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 Author: Papapetros Vaggelis [ Thu Feb 25, 2016 9:40 pm ] Post subject: Inequality 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$$\displaystyle{f(E(X))\leq E(f(X))}$$ .

 Author: ZardoZ [ Sat Mar 05, 2016 11:37 am ] Post subject: Re: Inequality $$\begin{eqnarray*}f\left(\mathbb{E}[\mathbb{X}]\right) &=& f\left(\sum_{i}x_{i}\cdot p(x_i)\right)\\&=&f\left(\sum_{i}\left(\frac{1}{2}x_{i}+\left(1-\frac{1}{2}\right)x_i\right)\cdot p(x_i)\right) \\&\leq& \frac{1}{2}\sum_{i} f(x_i)p(x_i) + \frac{1}{2}\sum_{i} f(x_i)\cdot p(x_i)\\ &=&\frac{1}{2} \mathbb{E}\left[f(\mathbb{X})\right]+ \frac{1}{2}\mathbb{E}\left[f(\mathbb{X})\right] \\&=&\mathbb{E}\left[f(\mathbb{X})\right]\end{eqnarray*}$$

 Author: Papapetros Vaggelis [ Sat Mar 05, 2016 6:20 pm ] Post subject: Re: Inequality Thank you Zardoz for your solution.Here is another one.SolutionSince $$\displaystyle{P(X\in I)=1}$$, 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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