Assume the standard situation, that is, let X1, . . . , Xn be independent and identically distributed with $X_k ∼ P_θ$, where$P_θ(x) = 1/(2θ^3).x^2.exp(−x/θ) , 0 < x < ∞$. This is a special case of the Gamma distribution, θ is a positive parameter to be estimated. We are given that the mean and variance of this distribution in terms of $θ$ are $E(X) = 3θ$ and $V(X) = 3θ^2$, respectively.
Calculate the Cramer-Rao lower bound for the variance of unbiased estimators of θ.
Here's what i've done but im not sure if im right:
$I(θ)= -E_θ[\partial^2/\partialθ^2[ log((1/(2θ^3).x^2.exp(−x/θ))]$ where $log((1/(2θ^3).x^2.exp(−x/θ)= log(x^2/(2θ^3)) - x/θ$ therefore; $\partial^2/\partialθ^2[log(x^2/(2θ^3)) - x/θ] = 3/θ^2 - 2x/θ^3$
so $-E_θ[\partial^2/\partialθ^2[ log(x^2/(2θ^3)) - x/θ]= 2x/θ^3 -3/θ^2$
And therefore the CR lower bound is $= 1/(nI(θ)) = θ^3/n(2x-3θ)$
Am i correct here? Would really appreciate the help.
You need to replace $x$ by random variable $X$ with given distribution and calculate expectation: $$I(\theta)=\mathbb E\left[\frac{2X}{\theta^3} -\frac{3}{\theta^2}\right]=\frac{2\cdot \mathbb E[X]}{\theta^3}-\frac{3}{\theta^2}=\frac{2\cdot 3\theta}{\theta^3}-\frac{3}{\theta^2}=\frac{3}{\theta^2}.$$
So CRLB is $$\frac{1}{nI(\theta)}=\frac{\theta^2}{3n}.$$