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The classical approach to deriving Ito's Lemma is to assume we have some smooth function $f(x,t)$ which is at least twice differentiable in the first argument and continuously differentiable in the second argument. We then perform a Taylor series expansion as follows: $$df = \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial x} dx + \frac{1}{2} \frac{\partial^2 f}{\partial t^2} dt^2 + \frac{1}{2} \frac{\partial^2 f}{\partial x^2} dx^2 + \frac{\partial^2 f}{\partial t \partial x} dt dx + \ldots $$

We then substitute $x=X_t$ where $X_t$ is a stochastic process such as an Ito process: $$dX_t = \mu dt + \sigma dW_t$$ where $W_t$ is a Wiener process. Realizing that $dX_t^2 = dt$ we obtained Ito's formula.

I have several questions regarding this procedure:

  1. How should we interpret differentials of stochastic terms e.g. $dW_t$ or derivatives with respect to stochastic processes like $\frac{\partial}{\partial X_t}$ which appear in the Taylor series expansion when we substitute $x=X_t$. This seems to be undefined since it's not a smooth function
  2. I am confused by what we mean when we say $f$ is smooth if it's a function of a stochastic process? I understand it's continuously differentiable in terms of its arguments, but as soon as we replace $x=X_t$ doesn't it become non-differentiable in time?
  3. How can we replace $x=X_t$ if $X_t$ is a function of $t$? Wouldn't this require us to define the time derivative of $X_t$, which by definition is non-differentiable? This is the same discussion as: https://math.stackexchange.com/questions/2252734/confusion-about-second-partial-derivative-term-in-itos-lemma-with-a-constraint

I understand that we're taking the Taylor series of $f$ (some ordinary function) and which has nothing to do with $X_t$. But treating the argument as $x$ and then substituting it with a time dependent argument $X_t$ seems a little bit un-intuitive. However, I do understand that substituting $X_t$ is the same as substituting any time-dependent process, regardless of it being non-differentiable or not in terms of time. It just seems that when we substitute $x=X_t$ the Taylor series just makes a bit less sense.

Edit: $d W_t^2 = dt$ not $d X_t^2 = dt$

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Just a few notes

  • How to make sense of $\text dW_t$ is the entire point of stochastic calculus. It's far beyond the scope of any answer here. You should read some introductory lecture notes/books on stochastic calculus. You could start here.

    • The idea: Riemann-Stieltjes integrals are of the form $\int_0^t f(s)\mathrm{d}g(s)$ and are well-defined if $f$ is continuous and $g$ has bounded variation, see also this answer. Brownian motion does not have finite variation. But Brownian motion has finite quadratic variation. We thus define a new integral, $I_t=\int_0^t X_s\text{d}W_s$ which converges in a (weaker) mean squared ($L^2$) sense. The construction is still the same: define this integral for step functions (which take random values over certain intervals) and approximate any well-behaved process $X_t$ by these step functions. The result is the Itô integral. One key property is that it is a martingale (e.g. $\text{d}I_t=X_t\text{d}W_t$ is driftless). I omitted many technicalities of course.
  • In the simplest case, the function $f$ needs to be smooth. Weaker conditions are possible, see this answer. You can take functions like $f(x)=x^2$, $f(t,x)=tx$ or indeed $f(t,x_1,...,x_n)$. These are ``standard'' functions. You then consider processes like $f(X_t)=X_t^2$ or $f(X_t)=tX_t$ by mechanically plugging in the process $X_t$ for the variable $x$.

    • It's a bit like algebra and polynomials: You have some general rule $p(X)=X+X^2$ and you can plug in elements from your ring/field (numbers) or for example fancier objects such as matrices and other linear maps.
    • The entire point of Itô's Lemma is that if you know the process $X_t$ but are interested in a process $f(X_t)$: for example, you have a model for variances $v_t$ but you're interested in volatilities $\sqrt{v_t}$ or you know a model for the stock price $S_t$ but are interested in the dynamics of futures prices. Itô's Lemma is thus some stochastic version of the chain rule.
  • $\text dX_t^2\neq \text dt$. Instead, $\text dW_t^2=\text dt$ and $\text dX_t^2 = \sigma^2(t,X_t)\text dW_t$

  • Derivatives like $W'(t)=\lim\limits_{h\to0}\frac{W_{t+h}-W_t}{h}$ do not exist, see here. Sample paths of Brownian motion are continuous but nowhere differentiable. Something like $\frac{\partial}{\partial W_t}$ does not make sense. In fact, the term ``$\text{d}W_t$'' technically doesn't make sense as a differential and is just shorthand notation for an integral, $\text{d}X_t=\sigma_t\text{d}W_t$ really only means $X_t=X_0+\int_0^t\sigma_s\text{d}W_s$. The differential notation is just shorter and handier.

Heuristic proof for Itô's Lemma

Consider a function $f(t,x)$ and an Itô process $\text{d}X_t=\mu(t,X_t)\text{d}t+\sigma(t,X_t)\text{d}W_t$. Taylor tells us \begin{align*} \text df(t,x) = f_t(t,x)\text dt+f_x(t,x)\text dx+\frac{1}{2}f_{tt}(t,x)\text dt^2+f_{tx}(t,x)\text dx\text dt+\frac{1}{2}f_{xx}(t,x)\text dx^2, \end{align*} where subscripts refer to partial derivatives. Now, we plug in mechanically $X_t$ for $x$ and obtain \begin{align*} \text df(t,X_t) = f_t(t,X_t)\text dt+f_x(t,X_t)\text dX_t+\frac{1}{2}f_{tt}(t,X_t)\text dt^2+f_{tx}(t,X_t)\text dX_t\text dt+\frac{1}{2}f_{xx}(t,X_t)\text dX_t^2 \end{align*} As $\text dt\to0$, we can ignore $\text dt^2$. In terms of magnitude, $\text{d}X_t\sim\sqrt{\text{d}t}$ and $\text{d}X_t^2\sim\text{d}t$. We can thus ignore $\text dX_t\text dt\sim \text{d}t^{3/2}$ but we cannot ignore $\text dX_t^2$ which is of order $\text{d}t$! This is the big difference for stochastic calculus from ordinary real calculus for which we can ignore such terms. Thus, \begin{align*} \text df(t,X_t) &= f_t(t,X_t)\text dt+f_x(t,x)\text dX_t+\frac{1}{2}f_{xx}(t,X_t)\sigma^2(t,X_t)\text dt \\ &= \left( f_t(t,X_t)+f_x(t,X_t)\mu(t,X_t)+\frac{1}{2}f_{xx}(t,X_t)\sigma^2(t,X_t)\right)\text{d}t+f_x(t,X_t)\sigma(t,X_t)\text{d}W_t, \end{align*} which is the standard formula you see in textbooks and on wikipedia.

Example for Itô's Lemma

We want to compute $\int_0^t W_s\text{d}W_s$. As it turns out, a clever way is to study $f(t,x)=x^2$ with $\mu(t,X_t)=0$ and $\sigma(t,X_t)=1$, i.e. $X_t=W_t$ is a standard Brownian motion. Then, \begin{align*} \text dW_t^2&=\left(0+0+\frac{1}{2}\cdot1\cdot2\right)\text{d}t+2W_t\text{d}W_t \\ \implies \int_0^t W_s\text{d}W_s&=\frac{1}{2}W_t^2-\frac{1}{2}t \end{align*}

The key difference to ``ordinary'' calculus, i.e. $\int x\text{d}x=\frac{1}{2}x^2$ is the term $-\frac{1}{2}t$ in the Itô integral. It comes from the mere fact that you can't ignore terms like $\text{d}X_t^2$ for stochastic processes (which have non-zero quadratic variation). In fact, it stems from the $\frac{1}{2}f_{xx}(t,X_t)\sigma^2(t,X_t)\text{d}t$ part.

Plugging in $X_t$ for $x$

This point is simple yet subtle. It's mainly due to notation. Consider $f(x)=x^2$. This function takes some input ($x$) and gives you some output ($x^2$). You can substitute anything for the variable (placeholder) $x$ for which you can define powers. For example,

  • if $(a_n)$ is a sequence of real numbers, then $f(a_n)=a_n^2$ is a new sequence of numbers
  • if $x$ is a real number, then $f(x)=x^2$ is another real number
  • if $A\in K^{n\times n}$ is a square matrix, then $f(A)=A^2$ is another square matrix
  • if $(X_t)_{t\geq0}$ is a stochastic process, then $f(X_t)=X_t^2$ is another stochastic process

Suppose $r_t$ is a process for the short rate. For example, Vasicek proposes $\text{d}r_t=\kappa(\theta-r_t)\text{d}t+\sigma\text{d}W_t$. The price of a zero-coupon bond is $e^{A(\tau)+r_tB(\tau)}$ for some functions $A,B$. You could now be interested in knowing the dynamics of the bond price, $\text{d}P$. You would thus use the function $f(t,x)=e^{A+xB}$ which, when you plug in $r_t$ for $x$ gives you the bond price.

It's confusing because it's often convenient to be a bit sloppy with notation. You often see the Black-Scholes solution being written as $V(t,S_t)=S_t\Phi(d_1)-Ke^{-rT}\Phi(d_2)$ where $$\frac{\partial V}{\partial t}+(r-\delta) S_t\frac{\partial V}{\partial S_t}+\frac{1}{2}\sigma^2S_t^2\frac{\partial^2 V}{\partial S_t^2}-rV=0$$ which is however nonsense. You should technically write something along the lines of the call option price is $V(t,S_t)$ where $V(t,x)=x\Phi(d_1)+Ke^{-rT}\Phi(d_2)$. The function $V$ satisfies $$\frac{\partial V}{\partial t}+(r-\delta) x\frac{\partial V}{\partial x}+\frac{1}{2}\sigma^2x^2\frac{\partial^2 V}{\partial x^2}-rV=0.$$ The difference is that $V(t,x)$ is a ``normal'' function which you can differentiate with respect to $x$. An expression like $\frac{\partial V}{\partial S_t}$ doesn't make any sense. Often, it's convenient to use this shorthand notation if your audience knows that you mean but it must be terribly confusing for students starting to learn about finance.

When deriving Itô's Lemma, you start with the Taylor expansion of the function $f(t,x)$. At this stage, $f$ is an arbitrary (real-valued) function. After computing the partial derivatives of $f$, you then simply plug in the stochastic process $X_t$ for the variable $x$. Remember: the variable $x$ is just a placeholder for something else (in our case: a stochastic process).

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  • $\begingroup$ Thanks, that makes a lot of sense. I guess one confusion is, what is the different in taking the Taylor series expansion of $f(t,x)$ vs $f(t,X_t,$. Ostensibly they are very different things as the latter time dependency in the second argument not in the former. I suppose we find the dynamics of the function $f$ itself regardless of the nature of its arguments and then find how that function behaves when we input the Ito process $X_t$? Is that the correct way to think about it? $\endgroup$ – Kevin M Sep 19 at 16:07
  • $\begingroup$ @KeSchn: I only checked out the first link but it looks really great at a glance. thanks for it. $\endgroup$ – mark leeds Sep 19 at 18:36
  • $\begingroup$ @KevinM I added a somewhat lenghty explanation with the difference between $f(t,x)$ and $f(t,X_t)$. But you're right: you start with a Taylor approximation for the function $f(t,x)$ and then after you computed its derivatives, you plug in the process $X_t$ for the variable (placeholder) $x$. $\endgroup$ – Kevin Sep 19 at 19:14
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    $\begingroup$ Ok. Then thanks to @vonjd. Sometimes I'm not sure whether I understand the fundamentals of stochastic calculus. That paper could possibly answer my question. $\endgroup$ – mark leeds Sep 20 at 18:23
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    $\begingroup$ I've noticed that too about the great contributors-yourself included. amazing. that paper will be the next thing I read. thanks. $\endgroup$ – mark leeds Sep 21 at 23:18

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