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Let $\varepsilon$ be a Gaussian distributed random variable with mean $\mu_0$ and standard deviation $\sigma_0$. Is it possible to compute/approximate the expected value

$$ \begin{eqnarray} & &\mathbb{E}\left[\Phi\left(\frac{\varepsilon-\mu_1}{\sigma_1}\right)\mid\varepsilon<c\right]=\cdots\\ & &\cdots =\int_{-\infty}^{c}d\varepsilon \frac{1}{\sqrt{2\,\pi\,\sigma_0^2}}\,\exp\left(-\frac{\left(\varepsilon-\mu_0\right)^2}{2\,\sigma_0^2}\right)\,\int_{-\infty}^{\frac{\varepsilon-\mu_1}{\sigma_1}}dt\,\frac{1}{\sqrt{2\,\pi}}\,\exp\left(-\frac{t^2}{2}\right) ? \end{eqnarray} $$

Note that while the integral

$$\int \exp\left(-x^2\right)\,\textrm{erf}\left(x\right)\,dx$$ exists (check on http://integrals.wolfram.com/index.jsp), any integral of the kind

$$ \int \exp\left(-(x-a)^2/b\right)\,\textrm{erf}\left((x-c)/d\right)\,dx $$ does not exist (at least according to Wolfram integrals).

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  • $\begingroup$ Are you sure that you have the conditioning $\varepsilon<c$ and not $\varepsilon=c$? $\endgroup$ – alexjo Nov 27 '13 at 11:47
  • $\begingroup$ Yes, pretty sure. $\endgroup$ – AlmostSureUser Nov 27 '13 at 11:50
  • $\begingroup$ Probably this could be useful en.wikipedia.org/wiki/… $\endgroup$ – AlmostSureUser Nov 27 '13 at 14:03
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Note that$$P\left[\varepsilon\lt c\right]=P\left[\sigma_0Y\leqslant c-\mu_0\right]=\Phi\left(\frac{c-\mu_0}{\sigma_0}\right), $$ and that $$ E\left[\Phi\left(\frac{\varepsilon-\mu_1}{\sigma_1}\right);\varepsilon<c\right]=P\left[\sigma_1X\leqslant\sigma_0Y+\mu_0-\mu_1,\sigma_0Y\leqslant c-\mu_0\right],$$ where $(X,Y)$ is i.i.d. standard normal. Some specific values of $(\mu_0,\mu_1,\sigma_0,\sigma_1,c)$ excepted, there is no general formula for the second RHS. Note that, if $c=+\infty$, the condition $\sigma_0Y\leqslant c-\mu_0$ disappears and $\sigma_1X-\sigma_0Y$ is centered normal with known variance hence one is left with $$ \Phi\left(\frac{\mu_0-\mu_1}{\sqrt{\sigma_0^2+\sigma_1^2}}\right). $$

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