Q: Determine mean and sigma from known CDF of normal at zero, and known linear payoff Suppose I have a normal distribution X with unknown mean $\mu$ and unknown standard deviation $\sigma$, but known CDF at x=0: $\phi(0)=k$. Now further suppose a consequence (payoff) function
$$ p(x) =
\begin{cases}
-mx, & \text{if x < 0} \\
0, & \text{if x >= 0}
\end{cases}
$$
with known m, and known expected value E[p(X)] of j. Given k, j, and m, is it possible analytically to determine $\mu$ and $\sigma$? Failing an analytical approach, what is the best numerical alternative?  

A: I assume you have told us $k=\mathbb P(X \le 0)=\Phi\left(-\frac{\mu}{\sigma}\right)$, so $-\frac{\mu}{\sigma}=\Phi^{-1}(k)$ using the cumulative distribution function of a standard normal $\Phi(x)$ and its inverse $\Phi^{-1}(y)$. 
You can also find $j=\mathbb E[p(X)] = -m \mathbb P(X \le 0)\left(\mu - \frac{\phi(-\frac{\mu}{\sigma})}{\Phi(-\frac{\mu}{\sigma})}\sigma\right) = -m k\left(\mu - \frac{\phi(\Phi^{-1}(k))}{k}\sigma\right)$ in terms of $k$ and the density function of a standard normal $\phi(x)$ (not the way you use this notation) 
With these you can solve for $\mu$ and $\sigma$.  Clearly $j$ and $m$, or more precisely $\frac jm$, are just scaling constants
You then get 
$$\mu =\dfrac{-\Phi^{-1}(k)}{(k\Phi^{-1}(k) +\phi(\Phi^{-1}(k)))}\dfrac{j}{m}$$ 
$$\sigma=\dfrac1{(k\Phi^{-1}(k) +\phi(\Phi^{-1}(k)))}\dfrac{j}{m}$$ 
For example with $k=0.1, j=3, m=2$ you get $\mu \approx 40.60411$ and $\sigma \approx 31.68355$
while with $k=0.9, j=3, m=2$ you get $\mu \approx -1.446561$ and $\sigma \approx  1.128758$
