Question. Let $X_{1}, X_{2}, \ldots$ be positive i.i.d. random variables with $\mathbf{E} X_{1}=a$ and $\operatorname{Var}\left(X_{1}\right)=\sigma^{2}$. Let $S_{n}=X_{1}+\cdots+X_{n}$. Show that $Y_n = \frac{\sqrt{n}(S_n – an)}{S_n}$ converges in distribution as $n \rightarrow \infty$.
My attempt:
Let $Z_n = \frac{S_n – an}{n^{1/2}}$ and $U_n = \frac{S_n}{n}$ so that $Y_n = \frac{Z_n}{U_n}$. Let $g(x,y) = x/y$ so that $Y_n = g(Z_n,U_n)$. If $(Z_n,U_n) \to_d (Z,a)$ then $g(Z_n,U_n) \to_d g(Z,a)$. But convergence in probability implies convergence in distribution and so it suffices to show that $(Z_n,U_n) \to_P (Z,a)$. Now, $Z_n \to N(0,1)$ by CLT and $U_n \to a$ by weak LLN, therefore
$$
\begin{aligned}
P(|Y_n – Y| > \varepsilon) &= P(|(Z_n – Z, U_n – a)| > \varepsilon) \\
&= P(\sqrt{|Z_n – Z|^2 + |U_n – A|^2} > \varepsilon) \\
\end{aligned}
$$
Let $A_{n,\varepsilon} := \{\sqrt{|Z_n – Z|^2 + |U_n – A|^2} > \varepsilon\}$, let $B_{n,\varepsilon} := \{|Z_n – Z| > \varepsilon\}$ and $C_{n,\varepsilon} := \{|U_n – a| > \varepsilon\}$. Then $A_{n,\varepsilon} \subseteq B_{n,\varepsilon} \cup C_{n,\varepsilon}$ and so
$$
P(A_{n,\varepsilon}) \leq P(B_{n,\varepsilon}) + P(C_{n,\varepsilon}) \to 0
$$
Therefore $P(A_{n,\varepsilon}) \to 0$ and so $g(Z_n, U_n) \to_P g(Z,a) = Z/a \sim N(0,\sigma^2/a^2)$.
Does this proof work? I tried to prove it directly but is it more appropriate to show that $Z_n + U_n \to_d Z + a$ by Slutsky's Theorem, so since the linear combination of the components of $(Z_n, U_n)$ converges in distribution, then the random vector itself must convergence in distribution?
Best Answer
I think your proof does not work because you need $P(|Z_n-Z|>\varepsilon)\to 0$, which means $Z_n\to_P Z$ and it's stronger than CLT. Maybe it's easier to compute the characteristic functions directly.