Moments of the Difference Between Correlated Geometric Brownian Motions

# Statistics of difference between two GBMs


# Statistics of difference between two GBMs












if I have two asset prices modeled separately as geometric brownian motions. How do i go about calculating the expected statistics of their difference? Like given the sigmas and mus of both processes, and their correlations, what would the standard deviation of that difference/sum be?

Is there an analytic solution for the case of two GBMs? Or are there even solutions for n>2?

I've done simulations with a large number of assets. But could intermarket spreads or similar be done without simulating?

I really thought this would be easy to find with google, but I was unable to.

## Answer by Brian B (score 2)

https://quant.stackexchange.com/a/7724

There is of course no closed-form formula for this. However, the community has long since worked out what all the distributional moments are. A common use is to get the equivalent lognormal (or sometimes shifted lognormal) distribution to a portfolio (such as your difference).

Here's a recent moments paper in which they go so far as to run a binomial tree for American basket options.

## Answer by Andrew (score 1)

https://quant.stackexchange.com/a/7711

$$ \frac{dS_{1t}}{S_{1t}}=\mu_1 dt + \sigma_1 dW_{1t} \to S_{1t} = S_{1,t=0}e^{\int^t_0 \mu_1 - .5\sigma^2_1 ds + \int^t_0 \sigma_1dW_{1s}}\\ \frac{dS_{2t}}{S_{2t}}=\mu_2 dt + \sigma_1 \rho dW_{1t} + \sigma_2 (1-\rho)dW_{2t} \to S_{2t} = S_{2,t=0}e^{\int^t_0 \mu_2 - .5 (1-\rho)^2\sigma_2^2 + \rho^2 \sigma_1^2 ds + \rho \int^t_0 \sigma_1dW_{1s} + (1-\rho)\int^t_0 \sigma_2dW_{2s}} \\ S_{2t} - S_{1t} = S_{1,t=0}e^{\int^t_0 \mu_1 - .5\sigma^2_1 ds + \int^t_0 \sigma_1dW_{1s}} - S_{2,t=0}e^{\int^t_0 \mu_2 - .5 (1-\rho)^2\sigma_2^2 + \rho^2 \sigma_1^2 ds + \rho \int^t_0 \sigma_1dW_{1s} + (1-\rho)\int^t_0 \sigma_2dW_{2s}} $$

Which as you can see, is not GBM but you can compute Expected value from here.