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 Post subject: InequalityPosted: Thu Feb 25, 2016 9:40 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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

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

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 Post subject: Re: InequalityPosted: Sat Mar 05, 2016 11:37 am

Joined: Wed Nov 11, 2015 12:47 pm
Posts: 13
$$\begin{eqnarray*}f\left(\mathbb{E}[\mathbb{X}]\right) &=& f\left(\sum_{i}x_{i}\cdot p(x_i)\right)\\&=&f\left(\sum_{i}\left(\frac{1}{2}x_{i}+\left(1-\frac{1}{2}\right)x_i\right)\cdot p(x_i)\right) \\ &\leq& \frac{1}{2}\sum_{i} f(x_i)p(x_i) + \frac{1}{2}\sum_{i} f(x_i)\cdot p(x_i)\\ &=&\frac{1}{2} \mathbb{E}\left[f(\mathbb{X})\right]+ \frac{1}{2}\mathbb{E}\left[f(\mathbb{X})\right] \\&=&\mathbb{E}\left[f(\mathbb{X})\right]\end{eqnarray*}$$

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 Post subject: Re: InequalityPosted: Sat Mar 05, 2016 6:20 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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

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

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