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, where is a standard Gaussian r.v..

The direct computation of 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) . Since and , we deduce, for ,

Now we can deduce using analytic continuation. I explain in details. Indeed, for , we know on , and we know a priori that it is analytic on the whole complex plane (using the "Analycity under thetheorem"). Therefore: since the analytic function coincides with on , it coincides on and we can take to deduce

.

We can turn to the variable . Let be fixed. We know for . And we know a priori that it is analytic on . We deduce, like above, the value for where : for any ,

,

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

- Another way: through integration in the complex plane. We have hence

.

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 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 that turns the integral on to an integralon a complex line, and finally using Cauchy formula (...), we can change to contour to integrate on again and use the above formula for the Gaussian...

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Did you have something else in mind?

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It comes up to my mind that one could probably generalize the differential proof of the value of : differentiate with respect to , integrate by parts, to find a first order ODE...

After further investigation : indeed it gives a "a bit" nicer solution. Fix , and consider . We get after integration by part. We deduce . What is ? Taking , we see that . If we know that is and the characteristic function of gamma distribution, then we conclude. Or we do the computation, which again implies some complex analysis.