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$X_1,X_2....X_n$~Norm($\mu,\sigma^2$).What are the distributions of those random variable that is following:
1.$Y_1=\frac{X_1-X_2}{\sqrt{X_3^2+X_4^2}}$
2.$Y_2=\frac{\sqrt{n-1}X_1}{\sqrt{\sum_{i=2}^n X_i^2}}$

I think that both $Y_1$and $Y_2$ obey Student Distribution. But I can't prove my answers. Anybody help me? Thanks in advance!

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  • $\begingroup$ I assume $X_1,X_2,...,X_n$ are identically independently distributed? $\endgroup$
    – benh
    Commented Dec 23, 2013 at 2:58
  • $\begingroup$ @benh: Yes.$X_1...X_n$ are identically independently. $\endgroup$
    – Jiabin He
    Commented Dec 23, 2013 at 5:30

1 Answer 1

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We obtain the following results by using linearity results on normal distributions: $$Z_1:=\frac{1}{\sqrt{2}\sigma}(X_1-X_2) \sim N(0,1).$$ Also, as $\frac{1}{\sigma}X_3,\frac{1}{\sigma}X_4\sim N(0,1)$ we have $$Z_2:=\frac{1}{\sigma^2}(X_3^2+X_4^2)\sim \chi^2_2.$$ As $Z_1$ and $Z_2$ are independent if $X_1,...,X_4$ are independent, by definition $$\frac{X_1-X_2}{\sqrt{X_3^2+X_4^2}}=\frac{Z_1}{\sqrt{Z_2/2}} \sim t_2.$$ So indeed $Y_1$ is $t_2$-distributed.

For 2) write $$Y_2 = \frac{\frac{1}{\sigma}X_1}{\sqrt{\frac{1}{\sigma^2}\frac{\sum_{k=2}^nX_k^2}{n-1}}}$$

and proceed as with $Y_1$ to obtain a $t_n$-distribution.

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  • $\begingroup$ I know $\frac{X_3-\mu}{\sigma}$~$N(0,1)$, but why $\frac1{\sigma}X_3$~$N(0,1)$? $\endgroup$
    – Jiabin He
    Commented Dec 23, 2013 at 5:33
  • $\begingroup$ Oh, I am sorry. That's a mistake, for some reason I was assuming $\mu = 0$. So then $\frac{1}{\sigma}X_3^2\sim \chi^2(\mu^2)$ only for 1). Hence $\sum_{k=2}^n X_k^2 = \chi^2_{n-1}((n-1)\mu^2)$. I don't think that this is a t-distribution in general. Are you sure, that not $\mu=0$ or that it is not actually $(X-\overline X)^2$? $\endgroup$
    – benh
    Commented Dec 23, 2013 at 12:39

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