Integral with respect to Brownian motion.

operationsres
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Suppose that \sigma(t,T) is a deterministic process, where t varies and T is a constant. We also have that t \in [0,T]. Also W(t) is a Wiener process.

My First Question

What is \displaystyle \ \ d\int_0^t \sigma(u,T)dW(u)? My lecture slides assert that it's equal to \sigma(t,T)dW(t) but I'm not sure why. So my question is "Why"?

My Second Question

What is \displaystyle \ \ d\int_a^t \sigma(u,T)dW(u), where a \in (0,t).

_________________________________

Thanks!
 
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operationsres said:
Suppose that \sigma(t,T) is a deterministic process, where t varies and T is a constant. We also have that t \in [0,T]. Also W(t) is a Wiener process.

My First Question

What is \displaystyle \ \ d\int_0^t \sigma(u,T)dW(u)? My lecture slides assert that it's equal to \sigma(t,T)dW(t) but I'm not sure why. So my question is "Why"?

My Second Question

What is \displaystyle \ \ d\int_a^t \sigma(u,T)dW(u), where a \in (0,t).

_________________________________

Thanks!

I've said it before and I will say it again: go back to the basics. What is meant by the stochastic integral? If Y(t) is a stochastic process, what do we mean by dY(t)? All this material is explained in books (admittedly, some almost unreadable), and in numerous web pages and the like. There is simply no substitute for getting the background first.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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