Let Xbe a random variable having finite expectation μand variance σ2, and let g(*)be a twice differentiable function. Show that

E[g(X)]g(μ)+g''(μ)2σ2

Hint: Expand g(·)in a Taylor series about μ. Use the first

three terms and ignore the remainder.

Short Answer

Expert verified

The random variable is showed as E[g(X)]g(μ)+g''(μ)2σ2having finite expectation.

Step by step solution

01

Given Information

The finite expectation of variance and twice differentiable function g(·).

02

Explanation

If n0is an integer and gis a function which is ntimes continuously differentiable on the closed interval [a,x]and n+1times differentiable on the open interval (a,x), then we have:

g(x)=g(a)+g'(a)1!(x-a)+g''(a)2!(x-a)2++g(n)(a)n!(x-a)n+Rn

The remainder term Rndepends on xand is small if xis close enough toa.

03

Explanation

Expand g(·)in Taylor polynomial:

g(x)=g(μ)+g'(μ)1!(x-μ)+g''(μ)2!(x-μ)2+Rn

Expected value of both side is,

[Eg(x)]=g(μ)+0+g''(μ)2!σ2+Rnσ2,E(x-μ)=0

[Eg(x)]g(μ)+g''(μ)2!σ2.

04

Final answer

The random variable is showed as [Eg(x)]g(μ)+g''(μ)2!σ2having finite expectation.

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