Integral Convergence for Expectation (Fisher Information Matrix)

Question

I have almost finished evaluating a Fisher Information matrix for my pdf where $x>0$. I am stuck with the following expression, and cannot seem to calculate the expectation below:

$$ I(\theta) = -E\left[ \frac{3}{\theta^2}-\frac{4}{(x+\theta)^2}\right]$$

Solving I have:

$$ I(\theta) = - \frac{3}{\theta^2}-E\left[\frac{4}{(x+\theta)^2}\right]$$

$$E\left[ \frac{4}{(x+\theta)^2} \right] = 4\int_0^\infty \frac{x}{(x+\theta)^2} \, dx = \left. \log(x+\theta)+\frac 1 {x+\theta} \right|^{x=\infty}_{x=0}$$

The above expression does not converge. Did I evaluate the integral incorrectly or did I set up the expectation wrong?

Answer 1

• You mentioned a pdf. Let's call it $f$. Then you should have $$ \operatorname{E}\left[ \frac 4 {(X+\theta)^2} \right] = \int_0^\infty \frac 4 {(x+\theta)^2} f(x)\,dx. $$
• Notice that I used capital $X$ for the random variable and lower-case $x$ for the variable with respect to which one integrates. They're two different things. Without distinguishing between them, how would one understand such an expression as $\Pr(X\le x)\text{?}$