# For the family, $f_\theta(x)=\frac{e^{-(x-\theta)}}{1+e^{-(x-\theta)})^2}$, compute the fisher information, is it an exponential family?

For the family, $$f_\theta(x)=\frac{e^{-(x-\theta)}}{1+e^{-(x-\theta)})^2}$$, compute the fisher information, is it an exponential family, $$x\in \mathbb{R},\theta \in \mathbb{R}$$?

I computed the fisher information as $$I(\theta)=E[(\frac{\partial log f_\theta(x)}{\partial\theta})^2]$$

$$\frac{\partial}{\partial\theta}lnf_\theta(x)=\frac{\partial}{\partial\theta}\ln\frac{e^{-(x-\theta)}}{(1+e^{-(x-\theta)})^2}=\frac{\partial}{\partial\theta}[\ln e^{-(x-\theta)}-\ln(1+e^{-x_\theta)})^2]=1-2\frac{1}{1+e^{-(x-\theta)}}\cdot e^{-(x-\theta)}$$

Then $$(1-2\frac{1}{1+e^{-(x-\theta)}}\cdot e^{-(x-\theta)})^2=1-4\frac{1}{1+e^{-(x-\theta)}}\cdot e^{-(x-\theta)}+\frac{e^{-2(x-\theta)}}{(1+e^{-(x-\theta)})^2}$$

Then $$E(1-4\frac{1}{1+e^{-(x-\theta)}}\cdot e^{-(x-\theta)}+\frac{e^{-2(x-\theta)}}{(1+e^{-(x-\theta)})^2})=1-4e^\theta E(\frac{e^{-x}}{1+e^\theta e^{-x}})+e^{2\theta}E(\frac{e^{-2x}}{1+e^{-(x-\theta)})^2})$$

$$E(\frac{e^{-x}}{1+e^\theta e^{-x}})=E(\frac{1}{\frac{1}{e^{-x}}+e^{\theta}})=\int_\mathbb{R}\frac{e^{-(x-\theta)}}{1+e^{-(x-\theta)})^2}\cdot\frac{1}{\frac{1}{e^{-x}}+e^{\theta}} dx$$

So my issue now is I don't feel I can compute this integral, so maybe I've made a mistake in computing this somewhere.

For showing whether it is or isn't an exponential family, I believe it is not, but I'm not sure how to prove something is not an exponential family, only how to show it is.

My definition of exponential family is:

if $$f_\theta(x)=h(x)e^{\sum_{i=1}^k c_i(\theta)T_i(x)-d(\theta)}$$ then $$f_\theta(x)$$ is a k-parameter exponential family.

• Use the alternative formula $I(\theta)=-E_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(X)\right]$. Mar 14, 2022 at 19:22

## 1 Answer

It's good advice to switch to an alternative definition of Fisher's information if one of them takes you to a dead end to no avail. Observe that $$\frac{d^2\ell}{d\theta^2} = -\frac{2e^{-(x-\theta)}}{(1+e^{-(x-\theta)})^2}.$$ Therefore, \begin{align} I(\theta) &= -E\left(\frac{d^2\ell}{d\theta^2}\right) = 2\int_{-\infty}^{-\infty}\frac{e^{-(x-\theta)}}{(1+e^{-(x-\theta)})^2}\frac{e^{-(x-\theta)}}{(1+e^{-(x-\theta)})^2}\,dx\\ &= 2\int_{-\infty}^{\infty}\frac{e^{-2(x-\theta)}}{(1+e^{-(x-\theta)})^4}\,dx \end{align}. Consider the transformation $$y = \frac{e^{-(x-\theta)}}{1+e^{-(x-\theta)}}$$ whose Jacobian is $$dx = \frac{1}{(y-1)y}\,dy$$. Note that $$y\to 0$$ as $$x\to\infty$$ and $$y\to 1$$ as $$x\to-\infty$$. Therefore, $$\int_{-\infty}^{\infty}\frac{e^{-2(x-\theta)}}{(1+e^{-(x-\theta)})^4}\,dx = -\int_1^0 \frac{y^2(1-y)^2}{y(1-y)}\,dy=\int_0^1 y(1-y)\,dy = B(2,2),$$ where $$B(a,b)$$ is the beta function. Using the relationship between the beta function and the gamma function, $$B(2,2) = \frac{\Gamma(2)\Gamma(2)}{\Gamma(4)} = \frac{1}{6}$$. Therefore, $$I(\theta) = \frac{1}{3}$$.

Now, does the logistic distribution in your question belong to an exponential family? No, because the density cannot be represented as $$h(x)\exp\{\eta(\theta)T(x)-A(\theta) \}$$. Observe that the density $$e^{-x}\exp\left[-2\log\{1+\exp(-(x-\theta))\} + \theta \right]$$ is impossible to be rewritten in terms of the product of $$\eta(\theta)T(x)$$.