Assume $(X_t)_{t\geq 0}$ follows an SDE of the form: $$dX_t = a(t, X_t) dt + b(t, X_t) dW_t$$ where $W$ is a standard $n$-dimensional Brownian motion, $a$ and $b$ are mappings from $\mathbb{R}_+\times\mathbb{R}^n$ to respectively $\mathbb{R}^n$ and $\mathcal{M}_{\mathbb{R}}(n,n)$ with all the usual regularity conditions, and $X_0=x_0$ for some deterministic $x_0\in\mathbb{R}^n$.

Now let $f$ be a mapping from $\mathbb{R}^n$ to $\mathbb{R}$. Assume we are interested in the sensitivity of $\mathbb{E}[f(X_t)]$ with respect to time ($f$ could be seen as a European payoff function and $X_t$ the value of the risk factors vector at maturity $t$), ie we want to compute $\partial_t \mathbb{E}[f(X_t)]$.

If we proceed via Monte-Carlo and a time-discretized scheme for our SDE (for example an Euler-Maruyama), how would one estimate the derivative with respect to time in a numerically stable way?

Assume $(\hat{X}_i)_{i\geq 0}$ is a time-discretized approximation of $X$ at discrete time-steps $0=t_0 < \dots < t_i = i h < \dots$ where $h>0$ is a constant step size. Then I see the following ways to approximate our differential (assuming $f$ is differentiable):

  • estimate $\frac{1}{h}\mathbb{E}[f(\hat{X}_{\lfloor\frac{t}{h}\rfloor})-f(\hat{X}_{\lfloor\frac{t}{h}\rfloor-1})]$ (ie look at the whole pricing function as a black box);
  • or estimate $\frac{1}{h}\mathbb{E}[(\hat{X}_{\lfloor\frac{t}{h}\rfloor}-\hat{X}_{\lfloor\frac{t}{h}\rfloor-1})^{\top} \nabla f(\hat{X}_{\lfloor\frac{t}{h}\rfloor})]$ (sort of a path-wise chain rule).

Unfortunately in both cases the resulting Monte-Carlo approximations have extremely high variance (I do use variance reduction techniques like common random numbers and antithetic variables, they only help marginally). Even in basic examples like a call price in a Black-Scholes with modest volatility ($30\%$) and even millions of paths, especially for small $h$ (I do take care to do everything in double precision to avoid round-off errors in the finite differences), when compared to sensitivities with respect to other diffusion parameters. For instance, the estimator of the sensitivity I get in both cases has a standard deviation that is of the order (or even bigger) of the ground truth value itself, meaning it's literally garbage.

Are there references about the calculation & numerical issues of the sensitivity with respect to time of value/price functions like the one introduced here using a time-discretization scheme for the SDE and Monte-Carlo simulations (no PDE based approaches)?


1 Answer 1


I would recommend to have a look at: https://www.youtube.com/channel/UC9RbRnYPhO9lpiY-6wWNHWg

It is quite long but worth it. It covers exactly the topic you are looking for.

More precisely, this playlist: https://www.youtube.com/playlist?list=PLJ9XZsVSloaTwgPWr4k0iYH7ahdM5WtEr

from video 23 onwards.

  • 1
    $\begingroup$ Simply a +1 for referencing Christian Fries. $\endgroup$ Commented Mar 11, 2022 at 12:39
  • $\begingroup$ It's really a great reference but, unless I missed something, it doesn't treat the specific problem of differentiating w.r.t time (as opposed to differentiating w.r.t any parameter of the SDE or its initial value), which I believe will be very sensitive to the numerical scheme itself. The variance issue I have it only with differentiation with respect to time, not the rest of the factors, which prompted me to ask whether (and it's probably the case) practitioners had already encountered this issue and developed specific variance reductions techniques for this specific sensitivity. $\endgroup$
    – BS.
    Commented Mar 11, 2022 at 16:27
  • $\begingroup$ Are you sure @B.S.? I recall him introducing several different discretization schemes, plus different variance reduction techniques. Maybe try searching also outside the playlist I have pointed you to. $\endgroup$ Commented Mar 13, 2022 at 10:30
  • 1
    $\begingroup$ @BS. The calculation of stable Greeks in Monte Carlo is far from being trivial. This paper is good to start reading. $\endgroup$
    – Kurt G.
    Commented Mar 14, 2022 at 10:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.