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I would like to evaluate the following integral. I suspect it just doesn't have an analytic form but thought I'd check to see if there's anything I'm missing,

$\int_{-\infty}^{\infty} \mathcal{N}(z;\mu,\sigma) \int_{z-a}^{z+a} \mathcal{N}(x;\mu,\sigma) \mathrm{d}x \mathrm{d}z.$

The reason for doing so is to determine $Pr(|x-z| < a)$ where (both independently drawn) $x\sim \mathcal{N}(\mu,\sigma)$ and $z\sim \mathcal{N}(\mu,\sigma)$.

I can determine the inner integral easy enough,

Integrate[PDF[NormalDistribution[mu, sigma], x], {x, z - a, z + a}]

which yields,

1/2 (Erf[(a + z - mu)/(Sqrt[2] sigma)] + Erf[(a - z + mu)/(Sqrt[2] sigma)])

Attempting to do the full integral using this intermediate expression results in returning unevaluated, however,

Integrate[1/2 (Erf[(a + z - mu)/(Sqrt[2] sigma)] + Erf[(a - z + mu)/(Sqrt[2] sigma)])
          PDF[NormalDistribution[mu, sigma], z], {z, -\[Infinity], \[Infinity]}]

Anyone got any idea on tricks to perform this integral?

Failing that, does anyone have any ideas for how to approximate it?

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Yes, I'm being stupid (no shock there!). There is a work around since things are Gaussian. If $z\sim \mathcal{N}(\mu, \sigma)$ and $x\sim \mathcal{N}(\mu, \sigma)$, then $x-z \sim \mathcal{N}(0, \sqrt{2} \sigma)$. Then the result follows easily,

$Pr(|x-z|<a) = \text{Erf}(a/(2 \sigma))$.

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    $\begingroup$ Using Mathematica: dist = TransformedDistribution[x - z, { x \[Distributed] NormalDistribution[\[Mu], \[Sigma]], z \[Distributed] NormalDistribution[\[Mu], \[Sigma]]}] // Simplify[#, DistributionParameterAssumptions[ NormalDistribution[\[Mu], \[Sigma]]]] &; Probability[Abs[y] < a, y \[Distributed] dist] $\endgroup$ – Bob Hanlon Jul 2 at 18:53

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