Inverting the expected value of a truncated normal

Let $X$ be a truncated normal random variable with original (pre-truncated) mean $\mu$, variance $\sigma^2$, and left and right truncation points $a$ and $b$ respectively. It's known that the expected value of $X$ is

$$E(X) = \mu + \frac{\phi\left(\frac{a-\mu}{\sigma}\right)-\phi\left(\frac{b-\mu}{\sigma}\right)}{\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)}$$

where $\phi$ is the pdf of the standard normal and $\Phi$ is the cdf of the standard normal.

But: if I know what $E(X)$ is equal to, as well as the values of $\sigma, a$, and $b$, how do I invert this expression for $\mu$? Is that even possible?

Intuition says that there should be a unique solution, but I can't quite seem to calculate what it is. Surely the answer will involve the inverse pdf and cdf of the standard normal, but I keep getting stuck. Any pointers?

• You are solving a non-linear equation involving $\phi$ and $\Phi$. So you are actually seeking a good numerical method to solve this equation. – BGM Nov 15 '17 at 10:46
• @BGM Ah, drat. I was hoping to get a solution in terms of the inverse PDF and inverse CDF. Can you suggest where I should look next? – gogurt Nov 15 '17 at 15:01