Inequality

[Papapetros Vaggelis]

Let \(\displaystyle{X}\) be a random variable and \(\displaystyle{I}\) an open interval in \(\displaystyle{\mathbb{R}}\) .

If \(\displaystyle{f:I\to \mathbb{R}}\) is a convex function, \(\displaystyle{P(X\in I)=1}\) and

\(\displaystyle{E(X)\,\,,E(f(X))}\) exist, then prove that

\(\displaystyle{f(E(X))\leq E(f(X))}\) .

[ZardoZ]

$$\begin{eqnarray*}f\left(\mathbb{E}[\mathbb{X}]\right) &=& f\left(\sum_{i}x_{i}\cdot p(x_i)\right)\\&=&f\left(\sum_{i}\left(\frac{1}{2}x_{i}+\left(1-\frac{1}{2}\right)x_i\right)\cdot p(x_i)\right) \\
&\leq& \frac{1}{2}\sum_{i} f(x_i)p(x_i) + \frac{1}{2}\sum_{i} f(x_i)\cdot p(x_i)\\ &=&\frac{1}{2} \mathbb{E}\left[f(\mathbb{X})\right]+ \frac{1}{2}\mathbb{E}\left[f(\mathbb{X})\right] \\&=&\mathbb{E}\left[f(\mathbb{X})\right]\end{eqnarray*}$$

[Papapetros Vaggelis]

Thank you Zardoz for your solution.

Here is another one.

Solution

Since \(\displaystyle{P(X\in I)=1}\), we have that \(\displaystyle{\mu=E(X)\in I}\).

The function \(\displaystyle{f}\) is convex, so it has a straight line at \(\displaystyle{x=\mu}\), that is,

there exists \(\displaystyle{u\in\mathbb{R}}\) such that \(\displaystyle{f(x)\geq f(\mu)+u\,(x-\mu)\,,\forall\,x\in I}\) .

Therefore, \(\displaystyle{f(X)\geq f(\mu)+u\,(X-\mu)}\) and

\(\displaystyle{E(f(X))\geq f(\mu)+u\,(E(X)-\mu)=f(\mu)=f(E(X))}\) .