1
$\begingroup$

I will first state the question I am trying to solve then I will demonstrate my thoughts!

Let $Y_1...Y_n$ be independent normal R.V.s each with mean $\mu$ and variance $\sigma^2$. We want to estimate $\sigma^2$ from a sample of size $n>2$ when $\mu$ is also unknown. We know that the ML estimator of $\sigma^2$ is $\hat\sigma^2 = X/n$ where $X = \sum_{i=1}^{n}(Y_i-\bar{Y})^2$. There are one thing we should note: $X/\sigma^2$ has a chi squared distribution with $n-1$ degrees of freedom. Find the bias and variance of $\hat\sigma^2$.

Okay, so to find the bias we need to solve:

$E(\hat\sigma^2)-\sigma^2$. Thus we need to find $E(\hat\sigma^2)$ which I have below:

$$E(\frac{\sum_{i=1}^{n}(Y_i-\bar{Y})^2}{n})$$ which is the name thing as:

$$(\frac{1}{n})E[{\sum_{i=1}^{n}(Y_i-\bar{Y})^2}]$$

Expanding the inside I get:

$$(\frac{1}{n})E({\sum_{i=1}^{n}(Y_i^2-2Y_i\bar{Y}+\bar{Y^2}})$$

Is this the correct approach? I am quite confused on how to go from here if it is correct? Help on the variance would also be greatly appreciated! Thanks so much :)

EDIT:

I actually figured out how to get the bias question actually (I believe it is $-\sigma^2/n$). Unfortunately, getting the variance of $\tilde\sigma^2$ has really frustrated me. Help on that aspect would be greatly appreciated! Thanks so much!

$\endgroup$
1
$\begingroup$

If you are given (or have established) that $X/\sigma^2 \sim \chi^2_{n-1}$, then knowledge of the mean and variance of the $\chi^2_\nu$ distribution gives you the answer easily, since $$\hat \sigma^2 = \frac{X}{n} = \frac{\sigma^2}{n} \frac{X}{\sigma^2}.$$ Hence $$\begin{align*} {\rm E}[\hat \sigma^2] &= \frac{\sigma^2}{n} {\rm E}\left[\frac{X}{\sigma^2}\right] = \frac{\sigma^2 (n-1)}{n}, \\ {\rm Var}[\hat\sigma^2] &= \frac{(\sigma^2)^2}{n^2} {\rm Var}\left[\frac{X}{\sigma^2}\right] = \frac{\sigma^4}{n^2} \cdot 2(n-1). \end{align*}$$

$\endgroup$
  • $\begingroup$ Ah, so we had to manipulate $\hat\sigma^2$ first, I see. Thanks so much, you've been a big help! $\endgroup$ – LearningIsPower Feb 25 '14 at 2:57

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.