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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?

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  • $\begingroup$ What is $E(f(x))$ in this case? Is it an Expectation? If so what is the probability distribution for $x$? Also the integral is not correct if we assume the 'expectation' is correct. The integral is $$\int \frac{1}{(x+\theta)^2}dx = \int-\dfrac{d}{dx}\frac{1}{x+\theta}dx=-\frac{1}{x+\theta}$$ $\endgroup$ – Chinny84 Nov 25 '16 at 17:02
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  • 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 $\Pr(X\le x)\text{?}$
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  • $\begingroup$ Correct, I did not multiply by my correct pdf f(x). $\endgroup$ – δ-Force Nov 25 '16 at 18:06

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