Let $y_1, \dots,y_n$ be i.i.d. random variables from $Exp(\theta)$, where $\theta$ is scale parameter. I know that $i_{\theta}(\theta)=\frac{n}{\theta^2}$ is the Fisher information. If I apply a reparametrization using the rate parameter $\lambda=\frac{1}{\theta}$ I get the new Fisher information $i_{\lambda}(\lambda)=\frac{n}{\lambda^2}$.

Now, based on this, I can get the relation between $i_{\lambda}(\lambda)$ and $i_{\theta}(\theta)$ by $$i_{\lambda}(\lambda)=i_{\theta}(\theta(\lambda))\left(\frac{d\theta}{d\lambda}\right)^2$$ so $$i_{\lambda}(\lambda)=\frac{n}{\lambda^2}=\frac{n}{(\frac{1}{\lambda})^2}\left(-\frac{1}{\lambda^2}\right)^2=\frac{n}{\lambda^2}$$

Is this correct?


1 Answer 1


Yes, it's correct. To check , I'll directly calculate the F.I. for $y$ from an exponential distribution with rate parameter $\lambda$; i.e. $f_y(x) = \lambda e^{-\lambda x}$ for $x \in [0,\infty)$.

Then, since our data is i.i.d., we get that the Fisher information $i_{\vec y}(\lambda) = n \cdot i_{y}(\lambda)$.

Here, $\ell(\lambda) = \ln( \lambda e^{-\lambda y}) = \ln(\lambda) - \lambda y \implies \frac{\partial}{\partial \lambda} \ell(\lambda)= \frac{1}{\lambda} - y \implies \frac{\partial^2}{\partial \theta^2} \ell(\lambda) = - \frac{1}{\lambda^2}$

\begin{align*} i_y(\theta) &= - E \left[ \frac{\partial^2}{\partial \lambda^2} \ell(\lambda) \right] = -E \left[ - \frac{1}{\lambda^2} \right] = \frac{1}{\lambda^2} \end{align*}

and multiplying by $n$ agrees with your answer.


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