My question is rather intuitive than formal and circles around the derivation of Ito's Lemma. I have seen in a variety of textbooks that by applying Ito's Lemma, one can derive the exact solution of a geometric Brownian Motion. So, if one has

$dS = a(S,t)dt + b(S,t)dZ$

the first part on the right side of the equation is the (deterministic) drift rate and the second term is the stochastic component. Now I have read that ordinary calculus cannot handle the stochastic integral because for example

$\frac{dZ}{dt} = \phi \frac{1}{\sqrt{dt}}\\ \rightarrow \infty \text{ as } dt \rightarrow 0$

But Ito came along and proposed something which culminated in his famous Ito Lemma which helped calculate the stochastic integral. What I don't understand is the step taken from realizing that ordinary calculus does not work in this context to proposing a function $G = G(S,t) \text{ of } S(t)$, Taylor-expaning it, etc.

Or more precisely: What did Ito realize and propose that helped solve the stochastic integral?


2 Answers 2


Baxter and Rennie say it better than me, so I will summarize them.

Suppose that $N_t$ is not stochastic and $f(.)$ is a smooth function then the Taylor expansion is $$ df(N_t) = f'(N_t)dN_t + \frac{1}{2}f''(N_t)(dN_t)^2 + \frac{1}{3!} f'''(N_t)(dN_t)^3 + \ldots $$ and the term $(dN_T)^2$ and higher terms are zero. Ito showed that this is not the case in the stochastic case. Suppose $W_t$ follows a Brownian motion and let $$ Z_{n, i} = \frac{W\left(\frac{ti}{n}\right) - W\left(\frac{ti}{n}\right)}{\sqrt{t/n}} $$ now consider $$ \int_0^t (dW_t)^2 \approx t \sum_{i=1}^n \frac{Z^2_{n,i}}{n}. $$ If $n \to \infty$ then $\sum_{i=1}^n \frac{Z^2_{n,i}}{n} \to 1$ and thus $\int_0^t (dW_t)^2 = t \neq 0$. Thus the second order term does not cancel (but higher order terms do) and we have an extra term in our derivative.

  • 1
    $\begingroup$ good explanation, upvoted $\endgroup$
    – Matt Wolf
    Jan 29, 2013 at 19:17
  • 2
    $\begingroup$ Why is if reading it in the book (however clear it was written) it seems to make more sense online :). $\endgroup$
    – Chinny84
    Aug 18, 2014 at 7:28
  • $\begingroup$ Please correct the definition of $Z_{n,i}$ $\endgroup$
    – Kuo
    Mar 10 at 16:39

The logic from Bob Jansen is correct. The problem is abuse of ideas and notation the integral symbol from the deterministic world gets sloppily applied to random variables. Unlike normal $dt$, which is always positive, $dW_t$ can go 'backwards'. Thus increments of terms like $W_t dW_t$ have a first element that goes up and down with the second element (which can go up and down) and the product is always positive. Mathematically it really comes down to the fact that, if you take a function of 2 variables in the normal (deterministic world) and do a Taylor expansion, all of the second order terms can be neglected when you try to integrate the expansion. But not if there is a Brownian motion...

I think the problem that gets everyone is the loose use of the $"\int"$ symbol. When I explain it I purloin the symbol $\oint$ for stochastic integrals (there is no connection to a deterministic line integral).

Quite simply, if $X_t(t,W(t))$ is a function of two variables (which it is), and $X_0=X(0)$, we can solve for $X_t$ by integrating: $X_t=X_0+\int_0^tX_t(t,W(t))$. Now, I myself am being sloppy, as we have no "$d?.$" in the integral.

What everyone does is substitute the relevant Taylor expansion terms so it starts to look like (and now I will use the other symbol):

$X_t-X_0=\oint_0^t[\frac{\partial X_t}{\partial t}dt+\frac{\partial X_t}{\partial Wt}dW_t+\frac{1}{2}\frac{\partial^2X_t}{\partial Wt^2}(dW_t)^2]$

Recall that the t on the limit of integration is the "real" t, and the ones inside the integration are all dummies (maybe using $\tau$ would help here).

Here is where it gets cute. Mathemeticians and finance people love the abstraction of things like $\frac{\partial^2X_t}{\partial Wt^2}(dW_t)^2$, whereas physicists often look at things like that and say "I don't know what is going on there but somehow it needs to turn into an integral with respect to time since the whole point of this is to model time evolution."

The first term is, indeed, done w.r.t. time, so we can use "$\int$" in place of "$\oint$" safely.

Let's jump to the third term. This term is in the stochastic case, and not the deterministic, because of the dynamics of what we call $dW_t$. We would have had $dt^2$ as well, but it got dropped. While a Taylor series is trivially correct at its 'anchor' point, we need potentially lots of terms to fit closely farther away. However, if we want to integrate a Taylor series at a neighborhood very close to its anchor point we need only $f(a)$ and the first derivative $f'(a)$ in the limit. Intuitively we need only the slope in our approximation for tiny $\Delta 's$. Imagine a $\Delta=10^{-30}$: if we add up (integrate) $10^{30}$ items on the order of this size they matter. But, even if we add up $10^{30}$ units on the order of the square of that, and thus of size on the order of $10^{-60}$ they accumulate so slowly that they are completely negligible. This is simply the usual order concept, and $o((dt)^2)<o(dt)$.

The sleight-of-hand that seems mysterious is this oddity $dW_t$. Imagine $X_t=3t+2W_t$. Now, plugging in above, we get

$X_t-X_0=\int_0^t\frac{\partial X_t}{\partial t}dt+\oint \frac{\partial X_t}{\partial Wt}dW_t$

$X_t-X_0=\int_0^t 3 dt+\oint 2 dW_t$

The first term can be seen, correctly, to be $3t$. The second by symbolic reasoning is $2W_t$. More structurally, $dW_t$ is a random variable that is the differential limit of an RV distributed as $N(0,1/k)$ as $k\rightarrow \infty$. It is, in a sense, a tiny increment of variance alone that accumulates at twice the rate of time in this instance of $X_t$.

The real oddball is $dW_t^2$. Even though it comes from something that has 'only variance' it is squared. When an RV distributed as $X\sim N(0,1)$, $X^2$ is a new RV distributed as a chi-squared distribution: $X^2\sim \chi_1^2$. Now, as we break up this one-period distribution into k sub-periods we get

$\sum _k \frac{1}{k} X^2 = \frac{1}{k}\sum _k X^2$ and $\sum_k X^2 \sim \chi_k^2$

but $E[\chi_k^2]=k \Rightarrow E[\frac{1}{k}X^2]=1$

And here is the rub. Breaking down the 'variance only' RV into smaller fragments and squaring it and summing over the pieces gives you something which doesn't shrink away any faster than $dt$ despite it seemingly (notationally) looking like $dt^2$ Thus, everyone says, $"dW_t^2=dt"$.

Deep down the problem is pretty simple, but the use of the integral sign with $DW_t$ differentials makes it mighty confusing. All that is really going on is that we have the repeated problem that a term like $W_tdW_t$ co-varies: a positive delta-tick in $W_t$, $dW_t$, means both of the items go up together in such an increment. And, if there is a negative $dW_t$ tick $W_t$ goes down so that product is also always positive. In contrast, when we write $\int t dt$ time is always going up and $dt$ is always positive as t increases; $\int -t dt$ still has $dt$ going up. $dW_t$, however, is allowed to go either way (hence a more Lebesgue-like integral varying over the range not the domain). You need to think about what $\int dW_t$ really means, which is why I like $\oint$: to minimize errant false conculsions.

My math is a little sloppy (why did I use $E[\chi^2]$? both the quadratic variation of a path and its expected value are the same for a random walk; see Shreve's book). But if you want to see it play out in front of your eyes try using $X_t=3t+2W_t^2$ instead of the above formula. Then the second derivative has a value that matters

  • $\begingroup$ can you explain how one period chi square random variable be break down into K sub periods ? $\endgroup$
    – Neeraj
    Aug 13, 2015 at 6:45

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