Plain English explanation of Ito's integral?

Question

I'm looking for a plain English explanation of Ito's integral. I don't need an exhaustive proof, derivation, etc. Just a simple ~this is effectively what it does and why it's better than a Riemann sum (or insert other numeric approximation.)

With Riemann sums, the function is divided into an arbitrarily large number of rectangles, the areas are summed, and you have a pretty good approximation of the function's area under the curve. I've read that this approach works best with "monotonic functions" and isn't ideal when the function jumps randomly. Hence, it performs poorly when approximating area under the curve of Brownian motion.

From various resources, I've pieced together that Ito's integral is still using an arbitrarily large number of small rectangles to approximate area. However, they are (A) of random width and (B) sometimes overlap with one another. Due to B, the function's area cannot be approximated as the summation of each rectangle. However, the function's area can be approximated in a probabilistic sense: The area can be seen as a random variable (and perhaps due to central limit theorem) can be conceived as the expectation of several random variables.

So essentially these random rectangles are averaged and we get a mean and standard deviation around our function's area estimate.

Good chance that I'm confused. Any chance someone could clarify this? Please don't rely heavily on LaTeX; again, I'm interested in a plain English summary.

Edit: If I am on the right track, then this method would work best "stationary" data, where the jumps generally cancel each other out and bounce around some mean value. However, if there is a general trend over time, performance might be negatively impacted..?

Answer 1

Ito's integral has nothing to do with the area under a curve and no connection with rectangles, random or otherwise.

Let me try to share a motivation in the simplest language I can manage.

There's really no better justification for Ito's formula than talking about stock prices, so let's suppose we have a stock, United Marshmallow, and an old-fashioned stock ticker. Let's suppose that, at each tick, the price of United Marshmallow goes up by 1 or down by 1.

Let's say the stock price starts at 10, and the next six prices are 11, 12, 11, 12, 11, 10. In other words, the fluctuations in prices are:

$$+1, +1, -1, +1, -1, -1$$

(I know that under this model, the stock price could go negative, but let's ignore that for now.)

Let's say we're a trader and we want to come up with a strategy for trading this stock. At time $t$, all we know is what's happened at that time, so whatever amount of stock we choose to buy (or sell) at time $t$ cannot depend on information after time $t$.

Let $W_t$ be the total fluctuation in the stock up to time $t$. We get this by adding up the fluctuations up to that time. So $W_1 = 1$, $W_2 = 2$, $W_3 = 1$, $W_4 = 2$, $W_5 = 1$, $W_6 = 0$.

One simple trading strategy is to buy $W_t$ shares at time $t$ and sell them at time $t+1$. The idea here is that, if the stock has gone up a lot, then we should be more inclined to make a bet on it going up again, and if it's gone down a lot, we should be inclined to bet on it going down again. Note that $W_t$ can be negative, which would correspond to selling the stock (short sales are allowed) and then buying it back again.

The profit from this strategy is just the sum over the amount we buy at time $t$ multiplied by fluctuation in price from time $t$ to time $t+1$

$$\sum_t W_t (W_{t+1} - W_t)$$

It looks a bit like a Riemann sum, but it's not, because the price can go down, so the "base" of the "rectangle" makes no sense. And also, our $W_t$ has to be strictly on the left-hand side of the interval from $t$ to $t+1$, or else we'd be using future information.

Let's look at the cumulative profit we would get if we did this strategy at each possible time point. After time $1$, we bet $W_1 = 1$ dollars, and the next fluctuation is $+1$, so our profit so far is $1 \times 1 = 1$. Next, we bet $W_2 = 2$ dollars, but then the stock goes down, so our "profit" is $2 \times -1 = -2$. The overall profit so far is $1 -2 = -1$.

Continuing in this fashion, we get the following table:

time $t$	2	3	4	5	6
cumulative profit by time $t$	1	-1	0	-2	-3

Now let's compare this with an integral. We are doing some sort of "integral" of the function $f(W) = W$, so maybe there's a relationship with the integral of this function?

time $t$	2	3	4	5	6
cumulative profit by time $t$	1	-1	0	-2	-3
$W_t$	2	1	2	1	0
$W_t^2/2$	2	1/2	2	1/2	0

Perhaps you notice that they match up exactly if you subtract $t/2$.

time $t$	2	3	4	5	6
cumulative profit by time $t$	1	-1	0	-2	-3
$W_t$	2	1	2	1	0
$W_t^2/2$	2	1/2	2	1/2	0
$W_t^2/2 - t/2$	1	-1	0	-2	-3

We get $\sum^t W_t(W_{t+1} - W_t) = W_t^2/2 - t/2$. You can do the same calculation with any other sequence of $+1$ and $-1$ as the price fluctuations and it will still be true!

This is the mathematical fact that underpins Ito calculus.

If you take some sort of limit, time becomes continuous, $W_t$ becomes Brownian motion, and you get the famous formula

$$\int_0^t W(t) dW = W(t)^2/2 - t/2$$

Here, both sides are random variables. But the reason why the formula is true is not because of some application of the central limit theorem. It's because the mathematical fact that we checked in the example above holds for any set of fluctuations we could have chosen.

This formula extends to functions of $W(t)$, and you get the chain rule of Ito calculus.

To summarise, Ito's integral is really a way to calculate the (random) profit from a strategy which someone would make if they were trading a stock which moved up and down randomly. It's not an attempt to compute the area under any kind of curve, and it doesn't have much to do with the Riemann integral, except for a superficial similarity in the definition, and absolutely nothing to do with the Lebesgue integral, except that Lebesgue integration is required to make the "passage to the limit" part work formally.

Answer 2

I think you're confusing the Lebesgue integral with Itô calculus. They are related concepts. I'll explain.

Lebesgue vs Riemann

The simplest explanation of the difference between Lebesgue and Riemann integration - that I know of - follows. Imagine a bunch of bank notes tossed on a carpet. Riemann would count the money by first drawing a rectangular grid on a carpet, then adding up bank notes row by row.

Lebesgue, would rather count number of 1-dollar bills, then 5-dollar bills, then 10-dollar bills etc. Then simply sum up: $1\times n_1+5\times n_5+10\times n_{10} \dots$

Why would you need Lebesgue integral if we already have Riemann? The reason is that the latter doesn't work on functions that are not flat at small scales. Consider any ordinary function you know of, e.g. $\exp$ or $\sin$: if you look at the small enough interval $\delta x$ the function will be flat between $x$ and $x+\delta x$. Some functions don't flatten when you take a magnifying glass and zoom in. They stay rough at any scale, then Riemann integration fails, and you need something else. Here comes the Lebesgue integral to save the day (in some cases).

Itô integration

Suppose you need to sum a value of fruit basket. Easy: $V=n\times p $, where $n,p$ - quantity and price of a fruit. If both $n$ and $p$ are stochastic, then you must apply Itô calculus because $dV\ne \frac{\partial V}{\partial n}dn+ \frac{\partial V}{\partial p}dp$. Here's an [almost] plain English explanation.

If we're control the amount of fruit ourselves, then $n$ is deterministic, i.e. we know in advance how much fruit we hold at any time $t$ in future. However, we're likely do not know the prices, they are stochastic.

The good thing is that we may know the parameters of the stochastic process, e.g. $p(t)-p(0)\equiv \Delta p(t)\sim\mathcal N(0,t)$. Now, to forecast $p$ we simply need to integrate it: $p(t)=p(0)+\xi_t$, where $\xi_t\sim\mathcal N(0,t)$

Now the value of the basket is simply: $V(t)=n\times p(t)$ or in Itô integral formulation: $dV(t)=n\times dp(t)$

So far so good, and it doesn't seem like this Itô integral is any different from Riemann.

Here's where it gets interesting: what if we're valuing someone else's fruit basket, where we don't know how much fruit is held at any time $t$ $n(t)=?$. It is a stochastic process though, and we may know its parameters, e.g. $dn(t)\sim\mathcal N(0,t)$

Can we get a process for the value of the basket $V(t)$? Because if we did, then we could integrate again: $V(t)=V(0)+\int_0^tdV(t)$

Riemann would say that it's easy: $dV=\left(dn\times p+n\times dp\right)$ - it's a usual full differential, and it's WRONG. That's where the Itô integral comes up: $dV=\left(dn\times p+n\times dp+dn\times dp\right)$.

WTH did the last term $dn\times dp$ come from?!
Consider the value of a basket at time 0 and time $t$:$$V(0)=n(0)\times p(0)\\V(t)= n(t)\times p(t)$$ Look at the difference: $$\Delta V(t)\equiv V(t)-V(0)= n(0)\times [p(t)-p(0)]+[n(t)-n(0)]\times p(0)+[n(t)-n(0)]\times[p(t)-p(0)] = \Delta n\times p(0)+n(0)\times\Delta p+\Delta n\times\Delta p$$ This corresponds to the Itô calculus $dV$ above.

How are these related?

If you have a function $V(t)$ where $t$ is a deterministic variable such as time, then usual calculus and Riemann integration works: $dV(t)=\frac{\partial V}{\partial t}dt$

If your function has stochastic variables $V(t,p)$ such as price of financial assets $p$, then Itô calculus and Lebesgue integration is in order: $dV(t)\ne\frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial p}dp(t)$

The example with a fruit basket was a very simple function $V(n,p)=n\times p$, where we can get to the answer without Itô calculus formalism, but if you apply Itô calculus you get the same answer.