# Can’t Find the Fisher Information of This Function

Can anyone help me find the Fisher information for this function:

$$f(x|\lambda) = \lambda\,x^{\lambda-1}\quad \text{ where } \lambda \in [0,1]\,.$$

• Please define the domain of $x$. – Glen_b Dec 8 '14 at 4:25
• This looks like bookwork. Is this for some class/subject/exam revision/textbook exercise etc? Please double check your definition of $f$ carefully, including the domain of $\lambda$ as well as the unstated domain for $x$. – Glen_b Dec 8 '14 at 4:39

Assuming you mean that $x$ is on the (open) unit interval, the distribution is a special case of the Beta distribution.

The Fisher information for the more general two parameter beta case seems at least potentially doable.

So to begin with, you might take the negative of the second derivative of the log-likelihood with respect to $\lambda$ and try to find the expectation of that quantity, and see if you can do it the 'standard' way.

If you run into difficulty, you'll need to explain where that difficulty lies so that we can give specific guidance (i.e. show what you tried and where things went wrong).

As suggested before the Beta trick is a neat way of solving this problem, but you have to know recurrence properties of trigamma functions. Also, this is probably your homework, I am not the homework police so here you go....

$f(x\mid\lambda)=\lambda x^{\lambda-1}$

$l=\log f(x\mid\lambda)=\log\lambda+(\lambda-1) \log x$

$\frac{\partial l}{\partial \lambda}=\frac{1}{\lambda}+\log x\quad\text{and}\quad \frac{\partial^2 l}{\partial \lambda^2}=-\frac{1}{\lambda^2}$

$\mathbb{E}\left[-\frac{\partial^2 l}{\partial \lambda^2}\mid \lambda\right]=\frac{1}{\lambda^2}$