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I was studying recently barrier options and found two traditional questions.

The first here: Let $S_t$ be a geometric Brownian Motion, i.e. $S_t = S_0 e^{\left(\mu - \frac{\sigma^2}{2} \right)t + \sigma W_t}$ and let $\alpha>0$ and $\tau = \inf \{t>0|S_t≥α\}$. Compute $\mathbb{P}(τ≤t)$, that is the probability that you it the barrier $\alpha$ upward.

The second here: Let $S_t$ be a geometric Brownian Motion, such as in question 1, then what is $\mathbb{E}[S_t]$?

My question is a mix of the two. I want to calculate:

$$\mathbb{E}[S_t | \tau \geq t]$$

Intuitively this is the expected value of the geometric brownian motion at time $t$ for all the paths that don't hit the barrier. I want to find this value, because I believe that the expected payoff for a up-and-out barrier option would be the above minus the strike and floored for a call. It is clear that I can not simply multiple $\mathbb{P}(\tau \leq t)$ and $\mathbb{E}[S_t]$, since many paths that cross the barrier contribute to the expected value. However I have no idea on how to start with this restriction. Any idea on how to start?

Johann Martens
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1 Answers1

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Note that if $S_t=xe^{(\mu-\sigma^2/2)t+\sigma W_t }$ then $M:=(e^{-\mu t}S_t)_{t\geq 0}$ is a martingale wrt the natural filtration of $W$. For simplicity assume $\alpha>x$. Optional stopping yields $x=E_x[M_{t\wedge \tau}]=E_x[e^{-\mu t}S_t\mathbf{1}_{\{\tau \geq t\}}]$ $+\alpha E_x[e^{-\mu \tau}\mathbf{1}_{\{\tau< t\}}]$. By rearranging and using the standard definition $E[Y|A]:=\frac{E[Y\mathbf{1}_{A}]}{P(A)},P(A)>0$ we get $$E_x[S_t|{\tau \geq t}]=\frac{e^{\mu t}(x-\alpha E_x[e^{-\mu \tau}\mathbf{1}_{\{\tau< t\}}])}{1-P_x(\tau < t)}$$ The expectation and the probability on the rhs can be computed at least numerically since the density of $\tau$ is well-known (you can find a reference in this answer).

Snoop
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