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Find the prior distribution for the natural parameter of an exponential family

Cross Validated Asked by xxtensionxx on February 17, 2021

Show that for the binomial likelihood $y$ ~$Bin(n, theta)$, $p(theta) propto theta^{-1} (1-theta)^{-1}$ is the uniform prior distribution for the natural parameter of the exponential family.

I am trying to simply understand the solution to the question given above. The solution goes as follows:

The binomial can be put in the form of an exponential family with (using the notation of
Section 2.4) $f(y)$ = $ {n}choose{k}$, $g(theta) = (1-theta)^n$ and $u(y) = y$ and natural parameter $phi(theta) = log(theta/(1-theta))$.

A uniform prior density on $phi(theta)$, $p(phi) propto 1$ on the entire real line, can be transformed to give the prior density for $theta = frac{e^{phi}}{1+e^{phi}}$:

(And here comes the part I do not understand in the solution)

$$q(theta) = p(frac{e^{phi}}{1+e^{phi}})|frac{d}{dtheta}log(frac{theta}{1-theta})| propto theta^{-1} (1-theta)^{-1}$$

Could anybody pleas help me understand how they get top the last part?

One Answer

The Jacobian or change-of-variable formula lets one find the density of a bijective transform of a random variable. If $theta$ is a random variable with density $q_1(cdot)$ and if the density of $phi=phi(theta)$ is denoted $q_2(cdot)$ then $$q_2(theta)=q_1(phi(theta)) times underbrace{left|frac{text{d}phi(theta)}{text{d}theta}right|}_{text{Jacobian}}$$

Answered by Xi'an on February 17, 2021

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