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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!

TimonSeldenrijk
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    $c-Y$ has mean $c-\mu$ and the same standard deviation as $Y$. You can see $\max(c-Y,0)$ as a censored normal distribution, related to a truncated normal distribution – Henry Oct 06 '23 at 11:08
  • @Henry So, this becomes a case of a new normal distribution $Z$ with mean $c-μ$, the same standard deviation, truncated at $0$ on left hand side which results in $E[Z|Z>0]$? – TimonSeldenrijk Oct 06 '23 at 11:27

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In general, $Eh(Y)=\int h(y)f_Y(y)dy$.

$E[Max(c-Y,0)]=\int_c^{\infty} 0 dy+\int_{-\infty}^{c} (c-y)\frac 1 {\sqrt {2\pi}\sigma_Y}e^{-(y-\mu)^{2}/2(\sigma_Y)^{2}}dy$.

Kavi Rama Murthy
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