Show that if $X\sim\text{Pois}(\theta)$, then $I(\theta;X) = 1/\theta$.

Question

The Fisher information matrix $I(\theta;X)$ about $\theta$ based on $X$ is defined as the matrix with elements $$I_{i,j}(\theta;X) = \operatorname{Cov}_\theta\bigg(\frac{\partial}{\partial \theta_i}\log f_X(X\mid\theta), \frac{\partial}{\partial\theta_i}\log f_X(X\mid \theta) \bigg).$$

Exercise: Let $X\sim \text{Pois}(\theta)$. Show that $I(\theta;X) = 1/\theta$.

What I've tried: If I'm not mistaken then $\dfrac{\partial}{\partial \theta_i} \log f_X(X \mid \theta) = \dfrac{x_i}{\theta} + \log e$. The Fisher information matrix (a $1\times 1$ matrix in this example) would be given by $\operatorname{Cov}_\theta\left(\dfrac{x_i} \theta + \log e, \dfrac{x_i} \theta + \log e\right)$. We know that $\operatorname{Cov}_\theta\left(\dfrac{x_i} \theta + \log e, \dfrac{x_i}{\theta} + \log e\right) = \operatorname{Var}_\theta\left(\dfrac{x_i}\theta + \log e\right) = \operatorname{Var}_\theta \left(\dfrac{x_i}{\theta}\right) = x_i^2 \operatorname{Var}_\theta \left(\dfrac 1 \theta \right).$ Obviously I'm doing something wrong as the $x_i^2$ before the variance is a problem. Besides that, I'm not sure if I can calculate $\operatorname{Var}_\theta\left(\dfrac 1 \theta \right)$.

Question: How do I show that $I(\theta;X) = 1/\theta$?

Thanks in advance!

Answer 1

The mistake is in the last step: $$ \operatorname{Var}(x_i/\theta) {{}\color{red}={}} x_i^2\operatorname{Var}(1/\theta). $$ Instead, it should be $$ \operatorname{Var}(x_i/\theta)=\operatorname{Var}(x_i)/\theta^2=\theta/\theta^2=1/\theta. $$