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  • For Exponential RVs $X \sim \exp(\beta)$, If the density is $$f(X|\beta) = \frac1\beta e^{-\frac1\beta X}$$ The Fisher information is $$\begin{align*} \mathcal I_{\beta} & = n\mathbb E_{\beta}\left( \left(\frac{\partial \log f(X| \beta)}{\partial \beta}\right)^2\right) \\ & = n\int_0^\infty \left(\frac{\partial \log f(X|\beta)}{\partial \beta}\right)^2 \, f(X|\beta) \, dx \\ &=n\int_0^\infty \left(\frac{\partial}{\partial \beta}\left(-\frac{1}{\beta}x+\log\frac{1}{\beta}\right)\right)^2 \, \frac{1}{\beta}e^{-\frac{1}{\beta}x} \, dx \\ & = n\int_0^\infty \left(\frac{x}{\beta^2} -\frac{1}{\beta}\right)^2 \, \frac{1}{\beta}e^{-\frac{1}{\beta}x} \, dx \\ & = n\int_0^\infty \left(\frac{x^2}{\beta^4} -2\frac{x}{\beta^3}+\frac{1}{\beta^2}\right) \, \frac{1}{\beta}e^{-\frac{1}{\beta}x} \, dx \\ &=n\left(\frac{2}{\beta^2}-\frac{2}{\beta^2}+\frac{1}{\beta^2}\right)\\ &=\frac{n}{\beta^2}\\ \end{align*}$$

  • For Exponential RVs $X \sim \exp(\lambda)$, If the density is $$f(X|\lambda) = \lambda e^{-\lambda X}$$ The Fisher information is $$\begin{align*} \mathcal I_{\lambda} & = n\mathbb E_{\lambda}\left( \left(\frac{\partial \log f(X| \lambda)}{\partial \lambda}\right)^2\right) \\ & = n\int_0^\infty \left(\frac{\partial \log f(X|\lambda)}{\partial \lambda}\right)^2 \, f(X|\lambda) \, dx \\ &=n\int_0^\infty \left(\frac{\partial}{\partial \lambda}\left(-\lambda x+\log\lambda\right)\right)^2 \, \lambda e^{-\lambda X}\, dx \\ & = n\int_0^\infty \left(-x +\frac{1}{\lambda}\right)^2 \, \lambda e^{-\lambda X} \, dx \\ & = n\int_0^\infty \left(x^2 -2\frac{x}{\lambda}+\frac{1}{\lambda^2}\right) \, \lambda e^{-\lambda X} \, dx \\ &=n\left(\frac{2}{\lambda^2}-\frac{2}{\lambda^2}+\frac{1}{\lambda^2}\right)\\ &=\frac{n}{\lambda^2}\\ \end{align*}$$

Am I correct? I am confusing why the different pdf get the same formula of Fisher information? And since the variance of the Exponential distribution is $\beta^2$ in the first formula and $\frac{1}{\lambda^2}$ in the second formula, what is the relationship between Fisher information and the variance of Exponential distribution?

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  • $\begingroup$ They should be the same p.d.f. The first density you wrote is that of a Exp($\beta^{\color{red}{-1}}$) r.v., not Exp($\beta$). $\endgroup$
    – nejimban
    Oct 5, 2021 at 18:07
  • $\begingroup$ Do you think they are both correct? $\endgroup$
    – Dan Li
    Oct 5, 2021 at 19:13

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