This exercise appears in K.L.Chung's A Course in Probability Theory, Chapter 7.
Ex.7.1-4
Let ${X_j}$ be independent r.v.'s such that $\max_{1\leqslant j\leqslant n} \frac{|X_j|}{b_n} \to 0$ in pr. and $(S_n - a_n)/b_n$ converges to a nondegenerate d.f. Then $b_n \to \infty$, $\frac{b_{n+1}}{b_n} \to 1$, and $\frac{a_{n+1} - a_n}{b_n} \to 0$.
I found it difficult, and I do not have any idea why this is put in the exercise of CLT. Anyone helps me solve this? Thanks.
The fact that $b_n\to \infty$ is quite easy to check: if not, there is a $M$ and a subsequence $(b_{n'})$ which remains below $M$, hence $\frac 1{b_{n'}}\max_{1\leqslant j\leqslant n'}|X_j|\geqslant \frac 1{M}\max_{1\leqslant j\leqslant n'}|X_j|$, hence $\max_{1\leqslant j\leqslant n'}|X_j|$ would converge to $0$ in probability, which is not possible.
Let us denote the convergence in distribution by $\Rightarrow$.
Theorem 7.6 in Durrett's book Probability: theory and examples, (second edition), provides an useful theorem here:
The convergence of types theorem. Assume that $Y_n\Rightarrow Y$ and there are constants $\alpha_n>0$, $\beta_n$ such that $Y'_n=\alpha_nY_n+\beta_n\Rightarrow Y'$, where $Y$ and $Y'$ are not degenerate. Then there are $\alpha>0$ and $\beta\in\Bbb R$ such that $\alpha_n\to \alpha$ and $\beta_n\to \beta$.
The proof uses the fact that in case of convergence in distribution, the sequence of corresponding characteristic functions actually converges uniformly on compact sets. Then we proves that for $\alpha$, there is an unique positive real number which can be a candidate, and the same for $\beta_n$.
Here, we use this theorem with $W_n:=\frac{S_n-a_n}{b_n}$, $\alpha_n:=\frac{b_n}{b_{n+1}}$ and $\beta_n:=\frac{a_n-a_{n+1}}{b_{n+1}}$. It works since $\frac{X_{n+1}}{b_{n+1}}$ goes to $0$ in probability.
Finally, we have to use independence in order to identify the obtained limits. We compute the characteristic function of $W_n$, and of $W_{n+1}$, and we take the limits.
