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Let $X$ have the logistic distribution with the PDF $$f(x) = \frac{\exp(-x-θ)}{(1+\exp(-x-θ))^{2}}$$ Does $f(x)$ belong to the exponential family?

My solution is $\exp[(-2)\cdot \ln(1+\exp\{-x-θ\})-x-θ]$. Since it does not have the form $Q(θ)T(x)$, it does not belongs to exponential family.

At the end of my book there is an answer, which is "Yes", meaning it belongs to the exponential family.

Is my solution right or wrong?

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    $\begingroup$ Your solution needs details. What are you looking at in your solution and why? You seem to have omitted starting from some fact, result or theorem. You need to explain what you're doing -- where did the expression in your answer come from? $\endgroup$
    – Glen_b
    Apr 25, 2017 at 21:33
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    $\begingroup$ Please add the self-study tag (this will help later readers looking for answers offering advice and hints rather than complete solutions). See its tag wiki here: [tag: self-study] . $\endgroup$
    – Glen_b
    Apr 25, 2017 at 21:41
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    $\begingroup$ $Q(\theta)T(x)$? What are these? This is not the exponential form. An exponential family can be written as: $h(x)c(\theta)exp[\sum_{i}^{k}w_i(\theta)t_i(x)]$ $\endgroup$ Apr 25, 2017 at 21:50
  • $\begingroup$ But your conclusion is still right, the logistic distribution family is not an exponential family. See the comment by @StatsStudent. Which book are you using which says "yes"? Maybe it has some other def than the usual one? $\endgroup$ May 26, 2018 at 20:05

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Summary: No, the logistic distribution is not an exponential family.

You are missing some parenthesis in your definition, so I repeat the logistic density is $$ f(x) = \frac{\exp(-(x-θ))}{(1+\exp(-(x-θ)))^{2}}. $$ One interesting way to see the result is to start with the centered logistic density $$ f_0(x)=\frac{e^{-x}}{(1+e^{-x})^2}, $$ construct an exponential family by exponential tilting, and then observe that it is not the logistic distribution we get. (The logistic distribution is a location-scale family.)

First we need the cgf (cumulant generating function, log of moment generating function) of $f_0$, which is (from Wikipedia) $$ k(t)=\log B(1-t,1+t),\qquad t\in (-1,1) $$ where $B$ is the Beta function. Then by exponential tilting define the density $$ f_\theta(x)=\frac{f_0(x)e^{\theta x}}{e^{k(\theta)}} $$ and this manifestly does not have the form of a logistic density.

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