Malliavin Monte Carlo Greeks: Weight Choice, Variance, and Gamma

# Derivation of a General Formula for the Expectation of the Delta and Gamma Using Malliavin Calculus


# Derivation of a General Formula for the Expectation of the Delta and Gamma Using Malliavin Calculus












My question is twofold and based on the paper "Applications of Malliavin Calculus to Monte Carlo Methods in Finance" by Fournié, Lasry etal (1999) (see https://www.researchgate.net/profile/Jean-Michel-Lasry/publication/24055612_An_Application_of_Malliavin_Calculus_to_Monte_Carlo_Methods_in_Finance/links/552e721b0cf2d49507184241/An-Application-of-Malliavin-Calculus-to-Monte-Carlo-Methods-in-Finance.pdf ).

First Question: In Proposition 3.2 it is stated that:

Consider a diffusion process $\{X(t)|0\leq t\leq T\}\subset\mathbb{R}^n$ of the form \begin{equation*} dX(t)=b(X(t))dt + \sigma(X(t))dW(t), \end{equation*} continuously differentiable in the initial condition $x,\ \forall (t,\omega)\in[0,T]\times\Omega$, with drift- and diffusion coefficients $b, \sigma$ that are continuously differentiable with bounded Lipschitz derivatives. Furthermore, we denote the first variation process associated to $\{X(t)|0\leq t\leq T\}$ as $\{Y(t)|0\leq t\leq T\}$, where \begin{align*} dY(t) &= b'(X(t))Y(t)dt + \sum_{i=1}^n\sigma_i'(X(t))Y(t)dW_i(t)\\ Y(0) &= \mathbb{I}_n, \end{align*} where $\mathbb{I}_n$ is the $n\times n$ identity matrix. Now consider a (discounted) payoff function $f:\mathbb{R}^{n\times m}\rightarrow \mathbb{R}$, with \begin{equation*} \mathbb{E}[f(X(t_1),\dots,X(t_m))^2]\leq \infty, \end{equation*} for $m\in\mathbb{N}\setminus\{0,1\}$ and $0\leq t_1\leq\dots\leq t_m\leq T$. The expectation value of interest reads \begin{equation*} u(x) = \mathbb{E}[f(X(t_1),\dots,X(t_m))|X(0)=x]. \end{equation*} Define now the following set: $\mathcal{A}_m = \left\{a \in L^2([0,T])| \int_0^{t_i} a(t)dt=1,\ \forall i = 1,\dots, m\right\}$. Then $\forall x \in \mathbb{R}^n,\ \forall a\in\mathcal{A}_m$, we have \begin{equation} \nabla u(x) = \mathbb{E}\left[f(X(t_1),\dots,X(t_m)){\int_0^T a(t)[\sigma^{-1}(X(t))Y(t)]^{*}dW(t)}| X(0) = x\right].\label{eq:General Malliavin Greek} \end{equation}

Now, if one takes \begin{equation*} a(t) = \frac{\mathbb{I}_{[0,t_1]}(t)}{t_1}, \end{equation*} where $\mathbb{I}_{[0,t_1]}(t)$ denotes the identity function on $[0,t_1]$, i.e. equal to $1$ if $t\in[0,t_1]$ and $0$ otherwise. Clearly, $a(t)\in L^2([0,T])$ and \begin{equation*} \int_0^{t_i}a(t)dt = \int_0^{t_1}\frac{dt}{t_1}=1,\qquad\forall i = 1,\dots,m, \end{equation*} so $a(t)\in\mathcal{A}_m.$ Plugging this into the above equation for $\nabla u$ in the case of a Black-Scholes model, i.e. $X(t) = S_t,\ \sigma(X(t))=\sigma S_t,\ Y(t)=S_t/S_0$, the equation reduces to \begin{align*} \frac{\partial u}{\partial S_0} &= \mathbb{E}\left[f(S(t_1),\dots,S(t_m))\int_0^T\frac{\mathbb{I}_{[0,t_1]}(t)}{\sigma S_0 t_1}dW_t\right]\\ &= \mathbb{E}\left[f(S(t_1),\dots,S(t_m))\frac{W_{t_1}}{\sigma S_0 t_1}\right]. \end{align*} But is it not the case that a whole lot information of the Brownian motion is lost, since we only account for $W_{t_1}$? Or is the necessary information contained in $S(t_1),\dots,S(t_m)$ and would this choice for $a(t)$ thus work for any such payoff function $f$?

Second Question: How would one use the Malliavin integration-by-parts formula to calculate the second derivative?

My current attempt follows the proof for the first derivative in the mentioned paper closely:

Define $v(t) = a(t)Y(t)/\sigma(X(t))$, then the previous result can be written as $$w(x) \equiv \nabla u(x) = \mathbb{E}\left[f(X(t_1),\dots,X(t_m)\delta(v)\right],$$ where $\delta$ denotes the Skorohod integral, i.e. $$\delta(v) = \int_0^T v(t)dW_t.$$ Then, under the right continuously differentiability assumptions, \begin{align*} \nabla w(x) &= \mathbb{E}\left[\sum_i \nabla [f(X(t_1),\dots,f(X(t_m))\delta(v)]Y(t_i)\right]\\ &= \mathbb{E}\left[\delta(v)\sum_i(\nabla f) Y_i\right] + \mathbb{E}\left[\sum_i f\nabla (\delta (v) )Y_i\right]\\ &= \mathbb{E}\left[\delta (v) \int_0^T D_t f v(t) dt\right] + (\text{second term})\\ &= \mathbb{E}\left[f(X(t_1),\dots, X(t_m))\delta(v)^2\right] + (\text{second term}). \end{align*} I am not completely sure about the last step and do not really see how to rewrite the second term. Ofcourse I could write \begin{align*} \nabla\delta(v) &= \delta(\nabla v)\\ &= \delta\left(\frac{a}{\sigma(X)}\nabla Y - \frac{a}{\sigma^2(X)}\sigma'(X)Y^2\right), \end{align*} but I don't see how I might proceed. Is there even a way to come up with a similar expression for the second derivative as for the first derivative in the case of such a general diffusion process?

Any help or hints are much appreciated!

## Answer by Andrea (score 1)

https://quant.stackexchange.com/a/81867

Using the parallelism between likelihood ratio (not to be confused with pathwise differentiation) and Malliavin calculus, one can get a good intuition on how it works (and why it is not a silver bullet).

For reference read Glasserman p401+

Take the simplest discretised model (lognormal, flat vol, no drift, constant timestep), 3 steps, and payoff function $f(x_1, x_2, x_3)$. What you want is

$ \frac{\partial}{\partial S_0} \int_{x_1, x_2, x_3} p(x_1 | S_0) p(x_2 | x_1) p(x_3 | x_2) f(x_1, x_2, x_3) \, dx$

I am not going to re-write the whole derivation, but you can immediately see that $S_0$ only enters the first term and why the apparent "throw away" mentioned in one of the comments (read the book for a more extensive discussion).

2 observations

- When the time of the first fixing ($x_1$) comes close to 0, the whole thing explodes (i.e. its variance grows very fast). This is a major limitation of these methods. Read in particular pages 407-9.

- You might have a $f$ not actually depending on $x_1$, but let's say only on $x_2$ and $x_3$. Would this change anything? Yes and no. More on this below.

Your question on Gamma: you can definitely get an expression for a multidimensional process, but the maths become very tedious and complicated. You could start simple and learn about symmetries.

For instance, start from 7.3.4 (page 405), which gives you a nice expression for a multidimensional lognormal delta, and derive it again.

What about stochastic volatility? There was recently a question about this: Delta of European Call using Malliavin Calculus

If I read it correctly, basically, one only needs to worry about the "path by path" volatility up to the point of the first fixing used in the payoff function (and not the first simulation point) (which temporarily pushes away that problematic $t_1$ in the denominator).

The intuition is that one can fold all the $p()$ into one that joins $S_0$ to the first required $x_i$ as in $p(x_2|S_0) = \int p(x_1|S_0) p(x_2|x_1) \, dx_1$, do some more maths on paper and reduce variance (from $t_1$ to $t_2$).

Likelyhood ratio vs Malliavin calculus: the statement "likelyhood ratio and Malliavin claculus are the same" is clearly incorrect and limited to the case of discretised SDEs where they produce similar results.





So, in a sense, as long as you can simulate (either exact or approximate), you do know the distribution of your discretised asset (typically as a function of Gaussians or other simple known distributions, see Example 7.3.5). You might hit the small step problem described above (e.g. Heston): a naive application will have a lot of variance, so you will try to do some maths to differentiate the effective discretised distribution at the first required fixing.