# Math Help - [SOLVED] Joint characteristic function

1. ## [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

2. 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.

3. 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}]
&=\int e^{itx} e^{ux^2} e^{-x^2/2} \cdot \frac{1}{\sqrt{2\pi}} ~dx \\
&=\int e^{itx} e^{-(1-2u)x^2/2} \cdot \frac{1}{\sqrt{2\pi}\sqrt{1-2u}} \cdot \sqrt{1-2u} ~dx
\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.
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

4. Originally Posted by Moo
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.

5. Originally Posted by Laurent
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!

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!).