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Let $(U , \mu)$ and $(V,\nu)$ be probability spaces. Let $f$ be a convex functional on $L^1(\mu)$, i.e.

$$f(tX + (1-t)Y) \leq t f(X) + (1-t)f(Y)$$

for all random variables $X$ and $Y$ in $L^1(\mu)$. Is there a version of Jensen's inequality for $f$? i.e., can we say that

$$E_\nu f(Z) \geq f(E_\nu (Z) )$$

for sufficiently well-behaved random variables $Z$ on $U\times V$?

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You seem to be looking for conditional Jensen inequality. –  Did Jun 10 '12 at 16:07
    
Any luck with this hint? –  Did Jun 18 '12 at 8:45
    
Yes, thank you. That is all I needed. –  user15464 Jun 18 '12 at 11:42
    
Doesn't a convex function satisfy $f(tX + (1-t)Y)\color{red}{\le} t f(X) + (1-t)f(Y)$? –  robjohn Jun 18 '12 at 12:26
    
Yes, sorry. I have corrected the typo. –  user15464 Jun 18 '12 at 21:28
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up vote 0 down vote accepted

You seem to be looking for conditional Jensen inequality.

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