1
$\begingroup$

Let $X_1,X_2,...$ be a sequence of random variables such that $P(X_n = \pm 2^n) = \frac{1}{2^{2n+1}}$ and $P(X_n = 0) = 1 - \frac{1}{2^{2n}}$. I'm supposed to show that this sequence satisfies WLLN, but isn't $Var(X_n) = E(X_n^2) = \sum_{n} \frac{2^{2n}}{2^{2n+1}} = \sum_{n} \frac{1}{2} = \infty$, so the WLLN can't be applied?

I already verified $E(X_n)=0$.

$\endgroup$
  • 2
    $\begingroup$ To check whether the WLLN holds, just check the definition of it and it probably is clear that it works (per definition of $X_n$, essentially since $2^{-2n}\to 0$). As I recall, the assumption of finite variance (in or outside of the limit) is not necessary for the WLLN to apply. $\endgroup$ – Stan Tendijck Jun 20 '18 at 23:11
  • $\begingroup$ that's odd, in my book one of the hypothesis is "variance uniformly limited" $\endgroup$ – creepyrodent Jun 21 '18 at 0:19
  • $\begingroup$ Two things: (1) the moment conditions for WLLNs generally depend on the strength of other assumptions such as stationarity and independence and (2) when you say WLLN, do you mean $X_n \to^p 0$ or $(X_1 + ... + X_n)/n \to^p 0$? $\endgroup$ – Galton Jun 21 '18 at 2:18
  • $\begingroup$ (1) and in this case there is stationarity or independence? I'm failing to see... (2) The latter $\endgroup$ – creepyrodent Jun 21 '18 at 17:20
  • $\begingroup$ (1) there is no stationarity since the variance changes but there is independence (since there is nothing stated that should suggest otherwise) $\endgroup$ – Stan Tendijck Jun 23 '18 at 9:30
2
$\begingroup$

$P\{X_n \neq 0\}=\frac 1 {2^{2n}}$. Hence $\sum_n P\{X_n \neq 0\}<\infty$. This implies that the $\limsup $ of the events $\{X_n \neq 0\}$ (which is set of points that belong to infinitely many of these events) has probability $0$. Hence $X_n \to 0$ almost surely which implies $\frac {X_1+X_2+...+X_n} n \to 0$ almost surely. This proves a much stronger result without using any major theorem.

$\endgroup$

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.