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The exponential distribution can be parameterized in two common ways: $$ f(x) = \lambda \exp(-\lambda x) $$ where $E[X] = \frac{1}{\lambda}$ $\text{Var}[X] = \frac{1}{\lambda^2}$, or as $$ f(x) = \frac{1}{\beta} \exp(-\frac{1}{\beta} x) $$ where $E[X] = \beta$ and $\text{Var}[X] = \beta^2$

When I calculate the Fisher Information using each of these parameterizations, I obtain $\dfrac{1}{\lambda^2}$ for the first parameterization and $\dfrac{1}{\beta^2}$ for the second.

Is there an intuitive reason why this is true? I would have assumed them to be inverses of each other.

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This is the impact of the Jacobian term for this specific transform (and only for this specific transform): denoting $I_1$ the information on $\beta$ and $I_2$ the information on $\lambda$ $$I_1(\beta)=I_2(\lambda(\beta)) \times \left(\frac{\text d\lambda}{\text d \beta} \right)^2=\frac{1}{\dfrac{1}{\beta^2}}\times\frac{1}{\beta^4}=\frac{\beta^2}{\beta^4}$$ Similar invariance properties can be found in other distributions by solving the differential equation $$I_2(\lambda(\beta)) \times \left(\frac{\text d\lambda}{\text d \beta} \right)^2=I_2(\beta)$$ See for instance the Poisson distribution where the Fisher informations are the same functions for $\lambda$ and $\beta=1/\lambda$.

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  • $\begingroup$ is this only true for exponential families? $\endgroup$
    – Carlos Llosa
    Feb 6 at 18:32
  • $\begingroup$ invariance of Fisher information to one-to-one transformations $\endgroup$
    – Carlos Llosa
    Feb 7 at 17:37
  • $\begingroup$ For any model with information matrix $I_2(\beta)$ such that the above equation allows for a solution in $\lambda$, this is true... $\endgroup$
    – Xi'an
    Feb 7 at 17:54

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