Let $y_1, \dots,y_n$ be i.i.d. random variables from $Exp(\theta)$, where $\theta$ is scale parameter. I have to find Fisher information $i(\theta)$.

The density function is $$f(y)=\frac{1}{\theta}e^{-\frac{y}{\theta}}$$

and the likelihood function $$L(\theta)=\frac{1}{\theta^n}e^{-\frac{\sum^{n}_{i=1}y_i}{\theta}}$$

The log-likelihood is


Now, the score function $$l_*(\theta)=\frac{dl(\theta)}{d\theta}=-\frac{n}{\theta}+\frac{1}{\theta^2}\sum^{n}_{i=1}y_i$$

given the MLE

$$\hat \theta=\frac{\sum^{n}_{i=1}y_i}{n}$$

I differentiate again to find the observed information


and Finally fhe Fisher information is the expected value of the observed information, so


Is everything correct?


Yes it's correct. Very well done.

This doesn't simplify the work a lot in this case, but here's an interesting result . . . In the case of $n$ i.i.d. random variables $y_1,\dots,y_n$ , you can obtain the Fisher information $i_{\vec y}(\theta)$ for $\vec y$ via $n \cdot i_y (\theta$) where $y$ is a single observation from your distribution.

Here $\ell(\theta) = \ln( \frac{1}{\theta} e^{-y/\theta}) = -y/\theta - \ln(\theta) \implies \frac{\partial}{\partial \theta} \ell (\theta) = \frac{y}{\theta^2} - \frac{1}{\theta} \implies \frac{\partial^2}{\partial \theta^2} \ell(\theta) = - \frac{2y}{\theta^3} + \frac{1}{\theta^2}$ \begin{align*} i_y(\theta) &= - E \left[ \frac{\partial^2}{\partial \theta^2} \ell(\theta) \right] = -E \left[ - \frac{2y}{\theta^3} + \frac{1}{\theta^2} \right] = \dfrac{2 \theta}{\theta^3} - \dfrac{1}{\theta^2} = \dfrac{1}{\theta^2} \end{align*} and multiplying by $n$ gives Fisher information $n/\theta^2$.


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