Expectation of Inverse Normal CDF Suppose a r.v. $\mu$ is distributed Normal $N(\theta,\sigma^2)$. Is there any way to derive the expectation $\mathbb{E}(\frac{\mu}{\Phi(\mu)})$ where $\Phi$ is the CDF of a standard Normal random variable.
 A: This diverges to $-\infty.$ Write
$$\mathbb{E}[\mu/\Phi(\mu)] = \mathbb{E}[1_{\{\mu\geq 0\}}\mu/\Phi(\mu)] + \mathbb{E}[1_{\{\mu< 0\}}\mu/\Phi(\mu)].$$
If $\mu\geq 0,$ we have $\Phi(\mu)\geq 1/2$ and hence $\mu/\Phi(\mu)\leq 2\mu.$ Hence, the first term can be bounded:
$$\mathbb{E}[1_{\{\mu\geq 0\}}\mu/\Phi(\mu)] \leq 2\mathbb{E}[1_{\{\mu\geq 0\}}\mu]<\infty.$$
For the second term we use the following bound for $t<0$:
$$\Phi(t) < \frac{1}{-t\sqrt{2\pi}}e^{-t^2/2}.$$
See for example here: https://www.johndcook.com/blog/norm-dist-bounds/.
Hence,
$$\mathbb{E}[1_{\{\mu< 0\}}\mu/\Phi(\mu)]\leq \mathbb{E}\left[1_{\{\mu< 0\}}\frac{\mu}{\frac{1}{-\mu\sqrt{2\pi}}e^{-\mu^2/2}}\right] = -\sqrt{2\pi}\cdot\mathbb{E}\left[1_{\{\mu< 0\}}e^{\mu^2/2}\right] = -2\sqrt{2\pi}\mathbb{E}\left[e^{\mu^2/2}\right].$$
The final two here comes from symmetry, allowing us to drop the indicator.
Now note that the RHS diverges to $-\infty.$ The expectation of the right hand side is the moment generating function at $1/2$ for $\mu^2.$ $\mu^2$ follows a $\chi^2_1$ distribution. As you can see here https://stats.stackexchange.com/questions/7278/finding-the-moment-generating-function-of-chi-squared-distribution, the moment generating function for a $\chi^2_1$ random variable diverges to infinity at $1/2$. Hence, we now obtain
$$\mathbb{E}[1_{\{\mu< 0\}}\mu/\Phi(\mu)] \leq -\infty.$$
In particular, 
$$\mathbb{E}[\mu/\Phi(\mu)] = -\infty.$$
