E(11+x2)E(\frac{1}{1+x^2}) under a Gaussian

This question is leading on from the following question.

Basically 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,τ1)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?


Let fσ(x)=12πσexp(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(xμ)=fσ(μx) (by symmetry), the expectation equals


This is the defining formula for the convolution (fπg)(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


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


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


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


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


A special case is μ=0,σ=1 for which this expression reduces to E1,0(11+X2)=eπ2erfc(12)=0.65567954241879847154.

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


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

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