# What is the CLRB of sigma^2?

I was doing this exercise and then I checked the solution, but I got the solution wrong.

$X_{1},...X_{n} \sim^{iid} N(0,\theta)$, i.e.

$f_{X_{i}} (x)= \frac{1}{\sqrt{2 \pi \theta}} -e^\frac{x^2}{2 \theta}$

b) find the CRLB for the class of unbiased estimators of $\theta$

So I compute the Likelihood and the logarithm, then I take first and second derivative and the calculation then should be:

$I =-E(\frac{1}{2\theta^2} - \frac{x^2}{\theta^3}$) and then CRLB = $\frac{1}{nI}$

However, doing this I came to the wrong result, can help me understand why?

• Precisely what result did you obtain? – whuber Sep 5 '18 at 11:52

$$I(\theta)=-\mathbb{E}(l^{''}\left(X\vert\theta\right))=-\mathbb{E}\left(\frac{1}{2\theta^2}-\frac{X^2}{\theta^3}\right)=\frac{1}{2\theta^2}$$
and therefore the CRLB is $$CRLB=\frac{2\theta^2}{n}$$