Finding the unbiased estimator of the given parameter.

Question

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 $r^2$ ?

Answer 1

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.