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I'm having trouble getting my head around the meaning of the stochastic Ito integral. Specifically: the intuitive meaning of "Stochastic Integral" to me is a function that takes a time $t$ and returns a PDF for the integral of a stochastic process on $[0,t]$. Is this indeed what Ito integrals model? If so, I'm confused about why Ito integrals depend on two stochastic processes (you integrate process A with respect to process B). How does the meaning of the integral change when process B changes?

Thanks for your help!

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2 Answers 2

I'm far from an expert here, but I've just been learning about this field so thought I might share! NB I'm not trying to be rigorous, just get the ideas across.

Let's first think of a 'standard' Riemann integral,

$\int\limits_{a}^{b} f(x)\,\text{d}x$

we interpret this as the limit of a sum of the areas of equally spaced boxes, each with width $\Delta x$ and height equal to the value of the function, then let $\Delta x\rightarrow 0$. So, as an approximation:

$\int\limits_{a}^{b} f(x)\,\text{d}x \approx \sum\limits_{i=0}^{n-1} f(t_i)(x_{i+1}-x_i)$,

where the $x_i$s are a partition of the interval $[a,b]$ and $t_i$ is a value somewhere in the interval $[x_i,x_{i+1}]$.

Now let's take a stochastic Ito integral of a non-random process, $f(t)$:

$\int\limits_{0}^{t} f(s)\,\text{d}B(s)$,

where $B(s)$ is a random process (normally Brownian motion). This isn't a PDF, it's actually a random variable itself. We can approximate this in the same way as a Riemann integral, i.e. a sum of the areas of boxes:

$\int\limits_{0}^{T} f(s)\,\text{d}B(s) \approx \sum\limits_{k=0}^{n-1}f(t_i)(B(t_{i+1})-B(t_i))$,

where the $t_i$s are a partition of the time interval $[0, T]$ into $n$ segments. Obviously, the larger $n$ is, the better the approximation. Looking at that expression, we can see that it's quite similar to the Riemann integral, but now the intervals themselves, whilst regularly spaced in $t$, are of random width. And a sum of random variables is itself a random variable.

It would be neat if there were a diagram showing these ideas (i.e. drawing the boxes) but I'm not sure how exactly to draw it.

If we now take the Ito integral of a random process, $X(t)$, then we have to replace $f(t_i)$ with a random variable $X(t_i)$ (I think there are several conditions you have to meet as well, I'm ignoring these here for the general picture), so the Ito integral becomes a sum of products of random variables, but again that's a random variable.

To answer your final question: changing the process $B$ means changing the process that generates our box widths, and hence changing the statistical properties of the integral.

Hope this helps, I await someone far more knowledgable to come along and tell me it's all wrong!

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Nice answer. I liked the idea of random width. Never heard that from my prof. Thanks –  triomphe Oct 17 '13 at 19:11

I'm confused about why Ito integrals depend on two stochastic processes (you integrate process A with respect to process B). How does the meaning of the integral change when process B changes?

Already (deterministic) Stieltjes integrals $\int\limits_0^tu(t)\mathrm dv(t)$ depend on two functions $u$ and $v$.

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