# Fisher Information invariant by a specific reparameterization of the Exponential Distribution

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.

## 1 Answer

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 $$\beta=1/\lambda$$.

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