If $X$ is integrable and $\phi$ is concave, how to show $\phi (X)$ is integrable

Question

If $X$ is integrable and $\phi$ is concave, how to show $\phi (X)$ is integrable:

My idea:

define $\phi (X)^{+}=\max \{ \phi (X),0\}$ and $\phi (X)^{-}=\max \{- \phi (X),0\}$

Then $E[\phi (X)]=E[\phi (X)^{+}]+E[\phi (X)^{-}]=E[\max \{ \phi (X),0\}]+E[\max \{- \phi (X),0\}]$

I am not sure whether I can use Jensen on functions of the form $\max \{ X,0\}$

Any help?

Answer 1

This is not true in general. What you actually have is that $\phi(X)_+ \in L_1$ (under the assumptions that $X\in L_1(\mathbb{P})$ and $\phi$ concave).

By changing $\phi$ to $-\phi$, we can work with convex functions instead. We show that if $X\in L_1(\mathbb{P})$, $\phi(X)_-\in L_1(\mathbb{P})$.

Further assume that $\phi$ has domain $(a,b)$ where $-\infty\leq a<b\leq\infty$ and $a<X<b$.

Suppose $x_0=\int X\,d\mathbb{P}$. If $\rho=D^+\phi(x_0)$ then

$$\phi(X)\geq \phi(x_0) + \rho(X-x_0)$$

since $x\mapsto -\min(x,0)=x_-$ is non-increasing and $x_-\leq|x|$, $$ (\phi(X))_-\leq|\phi(x_0)|+|\rho|(|X|+|x_0|) $$

From this, it follows that $\phi(X)_-\in L_1$. $\phi(X)_+$ may not be integrable, however. In any case this is not important for in such case, $\int\phi(X)\,\mathbb{P}$ is well defined and takes the value $+\infty$.

Answer 2

Counter-example: $X \sim exp(1)$ and $\phi (x)=-e^{x}$.