Introduction to Probability, Second Edition, 2nd Edition

where $Z~\mathcal{N}\left(0,1\right)$, $V~{\chi }_n^2$, and Z is independent of V. Then

$-T=\frac{-Z}{\sqrt{V∕n}},$

where $-Z~\mathcal{N}\left(0,1\right)$, so $-T~t_n$.

2.  Recall that the Cauchy distribution is defined as the distribution of $X∕Y$ where X and Y are i.i.d. $\mathcal{N}\left(0,1\right)$. By definition, $T~t_1$ can be expressed as $T=Z∕\sqrt{V}$, where $\sqrt{V}=\sqrt{Z_1^2}=\left|Z_1\right|$ with $Z_1$ independent of Z. But by symmetry, $Z∕\left|Z_1\right|$ has the same distribution as $Z∕Z_1$, and $Z∕Z_1$ is Cauchy. Thus the $t_1$ and Cauchy distributions are the same.

3.  This follows from the SLLN. Consider a sequence of i.i.d. standard Normal r.v.s $Z_1,Z_2,\dots$, and let

$V_n=Z_1^2+\cdots +Z_n^2.$

By the SLLN, $V_n∕n\to E\left(Z_1^2\right)=1$ with probability 1. Now let $Z~\mathcal{N}\left(0,1\right)$ be independent of all the $Z_j$, and let

$T_n=\frac{Z}{\sqrt{V_n∕n}}$

for all n. Then $T_n~t_n$ by definition, and since the denominator converges to 1, we have ...