How to find the Rao-Cramer lower bound of
$$f(x; \theta) = \dfrac{1}{\theta^2}x \exp(-x/\theta)$$
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I know I have to take the $\ln$ of $f(x;\theta)$. I got $$\ln(f(x; \theta)) = -2\ln(\theta) + \ln(x)-\frac{x}{θ}.$$
Then I took the derivative in order to get the score function as $$S(\theta) = \dfrac{\mathrm d \ln(f(x; \theta))}{\mathrm d \theta} = -\dfrac{2}{\theta} + \dfrac{x}{θ^2}$$
The Fisher information is equal to $$-\mathbb E\left[\dfrac{\mathrm dS(\theta)}{\mathrm d \theta} \right] = \mathbb E \left[-\dfrac{2}{\theta^2} - \dfrac{2x}{\theta^3} \right].$$
I know that I need to get the negative expectation of this, but I'm not sure how to do it. $-\mathbb E(2/\theta^2) = 2/\theta^2$ and $-\mathbb E(2x/\theta^3)$ I think should equal $-2 \theta /\theta^3 = -2/\theta^2$?
It's wrong. Please show me what I have done wrong. The answer should be $$\mathcal I(\theta) = 2/\theta^2.$$
Thank you!
$$ -\mathbb E\left[\dfrac{\mathrm dS(\theta)}{\mathrm d \theta} \right] = \mathbb E \left[-\dfrac{2}{\theta^2}\, {\bf\color{red}{+}}\, \dfrac{2X}{\theta^3} \right]=-\dfrac{2}{\theta^2}+\dfrac{2\mathbb E[X]}{\theta^3}=-\dfrac{2}{\theta^2}+\dfrac{2\cdot 2\theta}{\theta^3}=\dfrac{2}{\theta^2}. $$