Let $(X_i)_{i = 1}^{\infty}$ be a sequence of independent random variables (but not identically distributed) with $0\leq X_i\leq 1$ for all $i\ge1.$ Let $S_{n} = X_{1}+\dots +X_{n}$. Prove that $\frac{S_n}{\mathbb{E}[S_n]} \to 1\:\:\text{a.s.}$
Thoughts: I originally wanted to prove that $\mathbb{P}(|S_n-\mathbb{E}[S_n]|>\delta\:\:\text{i.o.}) = 0$. One would then need $\sum_{n} \mathbb{P}(|S_n-\mathbb{E}[S_n]|>\delta) < \infty$. However, I am unable to obtain an upper bound on the sum since the random variables are not identically distributed. Any hints?
Best Answer
Here is one way of solving this problem (there may be easier methods out there). We have $0\leq X_{i}^{2}\leq X_{i}\leq 1$ for each $i\in\mathbb{N}\setminus\{0\}$. Therefore, $$\text{Var}(S_{n}) = \sum_{i = 1}^{n}\text{Var}(X_{i})\leq \sum_{i = 1}^{n}\mathbb{E}[X_{i}^{2}]\leq\sum_{i = 1}^{n}\mathbb{E}[X_{i}] = \mathbb{E}[S_{n}]$$ The idea is to show that a subsequence of the original sequence converges $\text{a.s.}$ which we shall show is sufficient. Let $$n_{k} = \min\{n:\mathbb{E}[S_{n}]\geq k^{2}\}\:\:\:\:k = 1, 2, \dots$$ Since $\mathbb{E}[X_{i}]\leq 1$ for each $i\in\mathbb{N}\setminus\{0\}$ we have $k^{2}\leq \mathbb{E}[S_{n_{k}}]\leq k^{2}+1$. Let $\varepsilon > 0$ then by the Chebyshev inequality $$\mathbb{P}(|S_{n_{k}}-\mathbb{E}[S_{n_{k}}]| > \varepsilon\mathbb{E}[S_{n_{k}}])\leq\frac{\text{Var}(S_{n_{k}})}{\varepsilon^{2}\mathbb{E}[S_{n_{k}}]^{2}}\leq \frac{1}{\varepsilon^{2}\mathbb{E}[S_{n_{k}}]}\leq \frac{1}{\varepsilon^{2}k^{2}}$$ Therefore, $$\sum_{k = 1}^{\infty}\mathbb{P}\left(\left|\frac{S_{n_{k}}}{\mathbb{E}[S_{n_{k}}]}-1\right|> \varepsilon\right) < \infty$$ Using the First Borel Cantelli Lemma $$\mathbb{P}\left(\left|\frac{S_{n_{k}}}{\mathbb{E}[S_{n_{k}}]}-1\right|> \varepsilon\:\:\text{i.o.}\right) = 0\implies \frac{S_{n_{k}}}{\mathbb{E}[S_{n_{k}}]}\rightarrow 1\:\:\text{a.s.}$$ We now show that this is sufficient. Let $n_{k}\leq n\leq n_{k+1}$ then $$\frac{S_{n}}{\mathbb{E}[S_{n}]}\leq \frac{S_{n_{k+1}}}{\mathbb{E}[S_{n_{k}}]}\cdot\frac{{\mathbb{E}[S_{n_{k+1}}]}}{{\mathbb{E}[S_{n_{k+1}}]}}\leq \frac{S_{n_{k+1}}}{\mathbb{E}[S_{n_{k+1}}]}\cdot \frac{(k+1)^2+1}{k^2}$$ In a similar manner we obtain a lower bound $$\frac{S_{n}}{\mathbb{E}[S_{n}]}\geq \frac{S_{n_{k}}}{\mathbb{E}[S_{n_{k}}]}\cdot \frac{k^2}{(k+1)^2+1}$$ The result then follows by taking the $\limsup_{k\rightarrow\infty}$ and $\liminf_{k\rightarrow\infty}$ of the previous inequalities.