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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

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

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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    $\begingroup$ You seem to use "exist" in the sense of "has a conventional closed-form expression." ("Exist" in math and stats usually means that it has a well-defined value, whether or not that value can easily be written.) This raises the question of what you mean by "compute": are you looking for a closed-form expression, or would alternatives (such as a power series, and algorithm, or a numerical approximation) be acceptable? $\endgroup$ – whuber Nov 27 '13 at 15:37
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    $\begingroup$ Yes, this is exactly what I meant, the close-form expression seems to be not available, hence I am looking for something to approximate the integral, such as power series. In fact I have found something that can be helpful: en.wikipedia.org/wiki/… $\endgroup$ – AlmostSureUser Nov 28 '13 at 15:59

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