Asymptotic behavior of Posterior distributions

Question

I'm having a hard time understanding a passage from a famous paper, namely

Walker, A. M. “On the Asymptotic Behaviour of Posterior Distributions.” Journal of the Royal Statistical Society. Series B (Methodological), vol. 31, no. 1, 1969, pp. 80–88. link

Between formulas (7) and (8), there is a passage where the author moves on to proving point i). I will write the relevant passage, but have left the link for all the assumptions.

Quote:

Let $Z_i=\frac{log(f(X_i|\theta))}{log(f(X_i|\theta_0))}$. Then if $E(Z_i)$ is finite, it follows from the concavity of the logarithmic function that $$E(Z_i)<logE(e^{Z_i})=0$$

Now, I understand the use of the logarithmic concavity, but how is then the logarithm of the expected value of the exponential of the random variable equal zero? Does it have anything to do with the moment generating function of $Z_i$?

Answer 1

There are two steps here, the first is the use of Jensen's Inequality, and the second an expectation trick. First since $\log()$ is a concave function, by Jensen's Inequality for a random variable $X$, $$E(\log(X) ) \leq \log(E(X)).$$

Thus, \begin{align*} E(Z_i) & = E[ \log(e^{Z_i}) ]\\ &\leq \log(E(e^{Z_i})). \end{align*}

Next, lets calculate the expectation. \begin{align*} E(e^{Z_i}) & = \int \exp \left\{\log \dfrac{f(X_i|\theta)}{f(X_i|\theta_0)} \right\} f(X_i | \theta_0 ) \,dX_i\\ & = \int \dfrac{f(X_i|\theta)}{f(X_i|\theta_0)}f(X_i | \theta_0 ) \,dX_i\\ & = \int f(X_i | \theta) dX_i\\ & = 1. \end{align*}

Combining both results, we get, $$E(Z_i) \leq \log (1) = 0\,. $$