# Proving convergence in distribution of quotient of two random variables

probability theoryprobability-limit-theoremssolution-verification

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?

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.