2nd derivative of $\sigma^2$ in normal MLE

Question

Many times I differentiated the MLE of the normal distribution, but when it came to $\sigma$ I always stopped at the first derivative, showing that indeed:

$$\hat\sigma^2 = \frac{\sum(y_i-\bar y)^2}{n} $$

But I haven't seen anywhere a proof this is indeed a maximum point. I tried now to differentiate it again, and look at the the 2nd derivative on that point, and yet I get (unless I made a mistake) that for some small value of $\sum(y_i-\bar y)^2$ the 2nd derivative on this point can actually be positive.

So how do you prove that $\hat\sigma^2$ is indeed a maximum?

Answer 1

The calculation of the derivative of the log-likelihood is shown here.

From there, you can find the second derivative is $$\frac{n}{\sigma^2}(1-\frac{3}{\sigma^2} \hat{\sigma}^2)$$
If you plug in $\hat{\sigma}^2$ for ${\sigma}^2$, then you get $\frac{n}{\hat{\sigma}^2}(1-3)=-2 \frac{n}{\hat{\sigma}^2}$