All day, I've been busy with solving the following expectation: $E[Max(c-Y,0)]$ Where $c$ is a constant, $Y$ is normally distributed with mean $μ$ and standard deviation $σμ^β$. I tried using the fact that for a standard normal random variable $Z$, $E[(Z-z)^+] = φ(z)-z(1-Φ(z))$, but couldn't solve it.
Thanks in advance!
In general, $Eh(Y)=\int h(y)f_Y(y)dy$.
$E[Max(c-Y,0)]=\int_c^{\infty} 0 dy+\int_{-\infty}^{c} (c-y)\frac 1 {\sqrt {2\pi}\sigma_Y}e^{-(y-\mu)^{2}/2(\sigma_Y)^{2}}dy$.