Let $B_{k}$, $k \geq 1$ be independent Bernoulli random variables such that $P(B_k = 1) = \frac{1}{k} = 1 – P(B_k = 0)$. Note that these are also known as Poisson-Bernoulli random variables.
Now, I need to show that the random variables $X_{n,k} = \frac{B_k – \frac{1}{k}}{\log(n)}$ converge in distribution to the standard normal (i.e. that they obey the central limit theorem).
Since these are independent, but not identical random variables, I am trying to show convergence by showing that $X_{n,k}$ satisfy the Lyapunov Condition, which is as follows:
$$ \lim_{n \rightarrow \infty} \frac{1}{s_n^{(\delta+2)}} \sum_{k = 1}^{n} E[|X_{n,k}|^{\delta + 2}] = 0\,,$$
where $${s_n}^{2} = \sum_{k=1}^{n} \sigma_{n,k}^{2} = \sum_{k=1}^{n} E[X_{n,k}^{2}]$$
When I try to evaluate the condition for $\delta = 1$, the limit converges to 1, not 0! I know that $X_{n,k}$ do in fact converge in distribution to standard normal. So either there is something wrong with my derivation, assumptions or understanding. Any help would be appreciated.
Update — The following seems to work but I am uncertain if this is correct.
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Derived $$s_{n}^{3} = \left(\sqrt{\frac{\sum_{k=1}^{n}(\frac{1}{k})(1 – \frac{1}{k})}{\log(n)}}\right)^{3}$$
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Derived $$E[|X_{n,k}|^{3}] = \frac{\left(\frac{1}{k}-\frac{3}{k^{2}}+\frac{4}{k^{3}} – \frac{2}{k^{4}}\right)}{\left(\sqrt{\log(n)}\right)^{3}}$$
(I am fairly certain about the above two derivations. It's the part below I am unsure about.)
Now, substitute derivations in the Lyapunov Condition and simplify to obtain:
$$ \lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{3}{k^{2}}+\frac{4}{k^{3}} – \frac{2}{k^{4}}\right)}{\left(\sqrt{\sum_{k=1}^{n}(\frac{1}{k})(1 – \frac{1}{k})}\right)^{3}}$$
Observe that in the numerator, all the series are convergent, except for $\sum_{k=1}^{n} \frac{1}{k}$, which is the harmonic/divergent series. However, we know that this series asymptotically behaves like $\log(n)$ (i.e. $\frac{\sum_{k=1}^{n} \frac{1}{k}}{\log(n)} \rightarrow 1$ as
$n\rightarrow \infty$).
Similarly, the summation in the denominator is decomposed in two series, one convergent and the other divergent (harmonic) series. So for large n, the term in the limit behaves like:
$$\frac{\log(n)}{(\sqrt{\log(n)})^{3}} = \frac{1}{\sqrt{\log(n)}}\,,$$
which goes to $0$ as $n \rightarrow \infty$.
—-End of Update—-
- I am unsure about the last part where I say, "the term behaves like …". It feels like a lot of hand-waving there.
Best Answer
Actually it seems that there is no hand-waving: your argument show that $ \sum_{k=1}^{n}\left(\frac{1}{k}-\frac{3}{k^{2}}+\frac{4}{k^{3}} - \frac{2}{k^{4}}\right)$ is equivalent to $\log n$ and that $\left(\sqrt{\sum_{k=1}^{n}(\frac{1}{k})(1 - \frac{1}{k})}\right)^{3}$ is equivalent to $(\log n)^{3/2}$, using the fact that if $\sum_{k\geqslant 1}a_k$ diverges with $a_k>0$, $R_n^{-1}\sum_{k=1}^na_k\to 1$, then for all sequence $(b_k)$ such that $\sum_{k\geqslant 1}\lvert b_k\rvert$ converges, $R_n^{-1}\sum_{k=1}^n(a_k+b_k)\to 1$.