5
$\begingroup$

It seems to me that the Lyapunov CLT condition holds for any sequence of independent Bernoulli random variables $X_1,X_2,\dots,X_n$ no matter how they are distributed. Restating the condition says that we can apply the CLT if there exists a $\delta>0$ such that $$ \lim_{n\rightarrow\infty}\frac{1}{s^{2+\delta}} \sum_{i=1}^n E[ |X_i - E[X_i]|^{2+\delta}] = 0, $$ where $s = \sqrt{\sum_{i=1}^n Var[X_i]}$. For every $i$ and probability $p_i = Pr[X_i=1]$, setting $\delta=1$ shows that $$ E[ |X_i - E[X_i]|^{3}] = p_i |1 - p_i|^{3} + (1 - p_i)|(-p_i)|^3 ≤ p_i(1 - p_i) = Var[X_i], $$ and hence the sum is upper bounded by the sum of the variances. Since in the denominator we have $s^{2+\delta} = (\sum_{i=1}^n Var[X_i])^{3/2}$, the limit goes to $0$. What am I missing?

$\endgroup$
0

1 Answer 1

0
$\begingroup$

It is not true that the Lyapunov CLT condition holds for any sequence of independent Bernoulli random variables $X_1,X_2,\ldots$, or even that any such sequence converges in distribution to a normal distribution. A simple counterexample will suffice: Let $X_1$ be a Bernoulli random variable with mean 1/2, and let all subsequent random variables in the sequence $X_2, X_3, \ldots$ be independent Bernoulli random variables with mean 0. $\sum_{i=1}^n X_i\sim\mathrm{Bernoulli}(1/2)$ for any $n\geq 1$, showing there is no convergence in distribution to a normal distribution. As expected, the Lyapunov CLT condition with $\delta = 1$ does not hold:

\begin{equation*} \lim_{n\rightarrow\infty} \frac{\sum_{i=1}^n \mathbb{E}[|X_i-\mathbb{E}[X_i]|^3]}{(\sum_{i=1}^n \mathrm{var}[X_i])^{3/2}} = 1 \end{equation*}

Generalizing a bit, it is clear that the Lyapunov CLT condition with any $\delta>0$ can never hold for a sequence of random variables $X_1, X_2, \ldots$ if $\lim_{n\rightarrow\infty}\sum_{i=1}^n \mathrm{var}[X_i]$ is positive and finite. In the case of independent Bernoulli random variables $X_1, X_2, \ldots$ with probabilities $p_1, p_2, \ldots$, this means that the Lyapunov CLT condition will not hold if $\lim_{n\rightarrow\infty}\sum_{i=1}^n p_i(1-p_i)$ is positive and finite. It is clear that this restriction includes cases like the one above with only finitely many non-degenerate Bernoulli random variables. However, it also includes some sequences where all elements are non-degenerate Bernoulli random variables. As an example, the sequence of (non-degenerate) Bernoulli random variables $X_i\sim\mathrm{Bernoulli}(1/(i+1)^2)$ has a finite sum of variances and fails the Lyapunov CLT condition with $\delta=1$:

\begin{align*} \lim_{n\rightarrow\infty} \sum_{i=1}^n \mathrm{var}[X_i] &\approx 0.563 \\ \lim_{n\rightarrow\infty} \frac{\sum_{i=1}^n \mathbb{E}[|X_i-\mathbb{E}[X_i]|^3]}{(\sum_{i=1}^n \mathrm{var}[X_i])^{3/2}} &\approx 1.088 \end{align*}

That being said, you are correct that the Lyapunov CLT condition with $\delta=1$ holds if $\lim_{n\rightarrow\infty}\sum_{i=1}^n p_i(1-p_i)=\infty$, using the exact logic you state in your question:

\begin{align*} \lim_{n\rightarrow\infty} \frac{\sum_{i=1}^n \mathbb{E}[|X_i-\mathbb{E}[X_i]|^3]}{(\sum_{i=1}^n \mathrm{var}[X_i])^{3/2}} &= \lim_{n\rightarrow\infty} \frac{\sum_{i=1}^n p_i(1-p_i)^3+(1-p_i)p_i^3}{(\sum_{i=1}^n p_i(1-p_i))^{3/2}} \\ &= \lim_{n\rightarrow\infty} \frac{\sum_{i=1}^n p_i(1-p_i)((1-p_i)^2+p_i^2)}{(\sum_{i=1}^n p_i(1-p_i))^{3/2}} \\ &\leq \lim_{n\rightarrow\infty} \frac{\sum_{i=1}^n p_i(1-p_i)}{(\sum_{i=1}^n p_i(1-p_i))^{3/2}} \\ &= \lim_{n\rightarrow\infty} \frac{1}{(\sum_{i=1}^n p_i(1-p_i))^{1/2}} \\ &= 0 \end{align*}

$\endgroup$
2
  • $\begingroup$ Even though for $\delta = 1$ it does not hold, the existence of $\delta$ is enough in Lyapunov CLT. $\endgroup$
    – kurtkim
    Commented Jun 1, 2018 at 8:46
  • $\begingroup$ @kurtkim I'm not sure I understand your comment. The two counterexamples I provide show that the OP's statement "the Lyapunov CLT condition holds for any sequence of independent Bernoulli random variables $X_1,X_2,\ldots,X_n$ no matter how they are distributed" is not correct. $\endgroup$
    – josliber
    Commented Jun 1, 2018 at 14:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .