Find a sequence of random variables $(X_n)$ with $\lim E(X_n^2) = 0$ but not obeying SLLN

233 Views Asked by At

I am looking for some sequence of random variables $(X_n)$ such that

$$ \lim_{n \rightarrow \infty} E(X_n^2) = 0 $$

but such that the following almost sure convergence does NOT hold:

$$ \frac{S_n - E(S_n)}{n} \rightarrow 0$$

where the $S_n$ are the partial sums of the $X_n$.

Note: for any such sequence the convergence in probability will always hold; if the random variables are not correlated, so will the convergence almost surely. In particular, any counterexample must consist of correlated random variables.

Many thanks for your help.

1

There are 1 best solutions below

0
On

Here is an algorithm which gives you such a sequence. Let us work on the probabilised space $[0,1)$ with the Lebesgue measure.

For all $0 \leq k < n$, let $I_{k,n} := [k/n, (k+1)/n)$. Fix $\varepsilon \in (0,1)$

Start from $n = 1$, $k=0$, time $N=0$.

If $S_N < \varepsilon N$ on $I_{k,n}$, take $X_N = 1_{I_{k,n}}$.

Else :

  • if $k < n-1$ : increment $k$ by $1$.

  • if $k = n-1$ : increment $n$ by $1$, put $k=0$.

Rince and repeat, incrementing $N$ by $1$.

Now, for all $k,n$, we only need a finite time before $S_N \geq \varepsilon N$ on $I_{k,n}$ (the times at which these conditions are satisfied successively grow exponentially, though). Hence we will eventually increment $k$, and then $n$. Since any point in $[0,1)$ is in infinitely many $I_{k,n}$, that means that almost surely, $S_N \geq \varepsilon N$ for infinitely many $N$.

On the other hand, $\mathbb{E} (X_N) = \mathbb{E} (X_N^2)$ will converge to $0$. Hence, $\mathbb{E} (S_N)$ grows sub-linearly, so that almost surely, $S_N - \mathbb{E} (S_N) \geq \varepsilon N/2$ for infinitely many $N$s.