Discussion
Let \(\{X_{n}\}\) be a sequence of independent and uniformly bounded random variables, and \(E[X_{n}] = 0\), and \(\mathrm{Var}(X_{n}) = \dfrac{1}{(2n)}\) for each \(n\).
Let \(S_{n} = X_{1} + \cdots + X_{n}\) and \(\sigma_{n}^{2} = \mathrm{Var}(S_{n})\)
Then which of the following is not true ?
Answer: A
The statement that is not true is \(S_{n}/\sigma_{n}\) does not converge in distribution.
Here \(\sigma_{n}^{2} = \sum_{k \le n} 1/(2k) \to \infty\), since the harmonic series diverges. For independent, uniformly bounded variables (\(|X_{k}| \le M\)) with \(\sigma_{n} \to \infty\), Lyapunov's condition holds:
\[ \begin{aligned} &\frac{1}{\sigma_{n}^{3}}\sum_{k=1}^{n} E|X_{k}|^{3} \le \frac{M\sigma_{n}^{2}}{\sigma_{n}^{3}} \\ &= \frac{M}{\sigma_{n}} \to 0 \end{aligned} \]
So (C) is true, and the central limit theorem gives \(S_{n}/\sigma_{n} \to N(0, 1)\) in distribution (B). Then \(S_{n}/\sigma_{n}^{2} = (S_{n}/\sigma_{n})/\sigma_{n} \to 0\) in probability (D). Only (A) contradicts these.
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