Distribution of ∑ni=1XiYi∑ni=1X2i\frac{\sum_{i=1}^n X_iY_i}{\sum_{i=1}^n X_i^2} where Xi,YiX_i,Y_is are i.i.d Normal variables

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

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

I define Z=ni=1XiYini=1X2i

Then, Z(X1=x1,,Xn=xn)=ni=1xiYini=1x2iN(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 ZN(0,1)

Relating U and Z, I have U=Zni=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=1nZni=1X2in=Tn, 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?


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

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

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