Prove that $E[U(X)] \ge E[U(Z)]$

Question

Let U: $\mathbb R$ -> $\mathbb R$ be a concave function, let X be a random variable with a finite expected value, and let Y be a random variable that is independent of X and has an expected value 0. Define Z=X+Y. Prove that $E[U(X)] \ge E[U(Z)]$

I know that $E(X)=E(Z)$, and by Jensen's inequality $U[E(X)] \ge E[U(X)]$ but it gives me nothing so far.

Please help. Thanks a lot.

Answer 1

Rewrite the thesis as

$$E[u(X)] = \int u(x) dP^X \ge \int u(x + y) dP^X \otimes dP^Y = E[u(X + Y)]$$

Now let's concentrate on the right hand side; in particular we can use fubini tonelli to write

$$\int u(x + y) dP^X \otimes dP^Y = \int \left(\int u(x+y)dP^Y\right) dP^X$$

The inner integral is simply $E[u(x + Y)]$ for some constant $x$. Since $u$ is concave we know (by Jensen) that $$E[u(x + Y)] \le u(E[x + Y]) = u(x)$$

Therefore $$E[u(X + Y)] = \int \left(\int u(x+y)dP^Y\right) dP^X \le \int u(x) dP^X = E[u(X)]$$

which is what we wanted to prove.

Answer 2

Note that the following equality holds: $\mathbb{E}(Z|X = x) = \mathbb{E}(X +Y|X = x) = \mathbb{E}(x +Y|X = x) = x + \mathbb{E}(Y|X = x) =x + \mathbb{E}(Y) = x $,

We are given that $u$ is concave, so by Jensen's inequality: $$\mathbb{E}(u(X)|X = x) = u(x) \geq \mathbb{E}(u(Z)|X = x)$$ for all $x$. Therefore, we have $$\mathbb{E}(u(X)|X) \geq \mathbb{E}(u(Z)|X)$$ Taking expectation both sides: $$\mathbb{E}(\mathbb{E}(u(X)|X)) \geq \mathbb{E}(\mathbb{E}(u(Z)|X))$$ Therefore, $$\mathbb{E}(u(X)) \geq \mathbb{E}(u(Z))$$