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This is probably close to trivial for stats experts. Yet, although having searched the web for quite a while now, I have not been able to come up with a satisfactory answer.

Given two (approximately) Gaussian random variables, $u$ and $v$, what is the PDF and the expectation value of $u – v$ given the prior knowledge that $u – v \ge 0$? I remember vaguely having seen something along this lines based on Bayesian inference. However, now I seem unable to locate this resource.

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    $\begingroup$ Would more usually be described as a distribution conditional on $u–v \ge 0$. 'Prior knowledge' would be of a parameter. $\endgroup$ – Scortchi - Reinstate Monica Jul 2 '13 at 9:40
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If I understand your question, I don't think you need Bayesian stats here.

If $u \sim$ Normal($\mu_u, \sigma^2_u$) and $v \sim$ Normal($\mu_v, \sigma^2_v$) are independent random variables, then $w = u - v \sim$ Normal($\mu_u - \mu_v, \sigma^2_u + \sigma^2_v)$. If you are given that $w \ge 0$ (i.e. $u-v \ge 0 $), what you have is a truncated Normal $(\mu_u - \mu_v, \sigma^2_u + \sigma^2_v)$ from 0 to infinity (using the notation given in the Wiki article $a = 0$ and $b=\infty$).

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