Computing the Probability That One GBM Stock Outperforms Another

Summary

The document explains how to evaluate the probability that one of two correlated geometric Brownian motion stock prices will exceed the other at a future date. The key step is to take the ratio of the future prices: its logarithm is normally distributed, with its mean and variance determined by the stocks’ volatilities, their correlation, and the time interval. The desired probability can therefore be calculated using the cumulative normal distribution.

The answer expresses each stock’s future price using independent standard normal variables and gives the covariance terms needed to form the ratio distribution. It also writes a discounted probability as a price, which applies when valuing a unit cash-or-nothing claim under the relevant pricing assumptions. The question’s stated payoff is an undiscounted conditional expectation of an indicator, so that distinction matters. The excerpt does not finish the cumulative-normal formula or specify drift assumptions; readers should confirm the pricing measure and model inputs before applying it.

Key ideas

• The ratio of two correlated lognormal stock prices is itself lognormal.
• The log ratio’s variance depends on both volatilities and their correlation.
• The probability of one stock finishing above the other reduces to a cumulative normal calculation.
• Discounting the probability gives a claim price under suitable pricing assumptions.

Tags

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


# 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 $T$.

## Answer by Mark Joshi (score 2)

https://quant.stackexchange.com/a/34824

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.

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