Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Riemann-sums can e.g. be very intuitively visualized by rectangles that approximate the area under the curve. See e.g. Wikipedia:Riemann sum

The Ito integral has due to the unbounded total variation but bounded quadratic variation an extra term (sometimes called Ito correction term). The standard intuition for this is a Taylor expansion, sometimes Jensen's inequality.

But normally there is more than one intuition for a mathematical phenomenon, e.g. in Thurston's paper, "On Proof and Progress in Mathematics", he gives seven different elementary ways of thinking about the derivative.

My question
Could you give me some other intuitions for the Ito integral (and/or Ito's lemma as the so called "chain rule of stochastic calculus"). The more the better and from different fields of mathematics to see the big picture and connections. I am esp. interested in new intuitions and intuitions that are not so well known.

share|improve this question

9 Answers 9

up vote 19 down vote accepted

I find the intuitive explanation by Paul Wilmott particularly appealing.

Fix a small $h>0$. The stochastic integral $$\int_0^{h} f(W(t))\ dW(t)=\lim\limits_{N\to\infty}\sum\limits_{j=1}^{N} f\left(W(t_{j-1})\right)\left(W(t_{j})-W({t_{j-1}})\right),\quad t_j= h\frac{j}{N},$$ involves adding up an infinite number of random variables. Let's substitute every term $f\left(W(t_{j-1})\right)$ with its formal Taylor expansion. Then there are several contributions to the sum: those that are a sum of random variables and those that are a sum of the squares of random variables, and then there are higher-order terms.

Add up a large number of independent random variables and the Central Limit Theorem kicks in, the end result being a normally distributed random variable. Let's calculate its mean and standard deviation.

When we add up $N$ terms that are normal, each with a mean of $0$ and a standard deviation of $\sqrt{h/N}$, we end up with another normal, with a mean of $0$ and a standard deviation of $\sqrt{h}$. This is our $dW$. Notice how the $N$ disappears in the limit.

Now, if we add up the $N$ squares of the same normal terms then we get something which is normally distributed with a mean of $$N\left(\sqrt{\frac{h}{N}}\right)^2=h$$ and a standard deviation which is $h\sqrt{2/N}.$ This tends to zero as $N$ gets larger. In this limit we end up with, in a sense, our $dW^2(t)=dt$, because the randomness as measured by the standard deviation disappears leaving us just with the mean $dt$.

The higher-order terms have means and standard deviations that are too small, disappearing rapidly in the limit as $N\to\infty$.

share|improve this answer

Robert Anderson used nonstandard analysis to generate Brownian motion from a finite random walk obtained from coin tosses, where "finite" means indexed by an infinite, non-standard natural number. The corresponding random walk has bounded variation under a non-standard bound. One can then do everything in terms such an random walk, as has been done without rigorous justification before. The Itô-integral can be obtained from a Stiltjes-integral on the random walk, they differ only by an infinitesimal. An outline of the arguments can be found here. For the details, see:

MR0464380 (57 #4311) Anderson, Robert M. A non-standard representation for Brownian motion and Itô integration. Israel J. Math. 25 (1976), no. 1-2, 15--46.

share|improve this answer

I know this thread is already two years old, but, while preparing for a path integration exam, I arrived at an intuitive picture that sheds some light on the origin of the extra term. The picture represents an integral of a smooth function with respect to a concrete realization of Brownian motion. The sum of the areas of the green rectangles represents the difference between Ito (using the left point of each interval) and "anti-Ito" (using the right point of each interval) for sampling of the Brownian motion represented by the red line. Finer sampling leads to smaller rectangles, but they overlap more and more (because Brownian motion is not monotonic), so even if the area occupied by them tends to zero, the sum of their areas does not. This suggests (only suggests -- it is an upper bound on the difference, not a lower bound) that there is a "room" for Ito and "anti-Ito" to differ in their values. Stratonovich can be expected to lie somewhere in between.

Look at the following image:

https://lh6.googleusercontent.com/-bEPzm01WyGk/T-WplGQAc3I/AAAAAAAAACQ/mZr-5p0VUrw/s317/integral-wrt-brownian-motion.png


                 Brownian Motion

share|improve this answer
1  
(I included the image whose URL was cited.) –  Joseph O'Rourke Jun 24 '12 at 0:00
    
@Pavel: +1: Thank you, this is really interesting. Did you arrive at this picture yourself or did you see it in a paper, book etc? Then a reference would be great. Thank you again. –  vonjd Jun 24 '12 at 10:04
1  
I arrived at it myself, so I am not aware of any reference to provide. It is certainly very heuristic. For example, some of the rectangles have actually negative sign. OTOH, if we know from a rigorous proof that the extra term is nonzero, this picture attributes that term to the fact that the the error rectangles overlap. –  Pavel Bažant Jun 26 '12 at 10:00

Here is my two cents on an intuitive explanation of the Ito integral:

The Ito integral is $\int_S^T f(t,w) dB(t,w)$

We can thing of $B(t,w)$, the Brownian motion as the actual price (with mean subtracted) and $f(t,w)$ is a random trading action and its gain on the observable prices. As a result, $f(t,w)$ is $F_t$ adaptive, i.e., it can be dependent only on the history of the prices not future prices. Then, the Ito integration is the total gain from $S$ to $T$ using random trading action+gain $f(t,w)$.

share|improve this answer

Suppose you ask a physicist, if a particle is at x = t^2. Now you ask him where the particle is an arbitrarily small time just before t = 1.

Physicists always just do a Taylor expansion around the point t = 1, and then say we can ignore second order terms. Thus, dx/dt = 2t, and we know at t = 1 the particle is at x = 1. So, the logic goes, if you are a small z away from t = 1, the particle is at 1 + 2z.

That is the deterministic rule. The problem for the stochastic world is that the same logic does not hold. If you are arbitrarily close to x = 1 (loosely speaking) at t = 1 - z, I believe the challenge is that P(X will not move very far in the tiny time left) does not drop off as quickly as second order Taylor terms owing to the variation of a stochastic process, so it can't be ignored. The Ito formula adjusts for this little oddity.

share|improve this answer

One way to improve intuition is to work out a couple of

Discrete versions of Ito's lemma

  1. Øksendal (6th edition) Example 3.1.9: almost surely, $$ B_t^2 - t = \int_0^t 2B_s dB_s $$

This has a discrete version which holds everywhere: let $X_n=\pm 1$ and $S_n=\sum_{i=1}^n X_i$, then $$ S^2_n-n = 2\sum_{i=0}^{n-1} S_i X_{i+1} $$ To verify just note that both sides increase by $2S_{n-1}X_n$ when going from $n-1$ to $n$.

  1. Øksendal's exercise 4.2: $$ B_t^3 = \int_0^t 3B_s ds + \int_0^t 3B_s^2 dB_s $$

Here the discrete version is not a perfect analogue: $$ S_n^3 - S_n = 3\sum_{i=0}^{n-1} (S_i + S_i^2 X_{i+1}) $$ The extra term $S_n$ seems related to the fact that $(dB_t)^3 = 0$.

share|improve this answer

Yet another angle on this:

Normally, when we integrate a deterministic function, time always moves forward. Thus, in a question like $\int_0^t t dt$ or $\int_0^t t^2 dt$ t always moves forward and the t's are fully determined. By that I mean that we know (in the second case) at all time what the value of $t^2$ is - at 2 it has the value 4, at 3 it has the value 9, etc. We are doing a Riemann integral to get the area under the curve. $t$ is deterministic.

On the other hand, look at $\int_0^t W_t dW_t$. Here $W_t$ is not deterministic. But, for small $\Delta t$, $W_t$ changes by $dW_t$. But that means $W_t$ 'covaries' with $dW_t$. Now, if $dW_t$ is positive then $W_t$ goes up. But, if $dW_t$ is negative $W_t$ goes down. In both cases $W_t dW_t$ is positive and grows with time.

Thus, something like $\int_0^t W_t dW_t$ 'grows faster' than the deterministic $\int_0^t t dt$ as there is this positive addition each small $dt$. That is where the additional term comes in.

share|improve this answer

I found this explanation somewhere and wrote it down in my personal notes. I will explain with an example that I think exemplify why Riemann-Stieltjes will provide the wrong answer.

First, let's remember how we can define the Riemann-Stieltjes integral below

\begin{equation}\int_0^t Z(x)dZ(x)=\lim_{n\rightarrow\infty}\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))Z(t_{k-1})\tag{*}\end{equation}

where $0=t_1<t_2<...<t_n=t$, when $n\rightarrow\infty$, $(t_k-t_{k-1})$ goes to zero and $Z(x)$ is continuous and has bounded variation. The Ito integral can be defined in the same way (assuming $Z(t)$ to be any Brownian Path). So, in this elementar definition there is not really any difference, it is just that each is dealing with different kind of functions. But the integration rules will be different.

We can rewrite the terms in the sum as:

$$\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))Z(t_{k-1})=\frac{1}{2}Z(t)^2-\frac{1}{2}\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))^2$$

To exemplify, let's assume now that $Z(x)=x$. We know how to solve the integral in this case, this is $0.5t^2$, as in the first term of RHS above. What happens with this function is that we can ignore the second term above, since it will go to zero (try partioning $[0,t]$ with $t_k=kt/n$, for example). This comes from the fact that $Z(x)=x$ has bounded variation.

Why can't we simple do this if $Z(t)$ is a Brownian Path? The problem here is that $\frac{1}{2}\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))^2$ won't go to zero when $n\rightarrow \infty$. In fact,

$$\lim_{n\rightarrow\infty}\frac{1}{2}\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))^2=t$$

This is so because we are considering infinite realizations of a normal variable with variance $t_k-t_{k-1}$ as $n$ goes to infinity. The Brownian Path limit above does not go to zero because it fails to satisfy bounded variation.

That is why, even though Ito and Riemann-Stieltjes integration depart from the same definition (*) the results are very different. If $Z(x)$ is a Brownian Motion we get:

$$\int_0^t Z(x)dZ(x)=\frac{1}{2}Z(t)^2-\frac{1}{2}t$$

While if $Z(x)$ is a continuous function with bounded variation we get

$$\int_0^t Z(x)dZ(x)=\frac{1}{2}Z(t)^2$$

share|improve this answer

Intuition of Ito integral

Consider the stochastic process

$$f(B_t)$$

where $B_t$ is standard Brownian motion (or the Wiener process) and $f$ is a twice-differentiable function.

Ito’s lemma states that

$$df = f'(B_t) dB_t + \frac{1}{2} f''(B_t) dt$$

The first term is recognizable from the chain rule in classical calculus, but why the second term? If $dB_t$ is truly infinitesimal, it doesn’t even seem possible that $df \neq f'(B_t) dB_t$.

To understand Ito’s lemma intuitively, think of $dB_t$ as a little stochastic variable, specifying $B_t$‘s change during the next $dt$.

$$dB_t = B_t - B_{t+dt} \sim N(0,dt)$$

This models Brownian motion (or the Wiener process) completely. Now $df$ should be a little stochastic variable too, modeling the stochastic process $f(B_t)$.

The picture at the start considers an example $f(B_t)$ where $f'(B_t)=0$, thereby suppressing the “intuitive” or classical term in Ito’s lemma. The reason why $df>0$ in that picture is the reason why that “non-intuitive” term is needed.

Loosely speaking, wherever $f$ has curvature, $dB_t$ will diffuse around that curvature sufficiently to influence the expected result on the order of $dt$. (The specific example in the picture makes this trivial, as $f(\sqrt{t}) = t$.)

Why doesn’t $dB_t$ dominate away $(dB_t)^2$ (i.e. $dt$ )? Because $E[(dB_t)^2]$ does not come from $(E[dB_t])^2$ as happens with classical differentials. The strong law of large numbers implies that a stochastic differential’s expected value pushes its integral on a faster order than its deviation does.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.