# Question involving expectation and standard normal CDF

I am trying to solve this below problem:

Let $$Z \sim N(0,1)$$, and $$c$$ be a nonnegative constant. Find $$E(max(Z - c, 0))$$, in terms of the standard normal CDF $$\Phi$$ and pdf $$\varphi$$.

I am struggling with how to get started. Since $$Z$$ is standard normal, $$Z - c$$ is standard normal, with the one condition being that it is bounded below by $$0$$, but not bounded above. The expectation could simply be the integral from $$0$$ to $$\infty$$ of $$(z-c)$$ multiplied by the pdf of the distribution. But, if I integrate this to find the standard normal CDF, the pdf is no longer part of the result, though both are required in this answer. For this reason, I cannot quite figure out the correct approach.

I would greatly appreciate any insights.

• $c$ can be positive, zero or negative: the result ${\phi(c)}-c(1- \Phi(c))$ is still correct Oct 26, 2018 at 10:15

You might like to investigate the truncated normal distribution

You will find that $$E[Z \mid Z \gt c] = \dfrac{\phi(c)}{1- \Phi(c)}$$

and so $$E[Z-c \mid Z \gt c] = \dfrac{\phi(c)}{1- \Phi(c)}-c$$

and thus $$E[\max(Z-c,0)] = {\phi(c)}-c(1- \Phi(c))$$

It is $$\int_c^{\infty} (x-c)\phi (x)\, dx$$. To evaluate $$\int_c^{\infty} x\phi (x)\, dx$$ put note that the derivative of $$\phi (x)$$ is $$-x\phi(x)$$. Hence $$\int_c^{\infty} x\phi (x)\, dx=\phi (c)$$. The answer therefor is $$\phi (c) -c[1-\Phi (c)]$$.

• Could you elaborate on why you began the integral at $c$? I am also confused on the evaluation of that integral, if you would be willing to give more detail. Thanks for the answer.
– user465188
Oct 26, 2018 at 8:01
• $\max (Z-c,0)$ is $0$ when $Z <c$. So the integration is over the part $Z>c$. @Matt.P Oct 26, 2018 at 8:05
• Thank you. This makes sense. One last question: I am having difficult interpreting the maximum. How do we know, for example, that we are integrating $Z$ over $c$ to $\infty$ when, in the $Z < c$ case, we seem to be integrating $0$? I am sorry if this is a foolish question, but I am clearly misinterpreting the problem.
– user465188
Oct 26, 2018 at 8:37
• @Matt.P You split the integral into the portions $Z<c$ and $Z>c$ and note that the integral over the first part is $0$. Oct 26, 2018 at 8:41
• Ok. Then, because the integral over the second part is greater than $0$, it must be the maximum of the two and therefore it is the one we consider? Or, are there two separate answers, and only one of them is written in terms of the pdf and CDF? Just to be sure that I understand this correctly.
– user465188
Oct 26, 2018 at 8:43