This problem is practice for my qualifying exam and comes from Resnick, chapter 9. Could anyone comment on my solution(s)?

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


The condition $$ \max_{i\le n}\frac{\mu_i}{\sum^n_{j=1}\mu_j}\to 0 $$ means that for $\epsilon>0$, $\exists N\in\mathbb{N}$, such that $$ \max_{i\le n}\frac{\mu_i}{\sum^n_{j=1}\mu_j}<\epsilon $$ whenever $n>N$.

Then $\forall n>N$ $$ \max_{i\le n}\mu_i <\epsilon\sum_{j\le n}\mu_j\le\epsilon\sqrt{n}\left(\sum\mu_k^2\right)^{1/2}=\epsilon\sum_{j\le n}\mu_j\le\epsilon\sqrt{n}s_n $$ where the second inequality is due to Cauchy-Schwarz and $s_n=\sum_{j\le n}\mu_j^2$

Again, I check the Liapunov condition for $\delta=1$. So $\forall n>N$, $$ \frac{\sum_{k\le n}E|e_k|^3}{s_n^3}=6\sum \left(\frac{\mu_k}{s_n}\right)^3\\\le 6\sum \left(\frac{\max \mu_k}{s_n}\right)^3\\\le 6\sum \left(\frac{\epsilon\sqrt{n}s_n}{s_n}\right)^3\\=6\epsilon^3n^{5/2} $$

Since $\epsilon$ was arbitrary, we let $\epsilon^3\to0$ faster than $n^{5/2}\to\infty$, and we have the Liapunov condition, and therefore CLT.

Any help is appreciated. Thanks!

  • $\begingroup$ The Lyapounov's condition doesn't contain $\varepsilon$. $\endgroup$
    – Zhanxiong
    Commented Jul 16, 2015 at 2:56
  • $\begingroup$ @Zhanxiong what do you mean? I use $\epsilon$ in the bound. $\endgroup$ Commented Jul 16, 2015 at 2:58
  • 1
    $\begingroup$ @stats134711 If $\varepsilon$ is fixed in advance, you cannot argue like that "let $\varepsilon^3$ goes faster than...". Also, to check the Lyapounov condition, you need to compute moments for the centered random variables, not the original $e_i$. $\endgroup$
    – Zhanxiong
    Commented Jul 16, 2015 at 3:00
  • $\begingroup$ @Zhanxiong I see. Yes, that's where most of my concern is regarding which parts go to zero/infinity. Do you have any suggestions? $\endgroup$ Commented Jul 16, 2015 at 3:01
  • $\begingroup$ I guess checking the Lindeberg condition is the right track. If I got time, I can give some details. $\endgroup$
    – Zhanxiong
    Commented Jul 16, 2015 at 3:02

1 Answer 1


Set $$X_i = \frac{e_i - \mu_i}{\mu_i}, \quad i = 1, 2, \ldots, n$$ It is then easily seen that $X_1, \ldots, X_n$ are i.i.d., and we want to find the asymptotic distribution of $$\sum_{i = 1}^n \frac{\mu_i X_i}{\sqrt{\sum_{j = 1}^n \mu_j^2}} := \sum_{i = 1}^n Y_i.\tag{1}$$

We shall check the Lindeberg condition for sum $(1)$. Note $E(Y_i) = 0, i = 1, 2, \ldots, n$, and $s_n^2 = \sum_{i = 1}^n E(Y_i^2) = 1$. Given $\varepsilon > 0$, denote $\sqrt{\sum_{j = 1}^n \mu_j^2}$ by $A_n$, then \begin{align*} & \frac{1}{s_n^2} \sum_{i = 1}^n E(Y_i^2\text{I}(|Y_i| > \varepsilon s_n)) = \sum_{i = 1}^n E\left(\frac{\mu_i^2 X_i^2}{\sum_{j = 1}^n \mu_j^2} \text{I}\left(\left|\mu_iX_i\right| > \varepsilon A_n\right)\right) \\ \leq & \frac{1}{\sum_{j = 1}^n \mu_j^2}\sum_{i = 1}^n \mu_i^2 E(X_i^2\text{I}(\left|X_i\right| > \varepsilon A_n/\max_{1 \leq j \leq n}\mu_j)) \\ = & E(X_1^2\text{I}(\left|X_1\right| > \varepsilon A_n/\max_{1 \leq j \leq n}\mu_j)) \end{align*}

To here, it's time to use the condition. However, there is a little issue. The Lindeberg condition can be checked if $$\frac{\max_{1 \leq i \leq n}\mu_i}{\sqrt{\sum_{j = 1}^n \mu_j^2}} \to 0.$$ With the condition you provided, some difficulties seem exist (could you double check the condition or look for another way?).

  • $\begingroup$ Thanks for the details! I double checked the condition and it is the same as in the book. I will look for an alternative. $\endgroup$ Commented Jul 16, 2015 at 14:47
  • $\begingroup$ @stats134711 I see. I am looking forward to seeing the solution. This problem is harder than it looks to be. $\endgroup$
    – Zhanxiong
    Commented Jul 16, 2015 at 15:35
  • $\begingroup$ $ \lim\limits_{n\to\infty}\frac{\max\limits_{1\le i\le n}\mu_i}{\sum\limits_{j=1}^n \mu_j}=0 $ is not sufficient for the following CLT: $ \frac{\sum_{i=1}^n (X_i-\mu_i)}{\sqrt{\sum_{i=1}^{n}\mu_j^2}} \Rightarrow N(0,1). $ For example, $\mu_n=\frac1n$. $\endgroup$
    – JGWang
    Commented Jan 2, 2021 at 9:25

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