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Science Forum Index » Statistics - Math Forum » Characteristic Function...
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 Posted: Sat May 03, 2008 11:02 pm |
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Guest
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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. |
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| Jack Tomsky... |
| Posted: Sun May 04, 2008 4:40 am |
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Guest
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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 |
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| Posted: Mon May 05, 2008 6:43 pm |
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Guest
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Thanks for the reply, Jack. I'll have a look at it.
Cheers. |
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| Jack Tomsky... |
| Posted: Tue May 06, 2008 5:00 am |
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Guest
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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 |
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| Posted: Fri May 23, 2008 3:15 am |
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Guest
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Haven't had time to read through earlier (bogged down by work).
Jack, how did you get from E[E[exp(iuXY)|Y]] = (1/
sqrt(2*pi))*INT[exp(iuYx)*exp((-x^2)/2) dx ? I just can't see how the
exp((-x^2)/2) is derived from? Admittedly, my level of prob/stats
theory is really bad.
Another approach given by the solution is to express the
characteristic function as follows:
Since XY = [(X+Y)^2 - (X-Y)^2]/4
E(exp(iuXY)) = E[exp(iu((X+Y)^2)/4)]E[exp(-iu((X-Y)^2)/4)] = (1/sqrt(-
iu+1))(1/sqrt(iu+1)) = 1/sqrt(u^2+1)
The problem I've with the second approach is, I'm not sure how/why
E[exp(iu((X+Y)^2)/4)] = (1/sqrt(-iu+1))?
Any help is really appreciated. Thanks. |
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| Jack Tomsky... |
| Posted: Fri May 23, 2008 5:50 am |
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Guest
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Quote: Haven't had time to read through earlier (bogged down
by work).
Jack, how did you get from E[E[exp(iuXY)|Y]] = (1/
sqrt(2*pi))*INT[exp(iuYx)*exp((-x^2)/2) dx ? I just
can't see how the
exp((-x^2)/2) is derived from? Admittedly, my level
of prob/stats
theory is really bad.
Another approach given by the solution is to express
the
characteristic function as follows:
Since XY = [(X+Y)^2 - (X-Y)^2]/4
E(exp(iuXY)) =
E[exp(iu((X+Y)^2)/4)]E[exp(-iu((X-Y)^2)/4)] =
(1/sqrt(-
iu+1))(1/sqrt(iu+1)) = 1/sqrt(u^2+1)
The problem I've with the second approach is, I'm not
sure how/why
E[exp(iu((X+Y)^2)/4)] = (1/sqrt(-iu+1))?
Any help is really appreciated. Thanks.
Once it's conditional on Y, it's equivalent to treating Y as a constant. That takes care of the inner expectation in the expression. So what we have is an expectation of a function of the random variable X (which is N(0,1)).
The function we want the expectation of is f(x) = exp(iuXY). By definition, the expectation of f(x) is
E(f(x)) = INT[f(x)*(1/sqrt(2*pi))*exp((-x^2)/2) dx],
which is the integral of f(x) times the normal density of x.
Factor out of the integral the constant, 1/sqrt(2*pi), and set f(x) = exp(iuXY).
I hope this helps.
Jack |
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| Posted: Sat May 24, 2008 4:01 pm |
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Guest
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| Many thanks for the help, Jack. It makes more sense to me now. |
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