Skip to content
All library documents

Applying Adjoint Differentiation to Nondifferentiable Option Payoffs

Article Quant Q&A · Author: junhui

Summary

The document asks whether adjoint automatic differentiation (AAD) can estimate sensitivities when a pricing payoff has a nondifferentiable point, such as a range accrual payoff. It outlines approaches discussed in research: differentiate the recurrence in a time-stepping scheme rather than the payoff itself, use complex-step differentiation, and approximate a discontinuous component with a smooth function.

The response points to prior work on Monte Carlo sensitivity estimation and recommends examining a paper section for details. It gives no numerical comparison or worked range-accrual example, so it does not establish which approach is suitable in a particular implementation. The smoothing prescription is stated briefly, without enough detail here to assess its accuracy or convergence; practical use would require checking the estimator and approximation for the specific payoff and model.

Key ideas

  • AAD usually relies on differentiable operations in its computational graph.
  • A time-stepping recurrence may allow sensitivities to be calculated without differentiating the payoff directly.
  • Complex-step methods are presented as another way to estimate derivatives.
  • Smoothing a discontinuous component can make derivative calculations possible, with approximation error to consider.

Tags

Full text
# Applicability of AAD(Adjoint automatic differentiation) to an indifferentiable function at some point


# Applicability of AAD(Adjoint automatic differentiation) to an indifferentiable function at some point












Applicability of AAD(Adjoint automatic differentiation) to an indifferentiable function at some point.

I recently learned about Adjoint automatic differentiation(AAD) while studying Monte Carlo Simulation. Through the papers [1], [2], and [3], I reviewed the definition of AAD and examples of various financial engineering such as Basket option, Asian option, American option and CVA based on a fixed-for-floating IRS. Its basic idea is to decompose the process of differentiating into intrinsic functions and operations, thereby increasing the efficiency of the process. Therefore, this method must have differentiation at all steps. One question arises here. Can't this ADD method be applied to a pricing function with an in-differentiable payoff function? For example, I want to apply this method to a range accrual.

I think this method is really efficient. But can't it be applied to functions that are impossible to differentiate, as in the example I presented? I'm sorry for my poor English and expressions.

[1] Fast Greeks by algorithmic differentiation written by Luca Capriotti

[2] Financial Applications of Algorithmic Differentiation written by Chengbo Wang

[3] AAD and least-square Monte Carlo : fast Bermudan-style options and XVA Greeks written by Luca Capriotti, Yupeng Jiang and Andrea Macrina

## Answer by lehalle (score 3, accepted)

https://quant.stackexchange.com/a/69691

In 2017, I organised a Mini-symposium on automatic differentiation and its applications in the financial industry to share thoughts with other academics on the way AAD could be applied to financial questions. Olivier Pironneau presented a way to apply AAD to functions that are not everywhere differentiable. With Gilles Pagès, they show how to reuse a result in Giles (2008) that allow to rewrite the expression of $\mathbb{E}(V^\theta(t))$ such that its differentiate with respect to $\theta$ does not require to derivate $V$!

The trick is to write the Euler scheme of $V$ along $t$ and to differentiate the recurrence expressions of this scheme.

Another trick is to note that $${\rm Re}\left(\frac{f(a+i\delta a)-f(a)}{i\delta a}\right)={\rm Im}\left(\frac{f(a+i\delta a)}{\delta a}\right)=f'(a)+f^{(3)}(a)\frac{\delta a^2}{6}+o(\delta a^3)$$ where there is no more $\delta a$ on the denominator!

They also use replace non differentiable part of functions by a smooth version, replacing a Dirac in $x$ by $1/(a\sqrt{\pi})\exp -x^2/a^2$ and then throwing $a$ to infinity.

My advice it to read Section 4 of the paper and to look if you can use it for you specific problem.

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.