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:


Expanding the inside I get:


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 :)


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!


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 $$\begin{align*} {\rm E}[\hat \sigma^2] &= \frac{\sigma^2}{n} {\rm E}\left[\frac{X}{\sigma^2}\right] = \frac{\sigma^2 (n-1)}{n}, \\ {\rm Var}[\hat\sigma^2] &= \frac{(\sigma^2)^2}{n^2} {\rm Var}\left[\frac{X}{\sigma^2}\right] = \frac{\sigma^4}{n^2} \cdot 2(n-1). \end{align*}$$

  • $\begingroup$ 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

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