6
$\begingroup$

Let $X_1,X_2,...,X_n$ be a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$.
I showed that $(\bar X,S^2)$ is jointly sufficient for estimating ($\mu$,$\sigma^2$) where $\bar X$ is the sample mean and $S^2$ is the sample variance.

Then assuming that$(\bar X,S^2)$ is also complete I have to show that $$\sqrt{ n-1\over 2}{\Gamma ({ n-1\over 2})\over\Gamma (\frac n2)} S$$ is a Uniformly Minimum Variance Unbiased Estimator for $\sigma$.

I think I have to use Lehman Scheffe theorem as $(\bar X,S^2)$ is jointly sufficient and complete for $\sigma$. But how can I find a function which is unbiased for $\sigma$ that contains both $(\bar X,S^2)$.

I don't understand how to work when there's a joint sufficiency and completeness.

$\endgroup$
  • $\begingroup$ @Glen_b : I am studying UMVUE for a inference course and this was a question I came across in a note $\endgroup$ – clarkson Jul 7 '14 at 16:45
  • 2
    $\begingroup$ A reader notes this question has been cross-posted at math.stackexchange.com/questions/858677/…. Please decide where on SE you would like this to appear and delete the duplicate version(s). $\endgroup$ – whuber Jul 7 '14 at 17:12
  • $\begingroup$ @Glen_b : I included that tag.In my other questions I normally include it.But this time I forgot $\endgroup$ – clarkson Jul 8 '14 at 0:53
2
$\begingroup$

Although the question was posted almost 4 years ago, I would like to answer this question. English is not my mother tongue and I am learning it so please don't mind my awkward sentences.

To solve this problem, we notice that $(n-1)S^2/ \sigma^2$ has a chisquare distribution with $n-1$ degree of freedom, while $S^2= \sum^n_{i=1}{(X-\bar{X})^2\over n-1}={{\sum^n_{i=1}X^2}-n \bar{X}^2\over n-1}$ and $X$ has a normal distribution with mean $\mu$ and variance $\sigma^2$.

Note that $S$ contains ${\sum^n_{i=1}X^2}$ and ${\sum^n_{i=1}X}$.

Let's evaluate $E[S]$. To simplify let $q=(n-1)S^2/ \sigma^2$, then $S=\sqrt{q \sigma^2 /(n-1)}$. $$E[S]=\int^{\infty}_0 \sqrt{ \sigma^2 \over (n-1)} q^{1 \over2}f_q dq \\=\int^{\infty}_0 \sqrt{ \sigma^2 \over (n-1)} q^{1 \over2} { q^{{n-1 \over 2} -1} e^{-q \over 2} \over \Gamma({n-1 \over 2}) 2^{n-1 \over 2}} dq \\ = \sqrt{ \sigma^2 \over (n-1)} \int^{\infty}_0 { q^{{n \over 2} -1} e^{-q \over 2} \over \Gamma({n-1 \over 2}) 2^{n-1 \over 2}} dq \\= \sqrt{ \sigma^2 \over (n-1)} { \Gamma({n \over 2}) 2^{1 \over 2} \over \Gamma({n-1 \over 2}) } $$

After some rearranging you can get the desired result.

It would be appreciated if someone corrects any grammatical or mathematical mistakes.

$\endgroup$
  • $\begingroup$ Proper notation would be to use subscripts for X. $\endgroup$ – Michael Chernick Jun 25 '18 at 14:54
  • $\begingroup$ I made minor modifications to the text. But the grammar was not bad. $\endgroup$ – Michael Chernick Jun 25 '18 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.