|
Science Forum Index » Statistics - Math Forum » Characteristic Function...
Page 1 of 1
|
| Author |
Message |
| ... |
 Posted: Sat May 03, 2008 11:02 pm |
|
|
|
Guest
|
Hi all,
I would appreciate some help in understanding the solution to a
problem that I read. The problem ask to find the characteristic
function E[exp(iuXY)] where X & Y are standard normal distribution,
and Cov(X,Y) = 0.
The solution given is:
E[exp(iuXY)] = E[E[exp(iuXY)|Y]] = E[exp(-(u^2)*(Y^2)/2)] = ....
Should be pretty straight forward for some, but I can't quite
understand how I could arrive at the third expression from the second
expression.
Any help very much appreciated. Thanks. |
|
|
| Back to top |
|
| Jack Tomsky... |
| Posted: Sun May 04, 2008 4:40 am |
|
|
|
Guest
|
Quote: Hi all,
I would appreciate some help in understanding the
solution to a
problem that I read. The problem ask to find the
characteristic
function E[exp(iuXY)] where X & Y are standard normal
distribution,
and Cov(X,Y) = 0.
The solution given is:
E[exp(iuXY)] = E[E[exp(iuXY)|Y]] =
E[exp(-(u^2)*(Y^2)/2)] = ....
Should be pretty straight forward for some, but I
can't quite
understand how I could arrive at the third expression
from the second
expression.
Any help very much appreciated. Thanks.
E[E[exp(iuXY)|Y]] = (1/sqrt(2*pi))*INT[exp(iuYx)*exp((-x^2)/2) dx = (1/sqrt(2*pi))*INT[exp(-(x^2-2i(uY)x+(iuY)^2)/2)*exp((iuY)^2/2) dx
= (1/sqrt(2*pi))exp(-(uY)^2/2)*INT[exp((x-iuY)^2)/2 dx = exp(-(uY)^2/2).
The next to last step uses the fact that i^2 = -1.
Jack |
|
|
| Back to top |
|
| ... |
| Posted: Mon May 05, 2008 6:43 pm |
|
|
|
Guest
|
Thanks for the reply, Jack. I'll have a look at it.
Cheers. |
|
|
| Back to top |
|
| Jack Tomsky... |
| Posted: Tue May 06, 2008 5:00 am |
|
|
|
Guest
|
Quote: Thanks for the reply, Jack. I'll have a look at it.
Cheers.
Let me know if you need further details. The main idea is that when it's conditional on Y, then Y is treated as a constant, just as u.
Jack |
|
|
| Back to top |
|
| |
