Show that $(S_n)_{n\in\mathbb{N}}$ is tight

Question

Let $(X_n)_{n\in\mathbb{N}}$ be independent Bernoulli random variables such that $$\mathbf{P}(X_{n}=1)=1-\mathbf{P}(X_{n}=0)=1/n^2.$$ Set $S_{n}:=X_1+\cdots+X_n.$ Show that $(S_n)_{n\in\mathbb{N}}$ is tight.

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For each $n$,$X_{k}$ $(1\le k\le n)$ be independent random variables with characteristic function $\phi_{X_k}(t)=(1-\frac{1}{k^2})+\frac{1}{k^2}e^{it} (1\le k\le n).$ Then $$\phi_{S_{n}}(t)=E\exp(itS_{n})=\prod_{k=1}^{n}(1+\frac{1}{k^2}(e^{it}-1))$$ By Lévy’s continuity theorem, If $\phi_{S_{n}}(t)$ converges pointwise to a limit $\phi_{S_{\infty}}(t)$ that is continuous at $0$, then the asssociated sequence of distributions $S_{n}$ is tight. But how can I find the $\phi_{S_{\infty}}(t):=\displaystyle \lim_{ n\to \infty}\phi_{S_{n}}(t), t\in \mathbb{R}$,which is continuous at $0$?

Answer 1

Tightness can be obtained by observing that $$ \mathbb P\left(\lvert S_n\rvert>R\right)\leqslant \frac 1R\mathbb E\left[\lvert S_n\rvert\right]=\frac 1R\mathbb E\left[ S_n \right]=\frac 1R\sum_{k=1}^n\frac 1{k^2}\leqslant \frac 1R\sum_{k=1}^\infty\frac 1{k^2} $$ hence for a positive $\varepsilon$, we choose $R$ such that $\frac 1R\sum_{k=1}^\infty\frac 1{k^2}<\varepsilon$.

Your argument goes in the right direction and will give convergence in distribution. The product is of the form $\prod_{k=1}^n\left(1+p_k(t)\right)$ with $\sum_{k=1}^\infty\sup_{-1/2\leqslant t\leqslant 1/2}\left\lvert p_k(t)\right\rvert<\infty$ hence the convergence holds uniformly on $[-1/2,1.2]$ which is enough for continuity at $0$. See for instance Lemma~7.5 in these lecture notes by Jens Lorenz. For the convergence for each $t$, we use $\sum_{k=1}^\infty \left\lvert p_k(t)\right\rvert<\infty$.

Answer 2

By the second Borel-Cantelli Theorem and since $\sum_{k=1}^\infty P(X_k =1)=\sum_{k=1}^\infty 1/k^2 < \infty$, $P(X_k = 1, \mbox{ infinitely often })=0$. Hence, the compact set $K_M = [-M, M ]$ satisfies

$$ P(S_n \in K_M^c) \le P(X_k = 1, \mbox{ for some $k\ge M$}) \to 0, \; as \; M\to \infty. $$ Hence for each $\epsilon>0$, there must exist a $M_\epsilon$ so that $P(S_n \in K_{M_\epsilon}) \ge 1- \epsilon$ for all $n\in \mathbb{N}$. This shows that $S_n$ is a uniformly tight sequence.