Lindeberg CLT for exponential independent variables

Question

Crossposted in math.stackexchange: CLT for independent, but non-identically distributed exponential variables

This problem is self-study for my qualifying exam.

Problem

Suppose $(e_n)_{n\ge 1}$ are independent exponentially distributed random variables with $E(e_n)=\mu_n$. If $$ \max_{i\le n}\frac{\mu_i}{\sum^n_{j=1}\mu_j}\to 0 $$

then $$ \sum^n_{i=1}(e_i-\mu_i)/\sqrt{\sum^n_{j=1}\mu_j^2}\implies N(0,1). $$

I've attempted a solution using the Liapunov condition, but somehow get stuck at the last step in my justification.

In the link above, another user attempted an answer using the Lindeberg condition but somehow the conditions given in the problem do not conform to the solution's assumptions.

Does anyone have any hints on how to proceed?

Thank you!

Answer 1

When $\mu_j=1/j$ (which satisfies the assumptions of the book), it seems that the sequence $\left(\sum^n_{i=1}(e_i-\mu_i)/\sqrt{\sum^n_{j=1}\mu_j^2}\right)_{n\geqslant 1}$ converges to a constant plus a Gumbel distribution (see Subsection 5.3 in A uniform asymptotic expansion for weighted sums of exponentials by J.S.H. van Leeuwaarden and N.M. Temme.

Therefore, the condition $$\tag{C} \lim_{n\to \infty}\max_{1\leqslant i\leqslant n}\frac{\mu_i^ 2}{\sum_{j=1}^n \mu_j^2 } =0$$ seems to be the good one. It can be either derived by the method described in the link of the question, or in a more general way here (in particular, it seems that we do not need to assume the random variables have exponential distribution, but only a finite variance). In both cases, Lindeberg's central limit theorem is used.

In general, if we want to prove a central limit theorem for $s_n^{-1}\sum_{j=1}^nX_{n,j}$, where $(X_{n,j})_{j=1}^n$ are independent and centered, and $s_n^2=\sum_{j=1}^n\operatorname{Var}(X_{n,j})$, we can use Lindeberg's condition, namely, $$\forall \varepsilon\gt 0, \quad \lim_{n\to \infty} \frac 1{s_n^2}\sum_{j=1}^n\mathbb E\left[X_{n,j}^2\mathbf 1\{|X_{n,j}|\gt \varepsilon s_n \} \right] =0.$$
This implies that $s_n^{-1}\max_{1\leqslant j\leqslant n}\operatorname{Var}(X_{n,j})\to 0$. In our case, this is equivalent to condition (C).