Expected value of $E[(e^{X} - b)^{+}]$ where $X$ is a standard normal

Question

I took $f(x) = e^x - b$, which would give us $$E[f(X)] = \int_{-\infty}^{+\infty} f(x)h(x) dx$$ $$E[f(X)] = \int_{-\infty}^{+\infty} (e^x - b)(\frac{1}{\sqrt{2\pi}}e^{-x^2/2}) dx $$

But this doesn't take into account that we only want the positive portion, $E[(f(X))^+]$. I've been thinking about how I could convert the positive portion of the existing distribution into a true distribution (sum = 1) but I can't immediately visualize the existing distribution, since it's no longer a standard normal curve.

Answer 1

A normal distribution truncated to only take positive values is an example of a truncated normal distribution. In your case, I think the most straightforward approach is to take

$$ f(x)=\mathbb{1}[x>\ln(b)](e^x-b) $$ which is just a different way of writing the positive part in this particular setting. Now taking the expectation over the normal distribution, we get

$$ \int_{\ln b}^\infty e^x h(x)dx-b\int_{\ln b}^\infty h(x)dx $$ The second term is simply $b(1-H(\ln b))$ where $H(x)$ is the CDF of the normal distribution. The first term can be simplified by completing the square on the integrand, $$ e^xh(x)=\frac{e^{1/2}}{\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2}}=e^{1/2}h(x-1) $$ which is just an unnormalized normal distribution with mean $\mu=1$. The integral is then again simply computed using the CDF of the translated Gaussian, so that the whole expectation becomes

$$ \mathbb{E}[f(x)]=e^{1/2}(1-H(\ln(b)-1))-b(1-H(\ln(b))) $$

Answer 2

If you want to compute $E[(e^X-b)_+]$, you must simply take $f(x) = (e^x - b)_+$, which would give \begin{align*} E[f(X)] &= \int_{-\infty}^{+\infty} f(x)h(x) dx \\ &= \int_{-\infty}^{+\infty} (e^x - b)_+(\frac{1}{\sqrt{2\pi}}e^{-x^2/2}) dx \\ &= \int_{-\ln(b)}^{+\infty} (e^x - b)(\frac{1}{\sqrt{2\pi}}e^{-x^2/2}) dx \end{align*} (with the "convention" that $\ln(b)=-\infty$ if $b\leq 0$).