# Finding the maximum likelihood estimator of a two dimensional sample

Suppose that $(X_i,Y_i), i=1,...,n$ is a two-dimensional sample, with each $(X_i,Y_i)$ assumed to be uniformly distributed in a circle centered at the origin, of an unknown radius r. What is the MLE of r?

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I'm not sure how to begin attempting this problem. Any suggestions would be greatly appreciated! –  woaini Nov 19 '12 at 13:25

This is similar to the German Tank Problem. The MLE is the maximum of the observed values of $\sqrt{x_i^2 + y_i^2}$, where your sample is $(x_i, y_i)_{i=1}^n$. If that seems a bit suspicious, it is; in this case, the MLE is actually a biased estimator and maximum likelihood is not the best approach for inferring the value of $r$.

Here is a proof that the maximum likelihood estimator is $\mathrm{max}_i \sqrt{x_i^2 + y_i^2}$. The probability density of a single draw from a uniform distribution on a disc of radius $r$ is

$$f(x,y) = \begin{cases} 1/\pi r^2 &x^2+y^2 < r^2\\0 &x^2+y^2>r^2\end{cases}$$

we regard this as a function of $r$ as well, so that we have

$$f(x,y; r) = \frac{1}{\pi r^2} I(x^2+y^2 < r^2)$$ where $I$ is the Iverson bracket/characteristic function. ($I(X) =1$ if $X$ is true and $0$ if not.)

The density of $n$ independent draws $(x_i, y_i)$ is then

$$\prod_{i=1}^n f(x_i, y_i, r) = (1/\pi r^2)^n \prod_{i=1}^n I(x_i^2+y_i^2 < r^2)$$ which is the same as $$(1/\pi r^2)^n I(x_i^2+y_i^2 < r^2 \text{ for all } i)$$ This is your likelihood function, where you now regard the $(x_i, y_i)$ as fixed and allow $r$ to vary. It decreases with $r$, so it is maximised when $r$ is as small as possible. But the likelihood is only nonzero if $r \ge \mathrm{max}_i \sqrt{x_i^2 + y_i^2}$, so that value is your MLE, as claimed.

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