6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
8
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.