# Convergence a.s of the sequence $\frac{X_n}{S_n}$

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!

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

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