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I know the expectation preserves the concavity (or convexity), but I was wondering is it still true that the expectation of log-concave function still log-concave; to be more precise,

Let $g(x,Y)$ be log-concave function in $x$ where $Y$ be discrete-time random variable with density $f_Y$. Is it true that $$ E[g(x,Y)] $$be still log-concave in $x$?

I noticed there is one result called Prekopa theorem, which states that if $g(x,y): \mathbb{R}^{n+m} \to \mathbb{R}$ be (jointly) log-concave, then $$ h(x) = \int g(x,y) dy $$is log-concave. But I'm not sure how to apply properly on the expectation case, since I have to deal with the log-concavity of integrand function $g(x,y) f_Y(y)$ first; i.e., $$ E[g(x,Y)] = \int g(x,y) f_Y(y)dy $$

Any suggestion is appreciated. Thanks

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Not necessarily. Consider $Y$ such that $Pr[Y=0]=Pr[Y=1]=1/2$. Define $g(x,Y)=e^{Yx}$. Then $g(x,Y)$ is log concave in $x$ because $\log g(x,Y) = Yx$ is linear. But: $$ E[g(x,Y)] = \frac{1 + e^x}{2} $$ and $\log E[g(x,Y)] = \log(1/2) + \log(1 + e^x)$, which is no longer concave.

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    $\begingroup$ Thanks @Michael !! very helpful example. $\endgroup$ – Fianra Jun 28 '15 at 1:15
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It is true under stricter hypothesis.

Proposition. Let $g(x,y)$ be jointly log-concave and $Y$ be a log-concave random variable with density $f(y)$. Then $\mathbb{E}[g(x,Y)]$ is log-concave.

Proof. The function $g(x,y)f(y)$ is log-concave because is logarithm $\log g(x+y) + \log f(y)$ is the sum of two concave functions. Therefore, because log-concavity is preserved under marginalization, $$ \mathbb{E}[g(x,Y)] = \int g(x,y)f(y)\,dy $$ is log-concave.

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