Let $\{X_n\}_n$ be a sequence of equally distributed independent random variables with $\mathbb{E}[X_1]<\infty $ and $S_n=\sum_{i=1}^{n}X_i$. Show that if $\mathbb{E}[X_1]\neq0$ then $$\frac{X_n}{S_n} \rightarrow 0 \hspace{.5cm}a.s$$

I have tried to use the Borel-Cantelli lemma together with Chebyshov's inequality applying everything to the succession $Y_n=\frac{X_n}{S_n}$ but I've had problems to calculate the expectation of $Y_n$, I don't know if this technique is adequate, any advice or help to solve the problem would be very grateful. Happy 2022!

  • 2
    $\begingroup$ Does a.c. means almost surely convergence (i.e. probability of convergence is $1$)? $\endgroup$
    – mihaild
    Jan 1, 2022 at 0:44
  • $\begingroup$ I have corrected, it was a.s. convergence. $\endgroup$
    – Nick Weber
    Jan 1, 2022 at 0:59
  • $\begingroup$ It seems you are assuming the $X_n$ to be mutually independent. $\endgroup$ Jan 1, 2022 at 18:17

1 Answer 1


From the discussion in this question we see that $\mathbb E|X_1|<\infty\implies\sum_{n\geq1}\mathbb P(|X_n/n|>\varepsilon)<\infty$, for any $\varepsilon>0$.

So by Borel-Cantelli we deduce that $\mathbb P(|X_n/n|>\varepsilon\text{ i.o.})=0$, and hence $X_n/n\to0$ a.s.

By SLLN we know that $\frac{S_n}{n}\to\mathbb EX_1\neq0$ a.s. and so $\frac{n}{S_n}\to(\mathbb EX_1)^{-1}$ a.s. So combining the two results we deduce that $\frac{X_n}{S_n}=\frac{X_n}{n}\cdot\frac{n}{S_n}\to0$ a.s.

  • $\begingroup$ It seems to me that everything works fine as long as $Var[X_1]<\infty$ otherwise I would not be able to occupy SLLN. $\endgroup$
    – Nick Weber
    Jan 1, 2022 at 7:18
  • $\begingroup$ @Nick Weber: the SLLN does not require Var$[X_1]<\infty$, only $\mathbb{E}[|X_1|]<\infty$. $\endgroup$
    – bdx77
    Jan 1, 2022 at 9:36

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