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I was wondering whats the result of an Itō integral multiplied by a Riemann Integral. For example, what is $$\left(\int_0^T f(u)\ \mathsf dW_u\right)\left(\int_0^T g(v)\ \mathsf dv\right)$$ where $W$ is a standard Wiener process. Since $\mathsf dW_t\mathsf dt=0$, does that mean the multiplication of the two integrals is also zero? Is there a formal way of proving the result? Thanks in advance for any help.

Math1000
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xyz1010
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    Wait, you are asking if the product itself is $0$ or its expectation is 0? – Calculon Jul 31 '15 at 13:22
  • Thanks for taking your time to help! I was wondering about the product itself. For this instance, both $f$ and $g$ are deterministic functions. Is the product of the two integrals zero? If so, could you please provide some explanation as to why? Also, could you elaborate on what you meant by it is not correct to write $dW_tdt=0$? Thanks – xyz1010 Jul 31 '15 at 14:16
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    Since the Ito integral results in a stochastic process you need to be precise about what it means for the product to be equal to $0$. But let's fix $T$ first. When is the product of a random variable (the left integral) and a real number (the right one) equal to $0$ (almost surely, if you want)? The left integral is actually a normally distributed random variable so it is almost surely not $0$. That must mean the right integral must be $0$ then. Since $T$ was arbitrary, the integral must be $0$ at all times. What does that tell you about $g$? – Calculon Jul 31 '15 at 14:54
  • This is not going to be $0$, see also this related post. – dafinguzman Sep 29 '15 at 17:16

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Why it is not $0$:

The expression $\text{d}W_t\, \text{d}t = 0$ is valid when it is being jointly integrated with respect to the same variable $t$, thus, symbolically $$\int_{t=0}^T f(t) \text{d}W_t\, \text{d}t = 0.$$ In your case you have two different integrals with respect to different integration variables $u$ and $v$, so you can't say that $\text{d}W_u\, \text{d}v$ is null. You can interpret it as a "double differential" in a similar fashion than $\text{d}x\, \text{d}y$ in double integrals.

What is it then?

It is a good exercise to prove that an integral of the form $\int_0^T f(s)dW_s$ is normally distributed with zero mean and variance $\int_0^T |f(s)|^2ds $.

(Hint: write the Riemann sums and interpret them as a sum of independent normally distributed random variables.)

Thus, the factor to the left is a random variable. On the other hand, assuming that $g$ is a real valued function (non-random), the factor to the right is a number, call it $\sigma$. The product of those two then is a normally distributed random variable of mean $0$ and variance $\sigma^2\int_0^T |f(s)|^2ds$.

dafinguzman
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