This question is leading on from the following question.

https://math.stackexchange.com/questions/360275/e1-1×2-under-a-normal-distributionBasically what is the E(11+x2) under a general Gaussian N(μ,σ2). I tried rewriting 11+x2 as a scalar mixture of Gaussians (∝∫N(x|0,τ−1)Ga(τ|1/2,1/2)dτ). This also came to a halt, unless you folks have got a trick under your belt.

If this integral isn’t analytical any sensible bounds?

**Answer**

Let fσ(x)=1√2πσexp(−x22σ2) be the Normal(0,σ) PDF and g(x)=1π(1+x2)−1 be the PDF of a Student t distribution with one d.f. Because the PDF of a Normal(μ,σ) variable X is fσ(x−μ)=fσ(μ−x) (by symmetry), the expectation equals

Eσ,μ(11+X2)=Eσ,μ(πg(X))=∫Rfσ((μ−x)2)πg(x)dx.

This is the defining formula for the convolution (f⋆πg)(μ). The most basic result of Fourier analysis is that the Fourier transform of a convolution is the product of Fourier transforms. Moreover, characteristic functions (c.f.) are (up to suitable multiples) Fourier transforms of PDFs. The c.f. of a Normal(0,σ) distribution is

ˆfσ(t)=exp(−t2σ2/2)

and the c.f. of this Student t distribution is

ˆg(t)=exp(−|t|).

(Both can be obtained by elementary methods.) The value of the inverse Fourier transform of their product at μ is, by definition,

12π∫Rˆfσ(t)πˆg(t)exp(−itμ)dt=12∫Rexp(−t2σ2/2−|t|−itμ)dt.

Its calculation is elementary: carry it out separately over the intervals (−∞,0] and [0,∞) to simplify |t| to −t and t, respectively, and complete the square each time. Integrals akin to the Normal CDF are obtained–but with complex arguments. One way to write the solution is

Eσ,μ(11+X2)=√π2e−(μ+i)22σ2(e2iμσ2erfc(1+iμ√2σ)−erf(−1+iμ√2σ)+1)2σ.

Here, erfc(z)=1−erf(z) is the complementary error function where

erf(z)=2√π∫z0exp(−t2)dt.

A special case is μ=0,σ=1 for which this expression reduces to E1,0(11+X2)=√eπ2erfc(1√2)=0.65567954241879847154….

Here is contour plot of Eσ,μ (on a logarithmic axis for σ).

**Attribution***Source : Link , Question Author : sachinruk , Answer Author : whuber*