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, $\displaystyle \Phi_{(U_n,Z_n)}(t,u)=\Phi_{(X,X^2)}(t,u)^n=\mathb b{E}[e^{i(tX+uX^2)}]^n$ where $\displaystyle X=X_1$ is a standard Gaussian r.v..

The direct computation of $\displaystyle \mathbb{E}[e^{i(tX+uX^2)}]$ 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) $\displaystyle \mathbb{E}[e^{tX+uX^2}]=\int e^{tx+ux^2}e^{-x^2/2}\frac{dx}{\sqrt{2\pi}}$. Since $\displaystyle tx+ux^2-\frac{x^2}{2}=-\frac{1-2u}{2}\left(x-\frac{t}{1-2u}\right)^2+\frac{t^2}{2(1-2u)}$ and $\displaystyle \int e^{-\frac{(x-m)^2}{2\sigma^2}}\frac{dx}{\sqrt{2\pi}} = \sqrt{\sigma^2}$, we deduce, for $\displaystyle u<1/2$,

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

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

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

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

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

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

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

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

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 $\displaystyle \int e^{-\frac{(x-m)^2}{2\sigma^2}}\frac{dx}{\sqrt{2\pi}} = \sqrt{\sigma^2}$ 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 $\displaystyle \frac{z^2}{2}=-\frac{(x-m)^2}{2\sigma^2}$ that turns the integral on $\displaystyle \mathbb{R}$ to an integral *on a complex line*, and finally using Cauchy formula (...), we can change to contour to integrate on $\displaystyle \mathbb{R}$ 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 $\displaystyle \Phi_X(t)=e^{-t^2/2}$ : differentiate $\displaystyle \Phi_{X,X^2}(t,u)$ with respect to $\displaystyle t$, integrate by parts, to find a first order ODE...

After further investigation : indeed it gives a "a bit" nicer solution. Fix $\displaystyle u$, and consider $\displaystyle f(t)=\Phi_{(X,X^2)}(t,u)$. We get $\displaystyle f'(t)=-\frac{t}{1-2iu}f(t)$ after integration by part. We deduce $\displaystyle f(t)=C(u)e^{-\frac{t2}{2(1-2iu)}}$. What is $\displaystyle C(u)$ ? Taking $\displaystyle t=0$, we see that $\displaystyle C(u)=\mathbb{E}[e^{iuX^2}]$. If we know that $\displaystyle X^2$ is $\displaystyle \gamma_{1/2,1/2}$ and the characteristic function of gamma distribution, then we conclude. Or we do the computation, which again implies some complex analysis.