Q:

Truncated Mean Squared

Suppose that $X_{\sigma} \sim \mathcal{N}(\mu,\sigma^{2})$.   I am interested in whether
$f(\sigma)=\mathbb{E} (X_{\sigma}^2 1_{\{X_{\sigma}>0\}})$
is monotonic in $\sigma$ for all $\mu$.   
I ran a Monte Carlo and it does appear monotonic but I can't rigourous prove it. 

A:

Let $f_X(x)$ denote pdf of $\mathcal{N}(\mu, \sigma)$.
Using
$$
    \sigma^4 \frac{\partial^2}{\partial \mu^2} f_X + 2 \mu \sigma^2 \frac{\partial}{\partial \mu} f_X + \left(\mu^2 + \sigma^2\right) f_X = x^2 f_X
$$
and integrating over $x$ from zero to infinity:
$$ \begin{eqnarray}
    \mathsf{E}\left(X^2 \cdot 1_{X>0}\right) &=& \left[ \sigma^4 \frac{\partial^2}{\partial \mu^2}+ 2 \mu \sigma^2 \frac{\partial}{\partial \mu} + \left(\mu^2 + \sigma^2\right) \right] \circ \Pr\left(X > 0\right) \\
 &=& \left(\mu^2 + \sigma^2\right) \Pr(X>0) + \mu \sigma^2 f_X(0) \\
 &=& \sigma^2 h\left(\frac{\mu}{\sigma}\right)
\end{eqnarray}
$$
where 
$$
    h(x) = \left( \left(1 + x^2\right) \Phi\left(x\right) + x \phi(x) \right)
$$
and $\Phi(x)$ and $\phi(x)$ denote cumulative distribution function of the standard normal random variable and its pdf, respectively. 
Function $h(x)$ is strictly increasing since
$$
   h^\prime(x) = 2 \left( \phi(x) + x \Phi(x) \right) > 0
$$
where the last inequality is well-known (e.g., see here).
Furthermore:
$$
   f^\prime(\sigma) = \sigma \left( 2 h\left(\frac{\mu}{\sigma}\right) - \frac{\mu}{\sigma} h^\prime\left(\frac{\mu}{\sigma}\right) \right) = 2 \sigma \Phi\left(\frac{\mu}{\sigma}\right) > 0
$$
which proves your observation.

