Conditional expectation and indicator function

Question

Is there a simple way of computing the following \begin{equation}\tag{1} \mathrm{E}(S_T\mathbf{1}_{S_T>X}|S_t=s) \end{equation} (here $S_t$ follows a Geometric Brownian Motion) with $t<T$? My idea was to split the conditioned probabilities, but I guess it doesn’t make it simple, for $S_T$ and $\mathbf{1}_{S_T>X}$ aren’t independent.

Answer 1

Suppose there is no drift, then $$S_T= S_t\cdot \exp\left(-\frac{\sigma^2}{2}(T-t)+\sigma (W_T-W_t) \right)$$ We have $$\begin{align} L&:= \mathbb{E}\left( S_T \mathbf{1}_{\{S_T>X\}} | S_t\right) \\ &= \mathbb{E}\left( S_t\cdot \exp\left(-\frac{\sigma^2}{2}(T-t)+\sigma (W_T-W_t) \right) \mathbf{1}_{\left\{ W_T - W_t> \ln\left( \frac{S_t}{X} \right)+\frac{\sigma^2}{2}(T-t) \right\}} |S_t \right) \\ &= \mathbb{E}\left( S_0\cdot \exp\left(-\frac{\sigma^2}{2}(T-t)+\sigma \sqrt{T-t}\cdot\mathcal{N}(0,1) \right) \mathbf{1}_{\left\{ \mathcal{N}(0,1)> \frac{\ln\left( \frac{S_t}{X} \right)+\frac{\sigma^2}{2}(T-t)}{\sqrt{T-t}} \right\}} |S_t\right) \\ \end{align}$$ Denote $a=\frac{\ln\left( \frac{S_t}{X} \right)+\frac{\sigma^2}{2}(T-t)}{\sqrt{T-t}} $, then $$ L = S_t\cdot e^{-\frac{\sigma^2}{2}(T-t) }\int_{a}^{+\infty}e^{\sigma \sqrt{T-t}\cdot x }\frac{1}{\sqrt{2\pi}}\cdot e^{-\frac{x^2}{2}}dx $$ I let you do a change or variable and compute the integral. It's not difficult.