
0 & textotherwise
endcases \
&= frac1pi r^2 mathbf1_C_r(x, y),
endaligned
$$
where $mathbf1_A$ denotes the indicator function of the set $A$.

If we now have a sample $(X_1, Y_1), ldots, (X_n, Y_n)$ of random vectors independently distributed uniformly on $C_r$ for some unknown $r > 0$, then we can estimate $r$ by maximum likelihood.
By independence, the likelihood of the sample is
$$
beginaligned
L(r)
&= prod_i=1^n f_X_i, Y_i(X_i, Y_i mid r) \
&= prod_i=1^n frac1pi r^2 mathbf1_C_r(X_i, Y_i) \
&= frac1pi^n r^2 n prod_i=1^n mathbf1_C_r(X_i, Y_i) \
&= frac1pi^n r^2 n prod_i=1^n mathbf1_C_r(X_i, Y_i) \
&= begincases
frac1pi^n r^2n & textif $X_i^2 + Y^2 le r^2$ for all $i$ \
0 & textotherwise
endcases \
&= begincases
frac1pi^n r^2n & textif $maxlimits_i (X_i^2 + Y^2) le r^2$ \
0 & textotherwise
endcases \
&= frac1pi^n r^2n mathbf1_[0, r^2]left(maxlimits_i (X_i^2 + Y_i^2)right) \
&= frac1pi^n r^2n mathbf1_[0, r]left(sqrtmaxlimits_i (X_i^2 + Y_i^2)right) \
&= frac1pi^n r^2n mathbf1_[0, r]left(maxlimits_i sqrtX_i^2 + Y_i^2right)
endaligned
$$
(By the way, this shows that $max_isqrtX_i^2 + Y_i^2$ is a sufficient statistic!)
To summarize,
$$
L(r)
= frac1pi^n r^2n mathbf1_[0, r]left(maxlimits_i sqrtX_i^2 + Y_i^2right)
$$
Now to find the maximum likelihood estimator $widehatr_n$ of $n$, we have to maximize $L(r)$ with respect to $r$.
Notice that the indicator function in the final formula for $L(r)$ is either $0$ or $1$, and $1/(pi^n r^2 n) to infty$ as $r downarrow 0$.
I'll leave it to you to use these clues to figure out the formula for the MLE; you should get a relatively simple (and, in my opinion, intuitive) expression.

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 sphere we must have:

$$mathbbP(R_i leq t) = fracA(t)A(r) = fracpi cdot t^2pi cdot r^2 = Big( fractr Big)^2
quad quad quad textfor all 0 leq t leq r.$$

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:

$$L_mathbfr(r) = prod_i=1^n p_r(r_i) propto frac1r^2 cdot mathbbI(r geq r_(n)).$$

We can see that the likelihood function is strictly decreasing in $r$, so the maximum likelihood estimator (MLE) occurs at the boundary point $hatr_n = r_(n)$. 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:

$$beginequation beginaligned
mathbbE(hatR_n) = mathbbE(R_(n))
&= int limits_0^r mathbbP(R_(n) > t) dt \[6pt]
&= int limits_0^r (1 - mathbbP(R_(n) leq t)) dt \[6pt]
&= int limits_0^r Big( 1 - fract^2nr^2n Big) dt \[6pt]
&= Bigg[ t - frac12n+1 cdot fract^2n+1r^2n Bigg]_t=0^t=r \[6pt]
&= r Big( 1 - frac12n+1 Big) \[6pt]
&= frac2n2n+1 cdot r. \[6pt]
endaligned endequation$$

Thus, a bias-corrected scaled version of the MLE is:

$$tilder_n = frac2n+12n cdot r_(n).$$

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:

$$tilder_n = frackn+1kn cdot r_(n).$$