Suppose X1,…,Xn,Y1,…,Yn are i.i.d N(0,1) random variables.

I am interested in the distribution of U=∑ni=1XiYi∑ni=1X2i

I define Z=∑ni=1XiYi√∑ni=1X2i

Then, Z∣(X1=x1,…,Xn=xn)=∑ni=1xiYi√∑ni=1x2i∼N(0,1)

As this conditional distribution is independent of X1,…,Xn, the unconditional distribution should also be the same. That is, I can say that Z∼N(0,1)

Relating U and Z, I have U=Z√∑ni=1X2i

Now since I saw that Z∣(X1,…,Xn)d=Z, I can say that Z is independent of X1,…,Xn.

So I have

U=1√nZ√∑ni=1X2in=T√n, where T is distributed as a t distribution with n degrees of freedom.

I think conditioning is the easiest way to see the result here. But is this a perfectly rigorous argument and is there any direct/alternative way of finding distributions of such functions of linear combinations of i.i.d Normal variables?

**Answer**

Although this is a conditional argument as well, using the characteristic function is faster:

E[exp{ιt∑iYiXi/∑jX2j}]=E[E[exp{ιtYiXi/∑jX2j}]|X]=E[E[∏iexp{ιtYiXi/∑jX2j}]|X]=E[∏iE[exp{tYiXi/∑jX2j}]|Xι]=E[∏iexp{−t2X2i/2{∑jX2j}2}]=E[exp{−t2/2∑jX2j}]

Invoking Wolfram’s integrator, this expectation is equal to

\int_0^∞ \zeta^{n/2 – 1} \frac{\exp(-\zeta – t^2/\zeta)}{Γ(n/2)}\ \text{d}\,\zeta = \frac{2 t^{n/2} K_{-n/2}(2 t)}{Γ(n/2)}

where K_n is the modified Bessel function of the second kind. Hence, except for n=1 **this is not the characteristic function of the Cauchy distribution. This looks instead like the characteristic function of the Student’s t distribution.**

**Attribution***Source : Link , Question Author : StubbornAtom , Answer Author : Xi’an*