Expectation of a Standard Normal Variable

Question

Let Y be a standard normal variable, and $b$ a non-random real number. Calculate $E((Y − b)^+)$ , where $(Y − b)^+$ is the positive part of $Y − b$
So $(Y - b)^+ = (Y - b)$ if $Y ≥ b$ ; $0$ if $Y < b$

Answer 1

Let $Y$ be a random variable with density $f_Y$ then for each measurable function $g$ it holds: $$E[g(Y)] = \int_{-\infty}^\infty g(x)\,f_Y(x) dx$$

In your case $g(x) = (x-b)^+$ and $f_Y(x) = \frac{1}{\sqrt{2\pi}} e^\frac{-x^2}{2}$

So you have to calc:

$$E[(Y-b)^+] = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty (x-b)^+ \frac{1}{\sqrt{2\pi}} e^\frac{-x^2}{2} dx = \int_b^\infty (x-b) e^\frac{-x^2}{2}dx$$

You can solve that integral by your own…