I am having trouble with the following proof of Jensen's Inequality. I'll post the statement of the theorem, it's proof, and where I'm having problems:(adsbygoogle = window.adsbygoogle || []).push({});

Let [itex]X[/itex] be a random variable

with [itex]E(X) < \infty [/itex], and let [itex]f : \mathbb{R}\rightarrow\mathbb{R}[/itex] be a convex function. Then

[itex]\begin{equation*}f(E(X))\leq E(f(X))\end{equation*}[/itex] [1]

where a function is convex if [itex] \forall x_0\in \mathbb{R},\ \exists \lambda \in \mathbb{R}: f(x)\geq \lambda(x-x_0)+f(x_0)[/itex]

Proof: Let [itex]f[/itex] be convex and let [itex]\lambda \in \mathbb{R}[/itex] be such that

[itex] f(X)\geq \lambda(x-E(X))+f(E(X))[/itex] [2] then

[itex] E(f(x))\geq E(\lambda(x-E(X))+f(E(X))) [/itex] [3]Q.E.D

[itex]=f(E(X))[/itex] [4]

As is probably clear from my having problems with this, probability and dealing with expectations isn't my strong point but getting from [3] to [4] isn't looking obvious to me at all since if I expand RHS of [3] (and assume [itex]x[/itex] is meant to be [itex]X[/itex], a typo) then unless I'm wrong I get:

[itex]E(f(X))\geq E(\lambda(X-E(X))+f(E(X)))=E(\lambda(X-E(X)))+E(f(E(X)))[/itex] (using E(g(X)+h(X))=E(h(X))+E(f(X)))

[itex]=\lambda E(X)-E(X)+f(E(X))[/itex] (using E(aX+b)=aE(X)+b and E(a)=a (where a,b are constants))

[itex]=(\lambda-1)E(X)+f(E(X))[/itex]

and this isn't [4]

Where am I going wrong?

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# Jensen's Inequality

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