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The radius of a circle is measured with an error of measurement which is distributed $N(0,\sigma^2)$, $\sigma^2$ unknown. Given $n$ independent measurements of the $radius$, find an unbiased estimator of the area of the circle.

I tried as follows :

We are given $X=r+e$ where $r$ is the actual fixed radius , $e$ is the error term and $X$ is the radius after taking into account that error term.

Since , $e$~$N(0,\sigma^2)$ => $X$~$N(r,\sigma^2)$.

Also a sample of these radii(with error terms) is given as ($X_1$,$X_2$,$X_3$,.... ,$X_n$)

We need the unbiased estimator of Area($2 \pi r^2$) , if we are able to find the the unbiased estimator of $r^2$ we are done.

We know $\hat{r}=\frac{\sum x_i}{n}$ is an unbiased estimator of $r$ and $\hat{\sigma}^2=\frac{\sum (x_i -r)^2}{n}$ is an unbiased estimator of $\sigma^2$.

Writing the maximum likelihood function for ($r^2$):

$L(r^2)=\Pi_{i=1}^n f(X_i)$ where $X_i$~$N(r,\sigma^2)$

Taking $r^2=\mu$ , so we need the the M.L.E estimator of $\mu$.

Taking logs differentiating w.r.t $\mu$ and equating to zero yields $\hat{\mu}=\bar{x}^2$

But it isn't an unbiased estimator since , $E(\hat{\mu})=E(\bar{x}^2)=\frac{\sigma^2}{n}+r^2$

So , $E(\hat{\mu}-\frac{\sigma^2}{n})=r^2$

So can we take $(\hat{\mu}-\frac{\hat{\sigma^2}}{n})$ as an unbiased estimator of $r^2$ ?

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  • $\begingroup$ A similar (unanswered) question here. $\endgroup$
    – Mark
    Feb 9 '17 at 8:09
  • $\begingroup$ @Mark Not answered there. Already had a look ! $\endgroup$
    – User9523
    Feb 9 '17 at 8:23
  • $\begingroup$ @zoli Okay , I'll edit that. But besides that the idea ? $\endgroup$
    – User9523
    Feb 9 '17 at 8:24
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Let's consider the most natural estimate of the random area:

$$\pi\frac1n\sum_{i=1}^nX_i^2.$$

The question is if this is an unbiased estimate assuming that $X_i=R+e_i$ where $e_i$ are independent normal random variables with $0$ mean and $\sigma^2$ as variance.

So, we need to calculate the expectetion if our estimate above.

$$\pi E\left[\frac1n\sum_{i=1}^nX_i^2\right]=\pi\frac1nE\left[\sum_{i=1}^n(R+e_i)^2\right]=$$ $$=\pi R^2+2\pi RE[e_i]+\pi\sigma^2=\pi R^2+\pi\sigma^2.$$

So, it seems that

$$\pi\frac1n\sum_{i=1}^nX_i^2-\pi\sigma^2$$

is an unbiased estimate.

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  • $\begingroup$ Did you lose a $2$ in the squaring? (It doesn't matter because that term vanishes, but still...) Also, this doesn't fully handle the issue because $\sigma^2$ is not given. But that's easy enough to fix: just replace it with an unbiased estimator of $\sigma^2$ (because the role of $\sigma^2$ in the estimator you gave is linear). And we all know one of those, thankfully. $\endgroup$
    – Ian
    Feb 9 '17 at 14:46
  • $\begingroup$ I did not lose $2$. Is the area of a circle not $r^2\pi$? $\endgroup$
    – zoli
    Feb 9 '17 at 16:05
  • $\begingroup$ Oh, that $2$... $\endgroup$
    – zoli
    Feb 9 '17 at 16:37

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