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I have the negative binomial distribution and want to find the fisher information: $I(\theta) = V[\ell'(\theta)]$

How do i calculate this?

I know that the derivative of the log-likelihood is: $\ell'(\theta) = \frac{r}{\theta} - \frac{X - r}{1 - \theta}$

Also i know that $V[\ell'(\theta)] = \frac{V[X]}{(1-\theta)^2}$, but I don't understand how you get to that calculation.

I need to find the Fisher information to calculate Jeffreys prior, but now I am stuck.

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Ok, now I got it.

$$p(y) = {y-1 \choose r-1} \theta^r(1-\theta)^{y-r}$$

Let's denote $X$ as the random variable, then:

$$\ell(\theta) = C + r \ln(\theta) + (X-r) \ln(1-\theta)$$

$$\ell'(\theta) = \frac{r}{\theta} - \frac{X-r}{1-\theta}$$

$$\ell''(\theta) = -\frac{r}{\theta^2} - \frac{X-r}{(1-\theta)^2}$$

Now, we calculate the Fisher information, remember that $E[X] = \frac{r}{\theta}$

$I(\theta) = V[\ell'(\theta)] = -E[\ell''(\theta)]= -E\left[-\frac{r}{\theta^2} - \frac{X-r}{(1-\theta)^2}\right]$

Everything in the expression is constants except for $X$.

$-E\left[-\frac{r}{\theta^2} - \frac{(X-r)}{(1-\theta)^2}\right] = \frac{r}{\theta^2} + \frac{E[X]-r}{(1-\theta)^2} = \frac{r}{\theta^2} + \frac{\frac{r}{\theta}-r}{(1-\theta)^2} = r\left(\frac{1}{\theta^2} + \frac{\frac{1 - \theta}{\theta}}{(1-\theta)^2}\right)$

Now this gets simplified to:

$r\left(\frac{1}{\theta^2} + \frac{\frac{1 - \theta}{\theta}}{(1-\theta)^2}\right) = r\left(\frac{1}{\theta^2} + \frac{(1 - \theta)}{\theta(1-\theta)^2}\right) = r\left(\frac{1-\theta}{(1-\theta)\theta^2} + \frac{\theta}{\theta^2(1-\theta)}\right) = r\left(\frac{1}{\theta^2(1-\theta)}\right)$

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