I need to calculate the expected value of a modified normal distributed variable but i'm struggling. So maybe someone can help me.

Suppose we've got a normal distributed variable $X \sim \mathcal{N}(0,\sigma^2)$. Using this variable $X$ we define $Y$,

$Y:=-\left(a + \frac{X}{\sigma}\right), a\le 1 \in \mathbb{R}$.

Now i need help calculating the expected value of $Y^+$.

One can find similar terms in financial mathematics, e.g. the expected value of a put option $P = \left(K-S\right)^+$ is given by

$\mathbb{E}[P] = \int^K_0 \left(K-S\right)f\left(S\right)dS$,

with $f(S)$ being the probability density function of $S$.

But i just can't apply this result in order to find $\mathbb{E}[Y^+]$ and validate that it is positive for $a<1$. I would really appreciate any helpful comments. Thanks in advance!


1 Answer 1


Let us compute the value of $E((Z-z)^+)$ for every real number $z$, where $Z$ is standard normal, so that you will be able to deduce the exact result you are interested in. Consider the standard normal PDF $\varphi$ and the standard normal CDF $\Phi$, and note that, by definition, $$ E((Z-z)^+)=\int_z^\infty(u-z)\varphi(u)\mathrm du. $$ Since $u\varphi(u)=-\varphi'(u)$, $$ E((Z-z)^+)=\left.-\varphi(u)\right|_z^\infty-z\int_z^\infty\varphi(u)\mathrm du=\varphi(z)-z(1-\Phi(z)). $$ Recall that $$ \varphi(z)=\frac{\mathrm e^{-z^2/2}}{\sqrt{2\pi}},\qquad\Phi(z)=\int_{-\infty}^z\varphi(u)\,\mathrm du. $$


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