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.

## Best 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 $\eta(\theta)T(x)$.