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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!

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    $\begingroup$ To help other community members weigh in, it may be helpful to write up some of the approach. For instance, have you obtained the third and fourth moments for a single Exponential $\mu$ variable as well as for their sum? $\endgroup$ – AdamO Jul 20 '15 at 17:32
  • $\begingroup$ @AdamO, I have not. Would that be helpful since the Liapunov and Lindeberg conditions only deal up to second moments? $\endgroup$ – stats134711 Jul 20 '15 at 22:15
  • $\begingroup$ Lindeberg's condition implies that $\max_{1\leqslant j\leqslant n}\lambda_j^2\cdot\left(\sum_{l=1}^n\lambda_l^2 \right)^{-1/2} $ converges to $0$ as $n$ goes to infinity, which is not necessarily satisfied under the assumptions (for example if $\lambda_j=1/j$). $\endgroup$ – Davide Giraudo Jul 21 '15 at 10:16
  • $\begingroup$ (of course, $\lambda_j$ should be $\mu_j$) $\endgroup$ – Davide Giraudo Jul 21 '15 at 11:25
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    $\begingroup$ @AdamO Unfortunately, neither the Lindeberg nor the Liapounov conditions are implied by the assumptions of this problem. That suggests the statement may be incorrect and would inspire one to look for counterexamples. Such counterexamples must violate the Lindeberg conditions (by occasionally exhibiting a $\mu_n^2$ that is large relative to the sum of squares of all previous ones but for which $\mu_n$ is small compared to the sum of all previous ones). $\endgroup$ – whuber Jul 23 '15 at 12:09
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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).

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  • $\begingroup$ Giraudo (+1) Strong argument, swept away my confusions. $\endgroup$ – Zhanxiong Jul 23 '15 at 7:43

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