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

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: $V(t)=V(0)+n\times p(t)$ 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 $t$:$$V(0)=n(0)\times p(0)\\V(0)= n(t)\times p(t)$$ 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.