Estimating Initial-Value Sensitivity for Coupled Stochastic Processes

# Likelihood ratio and pathwise sensitivity method for coupled SDEs


# Likelihood ratio and pathwise sensitivity method for coupled SDEs












I have two coupled SDEs \begin{align*} dS_t=rS_tdt+V_tdW_t^{(1)},\\ dV_t=aV_tdt+b(V_t)dW_t^{(2)},\\ \end{align*} where $W_t^{(1)}$ and $W_t^{(2)}$ are independent Brownian motions, initial input data are $S_0$ and $V_0$, $a(\cdot)$ and $b(\cdot)$ are sufficiently well-behaved, and I use an Euler-Maruyama discretisation with $N$ timesteps. How exactly should one calculate the derivative of a payoff function $\mathbb{E}f(S)$ with respect to $S_0$ in this case?

In particular, I am confused as to how to apply the chain rule in this case due to the dependence of $S_t$ on $V_t$? For example, say with the likelihood ratio method, would my integral form of the probability density be formed by $2N$ integrals ($N$ with respect to the $S$ and $N$ with respect to $V$)? (Similarly for the pathwise sensitivity approach, I am confused as to how construct the chain rule.)

Any help is greatly appreciated!

## Answer by Animesh Saxena (score 1)

https://quant.stackexchange.com/a/63431

Are W1 and W2 independent? I would assume there is some correlation structure? Cholesky decomposition would help in generating the path. It's very similar to heston model where Vt is volatility.

https://en.wikipedia.org/wiki/Heston_model

## Answer by 123dogs (score 1)

https://quant.stackexchange.com/a/74372

So there are 3 ways that I know of to do this --- none of those, unfortunately, uses the likelihood ratio (when I tried to do the calculation with the likelihood ratio, eg. how policy gradients are calculated in reinforcement learning, I found that swapping expectation and derivative was giving me trouble and making things diverge to infinity)

- You can use finite differences, and run your monte-carlo procedure initialized at S_0, and S_0+eps, and then look at the difference divided by eps (or use a higher-order exact method for that). The issue here is that your sampling/approximation error will be off by more than 1/sqrt(number-of-monte-carlo-samples).

- You can use malliavin calculus, and the "so-called" integration-by-parts formula --- there are a number of references here (most are just for black-scholes but there are several that engage with stochastic volatility models). This is mathematically somewhat complex, but the final solution is just a weighted average of your payoff (eg. E[f(S)g(Z)] where Z, I believe, may have information about the path.

- You can just move the derivative inside the expectation and then look at df(S)/dS_0 = f'(S)dS/dS_0 for each given realization of your process. I am assuming here that dS/dS_0 is differentiable for almost every sample path (which is maybe non-trivial to show, but empirically I believe it is the case). f'(S) is simple; so the question is, what is dS/dS_0? In geometric brownian motion, for example, you get dS/dS_0 = S/S_0. In your case, for your discretized path, you should be able to empirically calculate this (eg. via finite differences --- fix your brownian motion increments, and plug in both S_0 and S_0+eps, and see what you get for S in each case, and then divide that difference by eps). I think you can also sit down and calculate this analytically: The important point here is that I don't think you need any Ito/Malliavin calculus at this point. When I played with this it was pretty simple and efficient (accuracy should be 1/sqrt(number-of-monte-carlo-samples))