Expected value of absolute value of sum of independent variables (Convergence in probability)

53 Views Asked by At

Given $X_1, X_2,..., X_n$ independent random variables with $P(X_n=k^n)=P(X_n=-k^n)=1/2$ (assuming $k$ is an arbitrary constant). Let $S_n = X_1 + X_2 +... + X_n$. Determine for which $\epsilon>0$, if any, that the $P(|S_n|/n>\epsilon)$ does not converge to $0$.

I was struggling for this question. I suppose we need to find the range of $\epsilon$ so that $P(|S_n|/n>\epsilon)$ does not converge to $0$. Using $n=2$, I manage to find $E(|S_n|)$ and think it does not converge, but not sure how to find a general term for $E(|S_n|)$ afterall... as well as determine a range for $\epsilon$.

1

There are 1 best solutions below

0
On

If $0 \leq k \leq 1 $, Chebyshev's (or Hoeffding's which is even simpler here) inequality tells you that $P(|S_n|/n > \epsilon) \rightarrow 0$ for any $\epsilon > 0$.

If $k > 1$, we may write $\sum_{i=0}^{N - 1} k^i = \frac{k^N - 1}{k-1}$. Let us note $N_0 = \lceil 1/(k-1) \rceil$. So, under the event $(X_n = k^n, X_{n-1} = k^{n-1}, \dots, X_{n-(N_0 +1)} = k^{n-(N_0 +1)})$, we have $|S_n| \geq k^{n}$. This event has a non null probability that only depends on $N_0$ and not $n$. So $P(|S_n|/n>\epsilon)$ does not converge to $0$ for any $\epsilon > 0$.