 napolean80
Topic Author
Posts: 87
Joined: January 2nd, 2008, 5:40 pm

### Wiener Integral

Why is a Wiener Integral of the form [Integral phi(u)*dW], a guassian process? What is its mean and variance. tqt
Posts: 14
Joined: June 3rd, 2008, 7:41 pm

### Wiener Integral

Assuming phi(u) is deterministic (otherwise the process is not gaussian), if you write the integral as the limit of \sum_i phi(u_{t_i}) * (W_{t_i} - W_{{t_i-1}})over finer partitions of the form 0=t0<t1<...<tn=T, then you see the process is a gaussian (because of independent increments), its expectation is zero, and its variance is given by \lim \sum_i phi(u_ti)^2 t_i-t_{i-1} which is \int \phi^2(t) dtYou can also get the variance directly by using Ito's isometry, since it says that the expected value of the square of your integral is the expected value of \int \phi^2(t) dt, which is deterministic. napolean80
Topic Author
Posts: 87
Joined: January 2nd, 2008, 5:40 pm

### Wiener Integral

QuoteOriginally posted by: tqtAssuming phi(u) is deterministic (otherwise the process is not gaussian), if you write the integral as the limit of \sum_i phi(u_{t_i}) * (W_{t_i} - W_{{t_i-1}})over finer partitions of the form 0=t0<t1<...<tn=T, then you see the process is a gaussian (because of independent increments), its expectation is zero, and its variance is given by \lim \sum_i phi(u_ti)^2 t_i-t_{i-1} which is \int \phi^2(t) dtYou can also get the variance directly by using Ito's isometry, since it says that the expected value of the square of your integral is the expected value of \int \phi^2(t) dt, which is deterministic. it is somewhat hard to understand the equation because it looks like you are typing in latex but with syntax error. but thank you v much.
Last edited by napolean80 on April 23rd, 2009, 10:00 pm, edited 1 time in total. repoman
Posts: 123
Joined: February 10th, 2009, 12:53 am

### Wiener Integral

QuoteOriginally posted by: tqtAssuming phi(u) is deterministic (otherwise the process is not gaussian), if you write the integral as the limit of \sum_i phi(u_{t_i}) * (W_{t_i} - W_{{t_i-1}})over finer partitions of the form 0=t0<t1<...<tn=T, then you see the process is a gaussian (because of independent increments), its expectation is zero, and its variance is given by \lim \sum_i phi(u_ti)^2 t_i-t_{i-1} which is \int \phi^2(t) dtYou can also get the variance directly by using Ito's isometry, since it says that the expected value of the square of your integral is the expected value of \int \phi^2(t) dt, which is deterministic.The expectation of any Wiener integral is 0, and as you mentioned Ito's isometry applies to any Wiener integral. Another way to check that the integral is normally distributed for a deterministic integrand, is to check the moment generating function, and this uses your observation that the for a deterministic integrand we are taking the expected value of a constant to compute the variance (see Shreve). For the direct way, writing out the definition of the Wiener integral as you did, see also this thread. napolean80
Topic Author
Posts: 87
Joined: January 2nd, 2008, 5:40 pm

### Wiener Integral

thank you repoman tqt
Posts: 14
Joined: June 3rd, 2008, 7:41 pm

### Wiener Integral

Quoteit is somewhat hard to understand the equation because it looks like you are typing in latex but with syntax error. but thank you v much. Good point. I'll try the real latex editor next time.
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