Expectation of truncated log-normal

Question

Let's assume that $y=e^x$, where $x\sim N(\mu,\sigma^2)$, that is, $y$ follows a lognormal distribution.

I'm interested in finding how $\mathbb{E}\left[y|y\geq a\right]$ varies with $\mu$ and $\sigma$. From https://en.wikipedia.org/wiki/Log-normal_distribution, it is to see that

$$ \mathbb{E}\left[y|y\geq a\right]=\frac{\int_{a}^{\infty}yf\left(y\right)dy}{\int_{a}^{\infty}f\left(y\right)dy}=\underbrace{e^{\mu+\frac{1}{2}\sigma^{2}}}_{\mathbb{E}\left[y\right]}\frac{\Phi\left(\sigma-\frac{\ln a-\mu}{\sigma}\right)}{1-\Phi\left(\frac{\ln a-\mu}{\sigma}\right)} .$$ So formally, if we define $h(\mu,\sigma) = \mathbb{E}\left[y|y\geq a\right]$, I want to understand $\frac{\partial h}{\partial\mu}$ and $ \frac{\partial h}{\partial\sigma} $, for any value of $a$. I've tried to work out the derivatives, but I always find that both can take any sign. I'm not surprised that $ \frac{\partial h}{\partial\sigma} $ can take any sign, but, shouldn't there be a way to prove that $$ \frac{\partial h}{\partial\mu} >0 ?$$

Is there a counterexample otherwise?

Answer 1

The claim $\frac{\partial h}{\partial \mu} > 0 $ looks fine to me, lets say \begin{align*} h = f_1 \cdot f_2 \cdot f_3, \end{align*} with \begin{align*} f_1 &= e^{\mu + \sigma^2/2} \\ f_2 &= \Phi(z_1) & z_1 = \frac{1}{\sigma}\left(1 + \mu - \ln a \right) \\ f_3 &= (1 - \Phi(z_2))^{-1} & z_2 = \frac{1}{\sigma}(\ln a - \mu) \end{align*} Now $f_1,f_2,f_3 > 0$ and clearly $\frac{\partial f_1}{\partial \mu} \cdot f_2 \cdot f_3 > 0$, so the only bit that might cause some concerns is the expression $\frac{\partial}{\partial \mu} f_2 \cdot f_3 $, but we have \begin{align*} \frac{\partial f_2 \cdot f_3 }{\partial \mu} &= \phi(z_1) \cdot \frac{\partial z_1}{\partial \mu} \cdot f_3 + f_2 \cdot (-\phi(z_2) )\frac{\partial z_2}{\partial \mu} \cdot (1 - \Phi(z_2) )^{-2} \end{align*} Now $\phi(\cdot)$ is positive, and $\frac{\partial z_1}{\partial \mu}$ is positive, and of course $ - \frac{\partial z_2}{\partial \mu }$ is positive, so again all of these functions are positive as hoped.