If $\omega$ is a Pólya-Gamma distributed random variable with parameters $b > 0$ and $c \in \mathbb{R}$, denoted $\omega \sim \text{PG}(b, c)$, then

$$ \omega \stackrel{\text{dist}}{=} \frac{1}{2 \pi^2} \sum_{k=1}^{\infty} \frac{g_{k}}{(k-1/2)^2 + c^2/(4\pi^2)} $$

In Bayesian inference for logistic models using Polya-Gamma latent variables, the authors prove the following identity

$$ \frac{(e^{\psi})^a}{(1 + e^{\psi})^b} = 2^{-b} e^{\kappa \psi} \int_{0}^{\infty} e^{- \omega \psi^2 / 2} p(\omega) \text{d}\omega $$

where $\kappa = a - b/2$ and $p(\omega) = \text{PG}(\omega \mid b, 0)$.

My problem: I do not understand the proof on page 7 at all. For example, the very first equality, when they write

$$ \frac{(e^{\psi})^a}{(1 + e^{\psi})^b} = \frac{2^{-b} \exp(\kappa \psi)}{\cosh^b(\psi/2)} $$

comes out of nowhere for me. I've asked around and it seems like people use the identity without understanding the proof. Can someone explain the proof of Theorem $1$ (page 7)?

(This question could also be posted on Cross Validated, but my hunch is that this proof leverages esoteric math rather than commonly known statistics.)


1 Answer 1


Start by plugging in $\kappa + \frac{b}{2}$ for a in equation (7):

$\frac{(e^{\psi})^a}{(1+e^{\psi})^b} = \frac{(e^{\psi})^{\kappa + \frac{b}{2}}}{(1+e^{\psi})^b} = \frac{(e^{\psi})^{\kappa}}{e^{-\frac{b\psi}{2}}(1+e^{\psi})^b} = \frac{(e^{\psi})^{\kappa}}{\left(\frac{1+e^{\psi}}{e^{\psi/2}}\right)^b}.$ (i)

Then manipulate the cosinus hyperbolicus function:

$cosh(x) = \frac{1}{2}(e^x+e^{-x}) \Rightarrow \hspace{.1cm} cosh(\frac{x}{2}) = \frac{1}{2}(e^{\frac{x}{2}}+e^{-\frac{x}{2}}) = \frac{1}{2} \left(e^{\frac{x}{2}}+\frac{1}{e^{x/2}}\right) = \frac{1}{2}\left(\frac{e^x+1}{e^{x/2}}\right).$

I.e. in the denominator of (i) $\frac{1+e^{\psi}}{e^{\psi/2}} = 2cosh(\psi/2).$

Plugging this in (i) we obtain: $\frac{(e^{\psi})^a}{(1+e^{\psi})^b} =\frac{(e^{\psi})^{\kappa}}{(2cosh(\psi/2))^b} = \frac{2^{-b}(e^{\psi})^{\kappa}}{cosh^b(\psi/2)}.$ (ii)

From equation (3) on page 4 we know that for a PG(b,0) variable $E[exp(-\omega t)] = \frac{1}{cosh^b\left(\sqrt{t/2}\right)}.$

As t is simply a constant we can set it to $t:= \frac{\psi ^2}{2}$.

This yields: $t:= \frac{\psi ^2}{2} \Rightarrow \frac{t}{2} = \frac{\psi ^2}{4} \Rightarrow \sqrt{\frac{t}{2}} = \frac{\psi}{2}$.

Plugging this into equation(3) we obtain: $E\left[exp\left(-\omega \frac{\psi ^2}{2}\right)\right] = \frac{1}{cosh^b(\frac{\psi}{2})}.$ (iii)

Plugging (iii) into (ii) we obtain

$\frac{(e^{\psi})^a}{(1+e^{\psi})^b} = \frac{2^{-b}(e^{\psi})^{\kappa}}{cosh^b(\psi/2)} = 2^{-b}e^{\psi \kappa}E\left[exp\left(-\omega \frac{\psi ^2}{2}\right)\right] = 2^{-b}e^{\psi \kappa} \int_0^{\infty}exp\left(-\omega \frac{\psi ^2}{2}\right)p(\omega)d\omega.$

As far as the second part of the theorem is concerned it amounts to plugging in $E\left[exp\left(-\omega \frac{\psi ^2}{2}\right)\right] = \int_0^{\infty}exp\left(-\omega \frac{\psi ^2}{2}\right)p(\omega)d\omega$ in equation (5) on page 5.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .