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$\left\{X_{n}\right\}_{n \geq 1}$ is a family of independent random variables such that $$ \mathbb{P}\left(X_{n} = 1\right) = \frac{n^2-1}{n^2+1}\quad \mbox{and}\quad \mathbb{P}\left(X_{n} = -n^{2}\right) = \frac{2}{n^{2} + 1} $$ $$ \mbox{Let}\ S_{n} = X_{1} + \cdots + X_{n} $$

  • The Three-Series Theorem tells me that $S_{n}$ does not converge almost surely.
  • But what about convergence in distribution or probability ?.

Intuitively I want to say it doesn't, but I don't know how to go about proving it.

Two things I've observed:

  • $X_{n}$ converges to $X \equiv 1$ in distribution ( and probability ).
  • $\mathbb{E}X_{n} = -1,\ \forall\ n;\quad$ hence $\quad\mathbb{E}S_{n} = -n \to -\infty$.

Can these facts be used to discuss convergence in distribution ?.

Felix Marin
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AlgebraBreaksMyBones
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1 Answers1

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By Borel-Cantelli Lemma, only finitely many of the events $A_n:=\{X_n=-n^2\}$ will occur (as $\sum_n \mathbb P(A_n)<\infty$). Hence for all large enough $n$ we have $X_n=1$. Therefore, $S_n/n\to 1$ a.s.

van der Wolf
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