[Math] Finding the unbiased estimator of the given parameter.

estimationparameter estimationsampling

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 $r^2$ ?

Best Answer

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.

Related Solutions

[Math] unbiased estimator of sample variance using two samples

Apart from the fact that it should be $m-1$ instead of $n-1$ in the right-hand denominator, your estimator for $\sigma^2$ looks fine. You can do slightly better on the variance of $\hat\mu$ (though the question didn't ask to optimize it): Consider a general convex combination

$$ \alpha\frac{X_1+\dotso+X_n}n+(1-\alpha)\frac{Y_1+\dotso+Y_m}{2m} $$

of the individual estimators for $\mu$. The variance of this combined estimator is

$$ n\left(\frac\alpha n\right)^2\sigma^2+m\left(\frac{1-\alpha}{2m}\right)^2\sigma^2=\left(\frac{\alpha^2}n+\frac{(1-\alpha)^2}{4m}\right)\sigma^2\;, $$

and minimizing this by setting the derivative with respect to $\alpha$ to zero leads to $\alpha=n/(n+4m)$, yielding the variance $\sigma^2/(n+4m)$. For $n=m$ the variance is $\frac15\sigma^2/n=0.2\sigma^2/n$, compared to $\frac5{16}\sigma^2/n\approx0.3\sigma^2/n$ for your estimator, and for $n$ fixed and $m\to\infty$ or vice versa, the variance of this estimator tends to zero whereas the variance of your estimator tends to a non-zero value.

You could optimize the variance of your unbiased variance estimator in a similar way, though the calculation would be a bit more involved.

[Math] unbiased pool estimator of variance

First, your notation for the sample variance seems to be muddled. The sample variance is ordinarily defined as $S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2,$ which makes it an unbiased estimator of the population variance $\sigma^2.$

Perhaps the most common context for 'unbiased pooled estimator' of variance is for the 'pooled t test': Suppose you have two random samples $X_i$ of size $n$ and $Y_i$ of size $m$ from populations with the same variance $\sigma^2.$ Then the pooled estimator of $\sigma^2$ is

$$S_p^2 = \frac{(n-1)S_X^2 + (m-1)S_Y^2}{m+n-2}.$$

This estimator is unbiased.

Because one says the samples have respective 'degrees of freedom' $n-1$ and $m-1,$ one sometimes says the $S_p^2$ is a 'degrees-of-freedom' weighted average of the two sample variances. If $n = m,$ then $S_p^2 = 0.5S_x^2 + 0.5S_Y^2.$

Note: Some authors do define the sample variance as $\frac{1}{n}\sum_{i=1}^n (X_i - \bar X)^2,$ but then the sample variance is not an unbiased estimator of $\sigma^2,$ even though it might have other properties desirable for the author's task at hand. However, most agree that the notation $S^2$ is reserved for the version with $n-1$ in the denominator, unless a specific warning is given otherwise.

Example: One common measure of the 'goodness' of an estimator is that it have a small 'root mean squared error'. If $T$ is an estimate of $\tau$ then $\text{MSE}_T(\tau) = E[(T-\tau)^2]$ and RMSE is its square root.

The simulation below illustrates for normal data with $n = 5$ and $\sigma^2 = 10^2 = 100,$ that the version of the sample variance with $n$ in the denominator has smaller RMSE than the version with $n-1$ in the denominator. (A formal proof for $n > 1$ is not difficult.)

set.seed(1888);  m = 10^6;  n = 5;  sigma = 10;  sg.sq = 100
v.a = replicate(m, var(rnorm(n, 100, sigma)))  # denom n-1
v.b = (n-1)*v.a/n                              # denom n
mean(v.a); RMS.a = sqrt(mean((v.a-sg.sq)^2)); RMS.a
[1] 100.0564  # 70.81563
[1] 70.81563  # larger RMSE
mean(v.b); RMS.b = sqrt(mean((v.b-sg.sq)^2)); RMS.b
[1] 80.0451   # biased   
[1] 60.06415  # smaller RMSE

Best Answer

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[Math] unbiased estimator of sample variance using two samples

[Math] unbiased pool estimator of variance

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