# Payoff of an odd indicator of one stock being greater than another

Suppose $S_t^1$ and $S_t^2$ are two stocks following GBMs and have current value $s_1$ and $s_2$ respectively. How can I explicitly compute the payoff $$V(t,s_1,s_2)\triangleq \mathbb{E}\left[ 1_{\{S_T^1>S_t^2\}} \mid S_t^1=s_1,\, S_t^2=s_2 \right],$$ where $T\geq t$ and $1$ is the indicator function of the event that $S_T^1$ will exeed the value of $S_T^2$ at time $T$.

• you need the correlation and the volatility data. – Gordon Jun 22 '17 at 15:07
• Suppose I know those – AIM_BLB Jun 22 '17 at 15:08
• Then it is a Bivariate Normal probability problem. – noob2 Jun 22 '17 at 17:38
• @noob2 if you give more details I'll accept your answer :) – AIM_BLB Jun 22 '17 at 17:42

The price is

$e^{-r(T-t)} \mathbb{P}(S_{T}^{1} > S_{T}^2) =e^{-r(T-t)} \mathbb{P}(S_{T}^{1} / S_{T}^2 >1)$

The crucial point is that the ratio of two log-normals is log-normal even when they are not perfectly correlated so it just comes down to a cumulative normal.

We assume vols are $\sigma_1$ and $\sigma_2$. Correlation between driving BMs is $\rho.$

Let $C_{11} = \sigma_{1}^{2}(T-t),C_{22} = \sigma_{2}^{2}(T-t), C_{12} = \rho \sigma_1 \sigma_2(T-t).$

We can write $$S_{T}^{1}= S_{t}^{1} e^{-0.5 C_{11} + \sqrt{C_{11}} Z},$$ $$S_{T}^{2}= S_{t}^{2} e^{-0.5 C_{22} + \sqrt{C_{22}} (\rho Z + \sqrt{1-\rho^2}W)},$$ with $W$ and $Z$ independent standard normals.

So $$S_{T}^{1} / S_{T}^2 = \frac{S_{t}^{1}}{S_{T}^{2}} e^{-0.5 C_{11} +0.5 C_{22} + \sqrt{C_{11}} Z- \sqrt{C_{22}} (\rho Z + \sqrt{1-\rho^2}W)}.$$ The rest is straightforward.