Effect of variance on truncated normal Suppose $x\sim \mathcal{N}(0,\sigma^2)$ and $c<0$. Consider the mean of the truncated normal: $\mathbb{E}[x|x>c]$. I would like to sign the derivative $\frac{\partial }{\partial \sigma}\mathbb{E}[x|x>c]$. I tried to take the derivative myself, but noticed it has both positive and negative terms, and it's not clear to me which dominates, and under what conditions. If there is no "closed form" answer, even some intuition for why we should expect the derivative to be positive or negative (as a function of $c$) would greatly help.
Thanks,
Dan
 A: $$\mathbb{E}[X|X>c] = \frac {\sigma \phi(\frac {c} {\sigma})} {1 - \Phi(\frac {c} {\sigma})}$$
Therefore
$$ \begin{align} \frac {\partial} {\partial \sigma} \mathbb{E}[X|X>c] &= 
\frac {[1 - \Phi(\frac {c} {\sigma})] [\phi(\frac {c} {\sigma})
+\sigma\frac {c} {\sigma}\phi(\frac {c} {\sigma})\frac {c} {\sigma^2}]
- \sigma \phi(\frac {c} {\sigma})[\phi(\frac {c} {\sigma})\frac {c} {\sigma^2}]} 
{[1 - \Phi(\frac {c} {\sigma})]^2} \\
& = \frac {\phi(\frac {c} {\sigma})} {[1 - \Phi(\frac {c} {\sigma})]^2}
\left\{\left[1 - \Phi\left(\frac {c} {\sigma}\right)\right]
\left(1+\frac {c^2} {\sigma^2}\right) - \frac {c} {\sigma}\phi\left(\frac {c} {\sigma}\right)\right\}
\end{align}$$
One good thing here is that it is solely in terms of $\frac {c} {\sigma}$, and thus the sign of the above derivative is solely depends on the sign of the function
$$ g(x) = [1 - \Phi(x)](1+x^2)-x\phi(x)$$
Note that when $x<0$, obviously $g(x) > 0$; $g(0) = \frac {1} {2}$; and $g(x) \to 0$ as $x \to +\infty$ (as all the moments of normal exist). It suffices to show that $g$ is monotonic decreasing, i.e. $g'(x) < 0$, and then we are done.
$$ g'(x) = [1 - \Phi(x)]2x - \phi(x)(1+x^2)+x^2\phi(x)-\phi(x)
= 2x[1 - \Phi(x)] - 2\phi(x)$$
Again $g'(x)$ is obviously negative when $x < 0$, and  $g'(0) = -2\phi(0) < 0$; $g'(x) \to 0$ when $x \to +\infty$. It suffices to show that $g''(x) > 0$. We differentiating once more,
$$ g''(x) = -2x\phi(x)+2[1-\Phi(x)]+2x\phi(x)
=2[1-\Phi(x)] > 0 $$
So we conclude that $\displaystyle \frac {\partial} {\partial \sigma} \mathbb{E}[X|X>c] > 0$ for any $\sigma > 0$.
Intuition: When you do one-sided truncation, with the lower end being truncated, those extremely low values are discarded, and the probability of all large values is scaled up accordingly. So the truncated mean is increased (as we see it is always positive in this case). The increase effect is diminishing when $c$ is increase to the positive sign, as a large part of the distribution is truncated already.
