$E[(ae^x-b)^+]$, $X$ is $\mathcal{N}(0,1)$ distributed

Question

I have to compute $E[(ae^x-b)^+]$ where $X$ is $\mathcal{N}(0,1)$ and $a,b$ are two real numbers. First of all I ruled out the trivial cases:

$$a=0,b=0 \to E[(ae^x-b)^+]=0$$ $$a=0,b<0 \to E[(ae^x-b)^+]=b$$ $$a=0,b>0 \to E[(ae^x-b)^+]=0$$

$$a>0,b<0 \to E[(ae^x-b)^+]=\nexists$$ $$a<0,b>0 \to E[(ae^x-b)^+]=\nexists$$ Then I computed $a>0,b>0$, it follows that $$E[(ae^x-b)^+]=a\sqrt{e}[1-F(\log{b/a}-1)]-b[1-F(\log{b/a})]$$

The problems comes with $a<0,b<0$...how I should deal with it? Is it the same as $a>0,b>0$?? It's really important, thanks to anyone who'll help me!

Answer 1

In general $\mathbb{E}h\left(X\right)^{+}=\mathbb{E}h\left(X\right)g\left(X\right)$ where $g$ is prescribed by $x\mapsto1$ if $h\left(x\right)>0$ and $x\mapsto0$ otherwise.

If $h\left(x\right)=ae^{x}-b$ then $g\left(x\right)=1$ iff $ae^{x}-b>0$.

If $a\neq0$ then this is the same as $e^{x}>\frac{b}{a}$.

If $\frac{b}{a}\leq0$ then this is always true so $g(x)=1$ for every $x$ so that $\mathbb{E}h\left(X\right)^{+}=\mathbb{E}h\left(X\right)$ .

If $\frac{b}{a}>0$ then this is true if $x>\ln\left|b\right|-\ln\left|a\right|$ so $g$ can be recognized as the characteristic funtion of $(\ln\left|b\right|-\ln\left|a\right|,+\infty)$ and $$\mathbb{E}h\left(X\right)^{+}=\int_{\ln\left|b\right|-\ln\left|a\right|}^{\infty}h\left(x\right)\phi\left(x\right)dx$$

Answer 2

For $a$ and $b$ positive, $$E((b-a\mathrm e^X)^+)=b\Phi\left(c\right)-a\sqrt{\mathrm e}\Phi\left(c-1\right)\qquad c=\log\frac{b}a$$

Answer 3

Hint: This can help too: $$ \left(a{\rm e}^X - b\right)^+ - \left((-a){\rm e}^X -(-b)\right)^+=a{\rm e}^X - b.$$