Solve Jensen's Inequality: Proof & Troubleshooting

[GregA]

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

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

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

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

f(X)\geq \lambda(x-E(X))+f(E(X)) [2]​

then

E(f(x))\geq E(\lambda(x-E(X))+f(E(X))) [3]
=f(E(X)) [4]​

Q.E.D




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 x is meant to be X, a typo) then unless I'm wrong I get:

E(f(X))\geq E(\lambda(X-E(X))+f(E(X)))=E(\lambda(X-E(X)))+E(f(E(X))) (using E(g(X)+h(X))=E(h(X))+E(f(X)))
=\lambda E(X)-E(X)+f(E(X)) (using E(aX+b)=aE(X)+b and E(a)=a (where a,b are constants))
=(\lambda-1)E(X)+f(E(X))
and this isn't [4]
Where am I going wrong?

 

[mathman]

E is a linear function. Therefore E(λ(x-E(x))=λ(E(x-E(x))=λ(E(x)-E(x))=0.

 

[GregA]

aggh...Now I see where I've pulled the lambda out incorrectly it's obvious! cheers :)