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Jensen's Inequality

  1. Mar 4, 2010 #1
    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:

    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] ​
    [itex] E(f(x))\geq E(\lambda(x-E(X))+f(E(X))) [/itex] [3]
    [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))
    and this isn't [4] :confused:
    Where am I going wrong?
    Last edited: Mar 4, 2010
  2. jcsd
  3. Mar 4, 2010 #2


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    E is a linear function. Therefore E(λ(x-E(x))=λ(E(x-E(x))=λ(E(x)-E(x))=0.
  4. Mar 4, 2010 #3
    aggh...Now I see where I've pulled the lambda out incorrectly it's obvious!!! cheers :)
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