Expected Value of a Ratio of Log-Normal Variables

Question

Suppose $a$ and $b$ are constants (with $a\neq b$) and $X$ is a log-normal random variable, i.e. $\ln(X) \sim \mathcal N(\mu,\sigma^2)$. Suppose that

$$ \mathbb E \left[ \frac{X-a}{X-b} \right]=0.$$

Is it possible to solve for $\mu$ as an explicit function of $a$, $b$ and $\sigma^2$?

Answer 1

I think that you'd need to restrict the potential values of $b$ such that $b \le 0$ as otherwise the expectation doesn't exist.

And I'm not convinced that there is a closed-form solution other than for $b=0$. Here is some Mathematica code to find $\mu$ in terms of $a$ and $\sigma^2$ when $b=0$:

b = 0
expectation = Integrate[E^(-((-\[Mu] + Log[x])^2/(2 \[Sigma]^2)))/(
   Sqrt[2 \[Pi]] x \[Sigma]) (x - a)/(x - b), {x, 0, \[Infinity]}, 
  Assumptions -> {a \[Element] Reals, \[Sigma] > 0, \[Mu] \[Element] 
     Reals}]

$$1-a e^{\frac{\sigma ^2}{2}-\mu }$$

This means that

$$\mu=\log (a)+\frac{\sigma ^2}{2}$$

where $a>0$.