Let $X_1, X_2, \dots$ be uncorrelated random variables with $\mathbb{E}[X_i]= \mu_i$ and $\displaystyle\frac{Var[X_i]}{i} \rightarrow 0$, when $i \rightarrow +\infty$. Show that $\displaystyle\frac{S_n}{n} - \frac{\mathbb{E}[S_n]}{n} \rightarrow 0$, as $n \rightarrow +\infty$ in $L^2$ and probability.

As convergence in $L^2$ implies convergence in probability, we only need to show convergence in $L^2$. So, we have to show that $\displaystyle\mathbb{E}\left[\left|\frac{S_n}{n} - \frac{\mathbb{E}[S_n]}{n}\right|^2\right] \rightarrow 0$,as $n \rightarrow +\infty$.

As $\displaystyle\frac{Var[X_i]}{i}$, given $\epsilon > 0$, there is $n_0$ such that for $n>n_0$, $\displaystyle\frac{Var[X_i]}{i} < \epsilon$.

Let $C = \displaystyle\max\left\{\frac{Var[X_1]}{1}, \dots, \frac{Var[X_{n_0}]}{n_0}, \epsilon\right\}$. Then, $\displaystyle\frac{Var[X_i]}{i} < C$, for all $i$.

$\displaystyle\mathbb{E}\left[\left|\frac{S_n}{n} - \frac{\mathbb{E}[S_n]}{n}\right|^2\right] = \frac{1}{n^2} \mathbb{E}[|S_n - \mathbb{E}[S_n]|^2] = \frac{1}{n^2}Var[S_n] = \frac{1}{n^2}\sum_{k=1}^n Var[X_k]$

Above, we used that $X_1, X_2, \dots$ are uncorrelated.


$\displaystyle\frac{1}{n^2}\sum_{k=1}^n Var[X_k] = \frac{1}{n}\sum_{k=1}^n \frac{Var[X_k]}{n} \leq \frac{1}{n}\sum_{k=1}^{n} \frac{Var[X_k]}{k} \leq \frac{1}{n} \sum_{k=1}^{n} C$ = C

And it gives me nothing... Perhaps I need to find another upper bound!

Can someone help me?


You simply threw away the fact that $\sigma_j^2:=Var(X_j)/j\to0$ and used only the fact that it's bounded by $C$. Do everything just the way you did til the very end. Then say $$\frac1n\sum_{j=1}^n\sigma_j^2\le\frac1n\left(\sum_{j=1}^{n_0}C+\sum_{j=n_0+1}^n\epsilon\right)\le\frac{C n_0}{n}+\epsilon.$$Now letting $n\to\infty$ shows that $$\limsup_{n\to\infty}\frac1n\sum_{j=1}^nv_j\le\epsilon.$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.