I would like some help with calculating the Fisher Information $I_o(\beta)$ and the expected information for a gamma distribution defined by

\begin{align*} f_X(x) = \frac{\beta^\alpha x^{\alpha - 1}e^{-\beta x}}{\Gamma(\alpha)} \; x > 0, \alpha >0, \beta > 0 \end{align*}

Where $\alpha$ is a known value and $\beta$ is the parameter of interest.


I have attempted to calculate a likelihood function as follows:

\begin{align} L(\beta | X_i) &= \prod_{i = 1}^{n}f(x_i | \alpha, \beta) \\ &= \prod_{i = 1}^{n}\left( \frac{\beta^{\alpha}}{\Gamma(\alpha)}x_{i}^{\alpha - 1}\mathrm{exp}{\{-\beta x\}}\right) \\ &= \left(\frac{\beta^{\alpha}}{\Gamma(\alpha)}\right)^{n}\prod_{i = 1}^{n} \left(x_{i}^{\alpha-1}\right) \mathrm{exp}\{-\beta\sum_{i = 1}^{n} x_i\} \\ L(\beta | X_i) &= \left(\frac{\beta^{\alpha}}{\Gamma(\alpha)}\right)^{n}\left(\prod_{i = 1}^{n} x_{i}\right)^{\alpha-1} \mathrm{exp}\{-\beta\sum_{i = 1}^{n} x_i\} \end{align}

Thus the log-likelihood would be the following:

\begin{align} \ell(\beta | x_i) &= \ln\left((L(\beta | X_i)\right)) \\ &= n\alpha\ln(\beta) - n\ln(\Gamma(\alpha)) + (na-n)\ln(x_i) - \beta \sum_{i = 1}^{n} x_i \\ \end{align}

I understand that the information is found by taking the 2nd derivative of any of the likelihood functions where $I_o(\beta) = -\frac{\mathrm{d}^{2}{\ell}}{\mathrm{d}{\beta}^{2}} $

The derivatives calculated were as follows: \begin{align} &= n\alpha\ln(\beta) - n\ln(\Gamma(\alpha)) + (na-n)\ln(x_i) - \beta \sum_{i = 1}^{n} x_i \\ &= \frac{n\alpha}{\beta} - \beta \\ &= - \frac{n \alpha}{\beta^2} - 1 \end{align}

Thus the information would be

\begin{align} I_0(\beta) = \frac{n \alpha}{\beta^2} + 1 \end{align}

I am unsure what to do once I get to the expectation.

\begin{align} \mathbb{E}\{I_0(\beta)\} &= \mathbb{E}\left(\frac{n \alpha}{\beta^2} + 1\right) \end{align}

Would I have made a mistake within the derivation process? Any insight would be very much appreciated.


1 Answer 1


You almost got it right ! You just made a tiny mistake when computing the derivative of the log-likelihood, you should have had : $$\frac{\partial\ell}{\partial \beta} = \frac{n\alpha}{\beta} -\color{red}{\sum_{i=1}^n x_i} $$

From which it follows that $$\frac{\partial^2\ell}{\partial \beta^2} = -\frac{n\alpha}{\beta^2} $$

Next, to compute the Fisher information, all you have to do is to take the expectation of $-\frac{\partial^2\ell}{\partial \beta^2} $. However, that expectation is with respect to the distribution of $x_i$, and there are no $x_i$'s in the expression of $\frac{\partial^2\ell}{\partial \beta^2}$, it is a constant ! Its expectation is thus equal to itself : $$ \mathbb E\left[-\frac{\partial^2\ell}{\partial \beta^2}\right] = \mathbb E\left[\frac{n\alpha}{\beta^2}\right] = \frac{n\alpha}{\beta^2}. $$

  • $\begingroup$ Thank you! I appreciate it. $\endgroup$ Commented Apr 10, 2023 at 3:55

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