Expected Value of a Log-Normal Random Variable

Question

I have this question for an assignment and I am having trouble with understanding what it is asking and how to go about solving it.

The question states: Consider a log-normal random variable $S=e^{a+bZ}$, with $Z \sim ~Norm(0,1)$. Find the formula for the following expected values.

1. $E[I_{S>K}]$ where $K$ is a real positive constant and $I$ is the identity function;
2. $E[SI_{S>K}]$

EDIT: I believe the professor meant that I is an indicator function, though this is how the question is given.

Thanks very much in advance.

Here is what I have done so far:

\begin{align} & P(S>K)=1-\varphi(K) \\[8pt] = {} & 1- \left[\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^K e^{\frac{-t^2}{2}} \, dt\right] \end{align}

Answer 1

The substitution $Z:=\frac{\ln S-a}{b}$ helps with both problems. Your first problem is to compute$$\int_K^\infty\frac{1}{bS\sqrt{2\pi}}\exp-\frac{(\ln S-a)^2}{2b^2}dS=\Phi\left(\frac{a-\ln K}{b}\right),$$where $\Phi$ is the $N(0,\,1)$ cdf. Working in integrals for problem 1 helps us with problem 2; we realise the integrand just needs an extra $S$ factor for that. But problem 1 can also be solved without integrals. We want$$P(S>K)=P\left(Z>\frac{\ln K-a}{b}\right)=\Phi\left(\frac{a-\ln K}{b}\right).$$

Your second problem is to compute$$\int_K^\infty\frac{1}{b\sqrt{2\pi}}\exp-\frac{(\ln S-a)^2}{2b^2}dS=\int_{\frac{\ln K-a}{b}}^\infty\frac{1}{\sqrt{2\pi}}\exp\left(a+bz-\frac{z^2}{2}\right)dz.$$Let's slightly adjust the substitution this time, viz. $w:=z-b$ so we want$$\int_{\frac{\ln K-a-b^2}{b}}^\infty\frac{1}{\sqrt{2\pi}}\exp\left(a+\frac{b^2}{2}-\frac{w^2}{2}\right)dw=\exp\left(a+\frac{b^2}{2}\right)\Phi\left(\frac{a+b^2-\ln K}{b}\right).$$Note how each problem's solution behaves in the $K\to0^+$ limit: since $-\ln K\to\infty$, each $\Phi$ factor $\to1$, recovering the results $1,\,\exp\left(a+\frac{b^2}{2}\right)$ we'd expect.

Answer 2

\begin{align} & \operatorname E[I_{S>K}] = \Pr(S>K) = \Pr(e^{a+bZ} >K) \\[8pt] = {} & \Pr\left(Z> \frac{(\log_eK) - b} b \right) \\[8pt] = {} & 1 - \Phi\left( \frac{(\log_eK) - a} b \right) = \Phi\left( \frac{a - \log_e K} b \right) . \end{align}