# Monte carlo methods for vanilla european options and Ito's lemma.

I understand that by applying Ito's lemma to the following SDE

$$dX=\mu\,X\,dt+\sigma\,X\,dW$$

one obtains a solution to the above SDE which is as follows:

$${X}\left( t\right) =\mathrm{X}\left( 0\right) \,{e}^{\sigma\,\mathrm{W}\left( t\right) +\left( \mu-\frac{{\sigma}^{2}}{2}\right) \,t}$$

I have been told that I can use either of these equations (SDE or its solution) for applying monte carlo simulations to vanilla European options although the second one converges faster that the first one.

Can someone confirm this statement?

Furthermore, one point remains unclear to me: What is Ito's lemma used for in quantitative finance?

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## 5 Answers

The difference between the two is that the first will lead you to a discretization scheme of the process.

So you will have to simulate a whole (approximate) trajectory of (meaning by that $X'_{t_0},...,X'_{t_n}$) up to time $T$ (the expiry of your vanilla option) to get to $X'_T$ which is then only an approximation of $X_T$.

The second method is exact and gives you the law of $X_T$ in one step only.

I simply don't get your second question.

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Thanks for your reply. Yes I could have formulated my second question in a simpler manner: what is Ito's lemma used for in quantitative finance? – balteo Jun 17 '11 at 8:05

If we have some function $f(a,b,c,...)$, where $a,b,c,...$ can be stochastic or otherwise, then Ito's lemma is used to find $df(a,b,c,...)$.

1)

You can simply do raw Monte Carlo. Consider a contingent claim maturing in $6$ months. Then for each $i$-th simulation you can calculate:

$S(T)_i = S(t)e^{(r-q-\frac12 \sigma^2)0.5 + \sigma \sqrt{0.5}z_i)}$

where $z_i \sim N(0,1)$, and $\sqrt{\Delta t}(z_i)$ is equal in distribution to $W_{\Delta t}$.

2)

You can use the above methodology but create a sample path. For example you might want to generate a 6-point sample path. That is;

$S(t_i) = S(t_{i-1})e^{(r-q-\frac12 \sigma^2)\frac{0.5}{6} + \sigma \sqrt{\frac{0.5}{6}}z_i)}$

for $i \in \{2,3,4,5,6,7\}$.

3)

You can discretize the SDE itself using Euler-Marayama.

$\Delta S(t_i) = a(t,S)\Delta t + b(t,S)\Delta W(t_i)$

4)

You can discretize the SDE using Milstein:

$\displaystyle \ \ \Delta S(t_i) = a(t,S)\Delta t + b(t,S)\Delta W(t_i) + 0.5b(t,S)\frac{\partial b(t,S)}{\partial S}\bigg((\Delta W(t_i))^2 - \Delta t\bigg)$

5)

Consider the above methodologies to be the function $f(v)$, where $v$ are your random numbers. You can use a control variate $g(v)$ in the following fashion:

$\frac1N \sum_{i=1}^N [ f(v_i) - g(v_i) ] + E[g(v)]$.

In practice you want to use correlations and other stuff to improve the outcome; however this is how it's presented to students. $E[g(v)]$ might be the closed-form BS price, $g(v_i)$ might be the monte carlo BS price for a simple instrument, $f(v_i)$ might be some really complicated instrument that's closely correlated with $g(.)$.

6)

You can use antithetic sampling. That is,

$\text{MC estimate} = \frac12 [ f(v) + f(-v)]$.

There are some technical conditions you need to satisfy to make this worth the extra computation.

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In quantitative finance, we sometimes find ourselves choosing a new stochastic model for what market variables are random, and how. For example, someone might decide that they like the SDE \begin{equation} dS = \mu\ S\ dt + \left( \frac{S_0}{S} \right)^{\frac32} \sigma\ S\ dW \end{equation} because they want to capture a leverage effect.

Now, this SDE may or may not have a closed-form solution. For example in your question, \begin{equation} X(t)=X(0)\ \exp{\left( σW(t)+(μ − \frac{σ^2}2)t\right)} \end{equation} is the solution to the Black-Scholes SDE. On the other had, I'm not even sure if the leverage SDE above has a solution.

Ito's Lemma is the mathematical tool we can use to prove that a potential solution to our SDE really satisfies the SDE. So in practice, that is where it is actively used in quantitative finance.

The lemma also underlies a lot of the stochastic calculus used in quantitative finance, for example the Girsanov theorem, but those uses are "hidden underneath" since the pure mathematics is reasonably mature by now.

To chime in on the question of convergence, if you possess a solution to your favorite SDE, then you can simulate terminal values $S_t$ of the process after macroscopic bits of time $t$. That then lets you quickly simulate, say, $M$ option values at expiration time $t$ and then average to form an estimate of their expected value at computational cost $O(M)$. If you do not possess a solution, then you must generate each $S_t$ by simulating a whole path of $S$ from time 0 to time $t$ in $J$ small increments of size $\Delta t$, using your SDE itself, at computational cost $O(J \times M)$.

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To answer the more general question that seems to be giving you trouble, Ito's lemma is the stochastic version of the chain rule of standard calculus.

What is it useful for? That's like asking what the chain rule is useful for. Calculus is useful in quantitative finance, and in particular, for stochastic processes, you need to use the stochastic version of calculus. To compute the derivatives in this stochastic calculus, you need a chain rule, and that's what Ito's lemma provides.

I suspect you never really thought out what the chain rule in normal calculus is useful for. Once you understand that, it's clear what you need Ito's lemma for.

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Difficult to determine what the main question really was, but good answer. – SRKX♦ Nov 26 '12 at 19:02

To add to the answer of TheBridge:

I understand your second question in the sense if you could use Ito's lemma for all stochastic processes. This is definitely not the case: It can also be used for processes with bounded quadratic variation (e.g. Wiener process) - you should google this term or look it up in wikipedia: http://en.wikipedia.org/wiki/Quadratic_variation

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Thanks for your reply too. My second question was not clearly formulated. I meant to ask: What is Ito's lemma used for in quantitative finance? – balteo Jun 17 '11 at 8:08
@balteo You should edit your question to be more clear about what you meant to ask. – chrisaycock♦ Jun 17 '11 at 14:24
@chrisaycock: done – balteo Jun 18 '11 at 11:50
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