Serving the Quantitative Finance Community

 
User avatar
Zakduka
Topic Author
Posts: 0
Joined: November 19th, 2002, 12:12 pm

Stochastic Integrals and Integration

November 20th, 2002, 1:27 pm

Could someone please explain the difference between integrating when the density is dW vs. integrating when the density is dt?Specifically : (Assume: ¦: integral sign from t to T)1- ¦ WdW2- ¦ Wdt3- ¦ tdW4- ¦ tdtIn general how does one treat W in integration? Is (2) a riemann like (4) so would the result be Wt?For (1) using Ito's integral (assuming f(0)=0) F(t) = ¦f'(x)dX + 1/2 ¦f''(x) dt assuming f'(x) = X --> F(x)=(X^2)/2 then replacing X with W ---> ¦ WdW = 1/2 (W^2 - t)But how do we treat (2) and (3) and in general how do we integrate a stochastic integral or an integral with a stochastic element in it (2)?Please HelpZakduka
 
User avatar
weare
Posts: 1
Joined: November 18th, 2002, 8:10 am

Stochastic Integrals and Integration

November 21st, 2002, 3:56 am

use the integration by part.u = W, dv = dt and du = dW, v = tthen, expression (2) and (3) can be used interchangeably by the IbP.
Last edited by weare on November 20th, 2002, 11:00 pm, edited 1 time in total.
 
User avatar
mj
Posts: 12
Joined: December 20th, 2001, 12:32 pm

Stochastic Integrals and Integration

November 21st, 2002, 9:03 am

Asking the question "how to do an integral" is a bit of a "how long is a piece of string" type question. After all there is no easy answer even in the non-stochastic case. When you are integrating against "dt" but the integrand is stochastic then the integral is a Riemann integral. However, it is a Riemann integral on a path by path basis. So once you have drawn a random path then it is Riemann. If you integrate Wdt from say 0 to 1, then the answer depends on how the path got from 0 to 1. If was positive all the way the answer will be positive even if W_1 = 0 and similarly if negative. So the answer is definitely not Wt. The easiest way to understand whats going on is to approximate the Riemman integral by a sum. So \int_{0}^{1} W_t dt = \sum_{j=0}^{N-1} \frac{1}{N} W_{j/N}so we see it is a sum of normals. You can further expand and work the variance. Hope this is some helpMJ
 
User avatar
brussels
Posts: 0
Joined: September 22nd, 2002, 2:22 pm

Stochastic Integrals and Integration

November 29th, 2002, 9:19 pm

I am not going to answer to the whole question, but let me at least explain what the integral of f.dW is: it shows that the Riemann-Stiltjes integral is particularly useful in financeSuppose a stock follows a process W(t). Suppose you decide beforehand how many shares of stocks you will hold at each moment. Then the sum of all your share gains and share losses will roughly beSum_i f(t_i)*(W(t_{i+1})-W_(t_i))(where t_i is a subdivision of the interval [0,T]) . It's just the number of shares at moment t_i multiplied by the gain or loss between moment t_i and t_{i+1} - summed over i. If you go to the limit where i->infinity, you get the integral (given that the sum above is not a number but a random variable, there are different meanings for 'limit'. In this case, it's the L^2 limit that works). This limit is of course itself a random variable.You can extend the idea to any function f(t,omega) as long as it is compatible with the filtration generated by W (i.e. you must decide the value of f(t) at the latest at time t / the value of f(t) can only depend on the path of W(.) up to time t). Such a function is called a 'strategy'I think that a way to summarize the need for this type of integral is to say that1) the natural formula for summing the gains generated by a strategy is Riemann-Stieltjes, which, in the classical case (i.e. with numbers and not random variables) only works with piecewise differentiable functions.2) efficient market hypothesis implies that W() should look like a brownian motion, and in particular will be nowhere differentiable (?)