Prove $\frac{\sum_{j=1}^nX_j}{\sqrt{\sum_{j=1}^nX_j^2}}$ converges to $\mathcal{N}(0,1)$ in distribution Suppose $\{X_n\}$ iid, $E(X_1)=0$, and $\operatorname{Var}(X_1)=1$. Prove that $\frac{\sum_{j=1}^nX_j}{\sqrt{}\sum_{j=1}^nX_j^2}$ converges to $\mathcal{N}(0,1)$ in distribution.
Actually up to now I have no feasible thoughts because I think the main difficulty is that the random operator appears in the denominator, then the CLT can't be directly used. And I have no extra thoughts to prove convergence in distribution.

Actually, this is true as long as $\operatorname{Var}(X_1) < \infty$. We'll use a fairly standard stochastic asymptotic toolkit:

*

*Strong law of large numbers (SLLN), although the weak law of large numbers would suffice.

*Continuous mapping theorem (CMT)

*Standard central limit theorem (CLT)

*Slutsky's theorem

*Multiplying by $1 = \sqrt n/ \sqrt n = \sqrt{\operatorname{Var}(X_1)} / \sqrt{\operatorname{Var}(X_1)}$ where convenient

We'll apply these with an excessive use of underbraces:
$$
 \frac{\sum_{j=1}^n X_j}{ \sqrt{\sum_{j=1}^n X_j^2 } } = \underbrace{ \underbrace{ \underbrace{ \bigg( \underbrace{\frac{1}{n} \sum_{j=1}^n X_j^2}_{\overset{\text{a.s.}}{\to} EX_1^2 = \text{Var}(X_1) \;\text{ by SLLN}} \bigg)^{-1/2} }_{\overset{\text{a.s.}}{\to} [\operatorname{Var}(X_1)]^{-1/2} \; \text{by CMT}}\cdot \sqrt{\operatorname{Var}(X_1)} }_{\overset{\text{a.s.}}{\to} 1 \; \text{by CMT}}\cdot \underbrace{ \frac{1}{\sqrt{n \cdot \operatorname{Var}(X_1)}} \sum_{j=1}^n X_j}_{\overset{\mathcal D}{\to} \mathcal N(0,1) \; \text{by CLT}} }_{\overset{\mathcal D}{\to} \mathcal N(0,1) \; \text{by Slutsky's theorem}} \overset{\mathcal D}{\longrightarrow} \mathcal N(0,1).
$$
