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Math Help - [SOLVED] Joint characteristic function

  1. #1
    Moo
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    [SOLVED] Joint characteristic function

    Hello

    So this is a problem I saw in the AoPs forum. Though they didn't solve it. My teacher assistant found a proof, which I found nice !

    Here is the problem.

    (X_n)_{n\geq 1} is a sequence of iid random variables, following a standard normal distribution \mathcal{N}(0,1)
    Let U_n=\sum_{i=1}^n X_i
    Let Z_n=\sum_{i=1}^n X_i^2

    Find the characteristic function of the couple (U_n,Z_n)


    As an extra, you can also find the joint pdf of the couple, by using the independence between U_n and V_n=\sum_{i=1}^n \left(X_i-\frac{U_n}{n}\right)^2
    And you can also try to prove that independence (that's a consequence of Cochran's theorem, but you can do it without it)


    Enjoy
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  2. #2
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    Hi,

    I did not find a "nice" proof, but the result doesn't seem really nice either, so...

    I'd first notice that, using independence and linearity, \Phi_{(U_n,Z_n)}(t,u)=\Phi_{(X,X^2)}(t,u)^n=\mathb  b{E}[e^{i(tX+uX^2)}]^n where X=X_1 is a standard Gaussian r.v..

    The direct computation of \mathbb{E}[e^{i(tX+uX^2)}] is not obvious to justify. I think a bit of complex analysis is needed any way.

    - One way: first find the Laplace transform (or moment generating function) \mathbb{E}[e^{tX+uX^2}]=\int e^{tx+ux^2}e^{-x^2/2}\frac{dx}{\sqrt{2\pi}}. Since tx+ux^2-\frac{x^2}{2}=-\frac{1-2u}{2}\left(x-\frac{t}{1-2u}\right)^2+\frac{t^2}{2(1-2u)} and \int e^{-\frac{(x-m)^2}{2\sigma^2}}\frac{dx}{\sqrt{2\pi}} = \sqrt{\sigma^2}, we deduce, for u<1/2,

    \mathbb{E}[e^{tX+uX^2}]=\frac{e^{\frac{t^2}{2(1-2u)}}}{\sqrt{1-2u}}.

    Now we can deduce \Phi(t,u)=\mathbb{E}[e^{i(tX+uX^2)}] using analytic continuation. I explain in details. Indeed, for u<1/2, we know t\mapsto \mathbb{R}[e^{tX+uX^2}] on \mathbb{R}, and we know a priori that it is analytic on the whole complex plane \mathbb{C} (using the "Analycity under the \int theorem"). Therefore: since the analytic function z\mapsto\frac{e^{\frac{z^2}{2(1-2u)}}}{\sqrt{1-2u}} coincides with t\mapsto \mathbb{R}[e^{tX+uX^2}] on \mathbb{R}, it coincides on \mathbb{C} and we can take t=it' to deduce

    \mathbb{E}[e^{it'X+uX^2}]=\frac{e^{-\frac{t'^2}{2(1-2u)}}}{\sqrt{1-2u}}.

    We can turn to the variable u. Let t'\in\mathbb{R} be fixed. We know u\mapsto\mathbb{E}[e^{it'X+uX^2}] for u<1/2. And we know a priori that it is analytic on \{u\in\mathbb{C}|\Re(u)<1/2\}. We deduce, like above, the value for u=iu' where u'\in\mathbb{R}: for any t',u'\in\mathbb{R},

    \mathbb{E}[e^{it'X+iu'X^2}]=\frac{e^{-\frac{t'^2}{2(1-2iu')}}}{\sqrt{1-2iu'}},

    where the complex square root is chosen with a positive real part.


    - Another way: through integration in the complex plane. We have i(tx+ux^2)-\frac{x^2}{2}=-\frac{1-2iu}{2}\left(x-\frac{it}{1-2iu}\right)^2+\frac{(it)^2}{2(1-2iu)} hence

    \mathbb{E}[e^{i(tx+ux^2)}]=e^{-\frac{t^2}{2(1-2iu)}}\int e^{-\frac{1-2iu}{2}\left(x-\frac{it}{1-2iu}\right)^2}\frac{dx}{\sqrt{2\pi}}.

    In the integral we have a kind of density of a Gaussian r.v. with complex mean and variance and if we apply the above formula \int e^{-\frac{(x-m)^2}{2\sigma^2}}\frac{dx}{\sqrt{2\pi}} = \sqrt{\sigma^2} we get the result (with a complex square root to be precised...). To make this argument rigorous, you can procede (with some care) to the change of variable \frac{z^2}{2}=-\frac{(x-m)^2}{2\sigma^2} that turns the integral on \mathbb{R} to an integral on a complex line, and finally using Cauchy formula (...), we can change to contour to integrate on \mathbb{R} again and use the above formula for the Gaussian...
    ---

    Did you have something else in mind?

    --

    It comes up to my mind that one could probably generalize the differential proof of the value of \Phi_X(t)=e^{-t^2/2} : differentiate \Phi_{X,X^2}(t,u) with respect to t, integrate by parts, to find a first order ODE...

    After further investigation : indeed it gives a "a bit" nicer solution. Fix u, and consider f(t)=\Phi_{(X,X^2)}(t,u). We get f'(t)=-\frac{t}{1-2iu}f(t) after integration by part. We deduce f(t)=C(u)e^{-\frac{t2}{2(1-2iu)}}. What is C(u) ? Taking t=0, we see that C(u)=\mathbb{E}[e^{iuX^2}]. If we know that X^2 is \gamma_{1/2,1/2} and the characteristic function of gamma distribution, then we conclude. Or we do the computation, which again implies some complex analysis.
    Last edited by Laurent; May 17th 2009 at 06:09 AM. Reason: Complement about the ODE method
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  3. #3
    Moo
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    Haha, I must tell you that I saw you were replying to this thread, for almost an hour. And that it's been hard to wait.
    It looks like it was worth it !

    What I had in mind is what my teacher did. And he used a slighty different way than the first method you mentioned.
    The answer is not beautiful, but I liked the way he did it (see below), and hence what you did. Maybe the correct word would have been "interesting", not "nice". That's a personal opinion. I'm a bit crazy sometimes


    So he did it this way :

    It's not the Laplace transform he studied, it's E[e^{itX+uX^2}]
    And then he directly uses the analytic continuation over u.

    \begin{aligned} E[e^{itX+uX^2}]<br />
&=\int e^{itx} e^{ux^2} e^{-x^2/2} \cdot \frac{1}{\sqrt{2\pi}} ~dx \\<br />
&=\int e^{itx} e^{-(1-2u)x^2/2} \cdot \frac{1}{\sqrt{2\pi}\sqrt{1-2u}}  \cdot \sqrt{1-2u} ~dx <br />
\end{aligned}

    But then we can recognize the pdf of \mathcal{N}\left(0,\tfrac{1}{1-2u}\right) and more precisely, its characteristic function !

    So :
    E[e^{itx+uX^2}]=\frac{e^{-\frac{t^2}{2(1-2u)}}}{\sqrt{1-2u}}

    f(u)=\mathbb{E}[e^{itX+uX^2}] and g(u)=\frac{e^{-\frac{t^2}{2(1-2u)}}}{\sqrt{1-2u}} are analytic and coincide over \{u ~:~ \text{Re}(u)<1/2\}
    So it coincides over \mathbb{C}
    By taking u=is, we get the result.


    He said that it could be using the two-variables analytic continuation (which I guess is what you did ?) but we never learnt it

    It comes up to my mind that one could probably generalize the differential proof of the value of \Phi_X(t)=e^{-t^2/2} : differentiate \Phi_{X,X^2}(t,u) with respect to t, integrate by parts, to find a first order ODE...

    After further investigation : indeed it gives a "a bit" nicer solution. Fix u, and consider f(t)=\Phi_{(X,X^2)}(t,u). We get f'(t)=-\frac{t}{1-2iu}f(t) after integration by part. We deduce f(t)=C(u)e^{-\frac{t}{1-2iu}}. What is C(u) ? Taking t=0, we see that C(u)=\mathbb{E}[e^{uX^2}]. If we know that X^2 is \gamma_{1/2,1/2} and the characteristic function of gamma distribution, then we conclude.
    I thought about this method too
    But didn't know how to deal with two variables... lol



    Anyway, I know it can look strange that I ask this question whereas I have the answer. But I wanted to know, among other things, if it was a reflex to use the analytic continuation. And it looks like it is. And I won 2 other methods
    But it's not easy at all to deal with the multivariate situations


    Thanks a lot, as always, and forever. And I really mean it
    Last edited by Moo; May 17th 2009 at 06:33 AM.
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  4. #4
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    Quote Originally Posted by Moo View Post
    So he did it this way :

    It's not the Laplace transform he studied, it's E[e^{itX+uX^2}]
    (...)
    He said that it could be using the two-variables analytic continuation (which I guess is what you did ?) but we never learnt it
    Yes this is what I did, but I took it in steps so technically I only used one-variable continuation twice.

    In fact \mathbb{E}[e^{itX}]=e^{-t^2/2} is proved in a natural way using analytic continuation: the Laplace transform is easy to compute (take u=0 in my proof), but the Fourier transform is not, so we jump from one to another.
    This is a general idea (the same works for gamma distributions, and I don't know an alternative proof), while the trick with the 1st order ODE is very specific to Gaussian r.v..

    And I admit this is a nice idea when you see it at first ; it confirms the wonderful power of complex analysis!

    I'm glad you liked it ('cause indeed I spent some time on it... ),

    Laurent.
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  5. #5
    Moo
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    Quote Originally Posted by Laurent View Post
    Yes this is what I did, but I took it in steps so technically I only used one-variable continuation twice.

    In fact \mathbb{E}[e^{itX}]=e^{-t^2/2} is proved in a natural way using analytic continuation: the Laplace transform is easy to compute (take u=0 in my proof), but the Fourier transform is not, so we jump from one to another.
    This is a general idea (the same works for gamma distributions, and I don't know an alternative proof), while the trick with the 1st order ODE is very specific to Gaussian r.v..

    And I admit this is a nice idea when you see it at first ; it confirms the wonderful power of complex analysis!

    I'm glad you liked it,

    Laurent.
    Yup, we studied both methods for \mathbb{E}[e^{itX}].
    Now I understand what you're saying. It's more natural to start from the Laplace transform, because what my teacher did was assuming one knows the characteristic function of a normal distribution.
    And yes... the way complex analysis is used is quite impressive. I will regret not paying attention to last semester's complex analysis course (though we didn't talk about the analytic continuation - or so little)

    Got it,

    Thanks

    ('cause indeed I spent some time on it... )
    And hence the few lines for the reason why I posted it. I didn't want to make you feel it was for a silly purpose (for me lol!).
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