# Fisher information for a single sampling of an exponential distribution

I am viewing an example of finding the Fisher information for a single sampling from an exponential distribution where: $$P(x|\theta) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$$ The score $S$ is $S(x|\theta) = \frac{\partial}{\partial\theta}logP(x|\theta) = -\frac{1}{\theta} + \frac{x}{\theta^2}$. Fisher information is the expectency of $S^2$ which is: $$E_x[S^2] = E_x\left[\frac{1}{\theta^2} - 2\frac{x}{\theta^3} + \frac{x^2}{\theta^4}\right]$$

I know this might sound strange, but I don't know how to calculate this expectation. Something is mixed for me here. I know that $$E[P(x)]=\int xp(x)dx$$

But I can't connect the two pieces of information.

In the book, they got: $$E_x[S^2] = E_x\left[\frac{1}{\theta^2} - 2\frac{x}{\theta^3} + \frac{x}{\theta^4}] = \frac{1}{\theta^2} - 2\frac{\theta}{\theta^3} + \frac{2\theta^2}{\theta^4}\right] = \frac{1}{\theta^2}$$

But I can't see how they got that.

Any information will be useful. Thanks.

Since $X$ is an exponential distribution with parameter $\frac{1}{\theta}$, that is, $X\sim\text{exp}(\frac{1}{\theta})$, its mean is given by

$$E[X] = \frac{1}{\frac{1}{\theta}} = \theta,$$

and the variance is given by

$$\text{var}(X) = \frac{1}{\left(\frac{1}{\theta}\right)^2} = \theta^2.$$

By other hand, we have that

$$E[X^2] = \text{var}(X) + (E[X])^2 = 2\theta^2$$.

In addition, you can use the linearity property of the expected value

\begin{align} E_x[S^2] &= E_x\left[\frac{1}{\theta^2} - 2\frac{X}{\theta^3}+ \frac{X^2}{\theta^4}\right]\\ &= \frac{1}{\theta^2} - \frac{2}{\theta^3}E[X]+ \frac{1}{\theta^4}E[X^2] \end{align}

where we have also used the fact that $\theta$ is a constant (a deterministic parameter) and therefore $E\left[\frac{1}{\theta^2}\right] = \frac{1}{\theta^2}$.

• This is so more clear now... Thank you! – MathBgu Nov 17 '15 at 13:32