Wald test for variance of normal distribution Let $y_1, y_2,...,y_t$ follow a $N(0,\sigma^2)$ distribution. [Note that the mean is zero and you know that it is zero]. Derive the LR, LM and Wald test of hypothesis $\sigma^2 = 1$.
I have got the MLE, the restricted and unrestricted estimates. I am in particular concerned about the Wald test. For it, we need the distribution of $\sigma^2$ right? Is it going to be normally distributed? I doubt it. I at least need the variance of $\sigma^2$ right? Or rather the variance of the MLE estimate of $\sigma^2$. What would that be?
Any help is appreciated.
Regards Ash
 A: The general Wald statistic for testing $H_0: \theta = \theta_0$ is given by
$$\frac{(\hat{\theta} - \theta_0)^2}{Var(\hat{\theta})},$$
so you want the variance of $\hat{\sigma^2}$, the MLE for $\sigma^2$.
You have already derived that
$$\hat{\sigma^2} = \frac{1}{t} \sum_{i = 1}^t (Y_i - 0)^2 = \frac{1}{t} \sum_{i = 1}^t Y_i^2$$
(normally we would have $\bar{Y}$ instead of $0$ in the middle but you told us that the mean was known to be $0$), so the variance of $\hat{\sigma^2}$ is
$$Var\left(\frac{1}{t} \sum_{i = 1}^t Y_i^2\right) = \frac{1}{t^2} Var\left(\sum_{i = 1}^t Y_i^2\right),$$
because $Var(aX) = a^2Var(X)$ for any constant $a$.
Each independent random variable $Y_i \sim N(0, \sigma^2)$, so $Y_i / \sigma^2 \sim N(0, 1)$. Hence
$$\frac{1}{\sigma^2} \sum_{i = 1}^t Y_i^2 = \sum_{i = 1}^t \frac{Y_i^2}{\sigma^2} \sim \chi^2(t)$$
So $$Var(\chi^2(t)) = 2t = Var\left(\frac{1}{\sigma^2} \sum_{i = 1}^t Y_i^2\right) = \frac{1}{\sigma^4} Var\left(\sum_{i = 1}^t Y_i^2 \right).$$
Putting this together we get
$$Var(\hat{\sigma^2}) = \frac{1}{t^2} 2t\sigma^4 = \frac{2\sigma^4}{t},$$
and so the Wald statistic would be
$$\frac{\left(\frac{1}{t} \sum_{i = 1}^t Y_i^2 - \sigma_0^2 \right)^2}{2\sigma_0^4/t},$$
and under $H_0$, this can be approximated by a $\chi^2(1)$ distribution.
