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Studying for a test. Couldn't answer this one.

Let $X_{1,i},X_{2,i},X_{3,i}, i=1,\ldots,n$ be iid $\mathcal{N}(0,1)$ random variables. Define

$W_i = (X_{1,i} + X_{2,i}X_{3,i})/\sqrt{1 + X_{3,i}^2}, i = 1, \ldots, n$,

and $\overline{W}_n = n^{-1}\sum_{i=1}^nW_i$,

$S_n^2 = (n-1)^{-1}\sum_{i=1}^n(W_i - \overline{W}_n)^2, n \ge 2.$

What is the distribution of $\overline{W}_n$, $S_n^2$?

How do I get an idea of the best method to use when starting a problem like this?

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Do you want the distribution for fixed $n$ or the asymptotic distribution? Are you interested in the marginal distributions of $\overline W_n$ and $S_n^2$ or their joint distribution? – cardinal Jan 13 '13 at 23:12
Sorry for the ambiguity. Keep $n$ fixed, and I am only interested in their marginals. They ask later on if the two statistics are independent, so I'm anticipating some use of Basu's theorem. – Taylor Jan 13 '13 at 23:51
up vote 7 down vote accepted

It's a trick.

Conditionally on $X_{3,i} = x$ we have that $W_i$ equals $$\frac{X_{1,i} + X_{2,i}x}{\sqrt{1 + x^2}} \sim \mathcal{N}(0, 1).$$ This follows from the fact that for fixed $x$ this is a simple linear transformation of the two independent $\mathcal{N}(0,1)$-distributed variables $X_{1,i}$ and $X_{2,i}$. Whence, $W_i \mid X_{3,i} = x$ has a normal distribution. The conditional mean is seen to be 0 and the conditional variance is (by the independence assumptions) $$V(W_i \mid X_{3,i} = x) = \frac{V(X_{1,i}) + V(X_{2,i})x^2}{1 + x^2} = \frac{1 + x^2}{1 + x^2} = 1.$$

Since the conditional distribution of $W_i \mid X_{3,i} = x$ doesn't depend upon $x$ we conclude that it is its marginal distribution as well, that is, $W_i \sim \mathcal{N}(0,1).$

The rest follows from standard results on averages and residuals for independent normal random variables. Basu's theorem is not needed for anything.

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very impressive! – Cam.Davidson.Pilon Jan 16 '13 at 18:30
well hey, look at that. thank you sir. – Taylor Jan 17 '13 at 2:02

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