Conditioning Correlated Geometric Brownian Motions on an Endpoint
Summary
The document proposes estimating the expected log value of one geometric Brownian motion conditional on the other reaching a specified endpoint. It writes each process in closed form, solves for the second Brownian motion’s terminal value implied by the conditioning event, and then uses the correlation decomposition to generate the first process’s terminal value. Repeated draws of the independent normal component provide a Monte Carlo estimate of the conditional expectation.
Because the quantity of interest depends only on the processes at the final time, the questioner argues that simulating terminal values directly may avoid interim time steps and their sensitivity analysis. They ask whether a Brownian bridge is needed and whether direct endpoint simulation is more accurate than a discretized path simulation. The document contains no answer or validation, so it does not settle those questions. The proposed method applies to this endpoint-conditioned quantity under the stated Gaussian model; path-dependent payoffs would require information about the trajectory, and the conditioning value’s notation and dimensions should be checked carefully before implementation.
Key ideas
- Closed-form geometric Brownian motion solutions connect terminal log values to Brownian motion endpoints.
- Conditioning one process on the other’s endpoint fixes the latter’s Brownian terminal value.
- Correlation allows the first Brownian endpoint to be generated from the conditioned endpoint and an independent normal draw.
- Direct terminal simulation may suffice for endpoint-only expectations, while path-dependent quantities require path information.
- The document poses, but does not answer, questions about Brownian bridges and simulation accuracy.
Tags
Full text
# Simulate correlated Brownian motions conditioned on future state(s)
# Simulate correlated Brownian motions conditioned on future state(s)
Consider a model defined by 2 geometric Brownian motions
$$dY_{1}(t) = \sigma_{2} Y_{1}(t)dW_{1}(t)$$
$$dY_{2}(t) = \sigma_{2} Y_{2}(t)dW_{2}(t)$$
with $Y_{1}(0) = y_{1}$, $Y_{2}=y_{2}$ and $dW_{1}(t)dW_{2}(t)=\rho$. By performing a Monte Carlo simulation I want to compute
$$\mathbb{E}[\log(Y_{1}(T))|\log(Y_{2}(T))=a\cdot y_{2}]$$
for some future time $T\geq0$ with $t\in[0,T]$ and constant $a\in\mathbb{R}$. Additionally, I want to do a sensitivity analysis of the Monte Carlo results on the amount of paths and time steps. First, both geometric Brownian motions can be solved as following
$$Y_{1}(T) = y_{1}\exp(\frac{-\sigma_{1}^{2}}{2}T+\sigma_{1}W_{1}(T))$$
$$Y_{2}(T) = y_{2}\exp(\frac{-\sigma_{2}^{2}}{2}T+\sigma_{2}W_{2}(T))$$
Second, by solving $W_{2}(T)$ for $\log(Y_{2}(T))=a\cdot y_{2}$ I find
$$W_{2}(T)=\frac{1}{\sigma_{2}}(a\cdot y_{2}-\log(y_{2})-\frac{\sigma_{2}^{2}T}{2})$$
Third, the correlated Brownian motion up to time $T$, $W_{1}(T)$, can be computed as following
$$W_{1}(T)=\rho W_{2}(T)+\sqrt{1-\rho^{2}}Z(T)$$
with $Z(T)\sim\mathcal{N}(\mu=0,\sigma^{2}=T)$. Finally, to compute the expected value of $\log(Y_{1}(T))$, I draw $N$ times a random normal variables with mean $0$ and variance $T$ and compute $W_{1}(T)$ accordingly. By filling $W_{1}(T)$ in to the exact solution for $Y_{1}(T)$, I can compute $N$ times the value of $\log(Y_{1}(T))$ and take the mean to approximate the expectation.
My question is i) is this approach correct or should I look into simulating a Brownian bridge, since the problem is perhaps path dependent? And ii) If correct, is the approach of directly simulating $W_{1}(T)$ (always) more accurate compared to simulating a Brownian motion with multiple interim timesteps? Therefore, this problem should not necessarily need a time step sensitivity analysis as there is no dependance if the simulation is done optimally.
Thank you in advance.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.