Let $X$ have a gamma distribution with $\alpha = 4$ and $\beta = \theta > 0$. Find the Fisher information $I(\theta)$. I have found the second derivative of the log of the likelihood function and then to find the information, I did:$E((-4/o^2)+(x/o^2))^2$. The answer is $\frac{4}{o^2}$ but I don't know how to get here.

  • $\begingroup$ Do you know that $E[(a-bX)^2] = a^2 -2abE[X]+b^2E[X^2] = a^2 -2ab\mu + b^2(\sigma^2+\mu^2)$? Can you apply it to your problem? $\endgroup$ – Dilip Sarwate Feb 11 '14 at 2:19

I'm doing this to work through this myself as much as help you. Lets give it a go.

PDF of a Gamma = $\frac{X^{\alpha-1}}{\Gamma(\alpha)\theta^{\alpha}}e^{\frac{X}{\theta}}$. Log likelihood is then:

\begin{align} L(\theta) &= (\alpha - 1) \Sigma \log X_i - n \log(\Gamma (\alpha)) - n\alpha \log(\theta) - \frac{1}{\theta} \Sigma X_i \\[5pt] \frac{\partial}{\partial \theta} &= -\frac{n\alpha}{\theta} + \frac{\Sigma X_i}{\theta^2} \\[5pt] \frac{\partial^2}{\partial \theta^2} &= \frac{n\alpha}{\theta^2} - \frac{2\Sigma X_i}{\theta^3} \end{align}

What is the expectation of a gamma dist? (looks like $\alpha \theta$)

\begin{align} -E \frac{\partial^2}{\partial \theta^2} &= -\frac{n\alpha}{\theta^2} + \frac{2\alpha n}{\theta^2} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \\[5pt] &= \frac{n\alpha}{\theta^2} \\[5pt] \alpha &= 4 \text{ so;} \\[5pt] &= \frac{4n}{\theta^2} \end{align}

so if $n = 1$ (i.e., a single observation from a gamma distribution, like this problem seems to be asking), then in fact the answer is:

$$ = \frac{4}{\theta^2}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$

Feel free to correct/critique my errors.

  • 1
    $\begingroup$ @RibD, despite the OP's comment, it is better for you to submit your own answer than edit someone else's to correct it. $\endgroup$ – gung May 3 '16 at 1:45
  • 3
    $\begingroup$ The expectation of $\sum_{i=1}^{n}X_{i}$ is not $\alpha\theta$. It should rather be $n\alpha\theta$. $\endgroup$ – Daeyoung Lim Jul 31 '16 at 7:26

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