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I am asked to construct a sequence of independent random variables $X_n$ with $E(X_n)=0$ for any $n$, but the sample mean of the $X_n$’s diverges almost surely to minus infinity.

I understand that the only condition not mentioned here to fall into Strong Law of Large Numbers is the « identically distributed » condition. So I know I need to build a sequence of independent but not identically distributed random variables.

But I get stuck…. Would appreciate any leads

user773458
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JenniferK
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    Consider something like $P(X_n = -n) + P(X_n = a_n) = 1$, so that $\sum_n P(X_n = a_n) < \infty$. – Brian Moehring Feb 08 '24 at 08:05
  • Thank you! I can see how this goes: the variance grows very fast with n and so the behavior is erratic and it does not converge. But how would you go about proving that the sample mean diverges to minus infinity almost surely? – JenniferK Feb 08 '24 at 18:51
  • Use the Borel-Cantelli lemma. – Brian Moehring Feb 08 '24 at 18:58
  • I see so Xn takes values -n with probability p and a a_n with probability (1-p) such that the n(1-p) converges so (1-p) could be 1/n and thus p should be (1-1/n). For E(Xn) to be 0 I need a_n= n^2-n …

    Then by the first BC lemma since the sum if P(Xn=a_n) is finite, we have that the probability that infinitely many of these events occur is 0.

    Thus the probability that infinitely many of the event Xn=-n occurs is 1.

    And as a result the sample mean will diverge to minus infinity. Is that a correct way to prove it?

     – JenniferK Feb 08 '24 at 19:56
  • You have a few errors (for instance, it's not enough to say that $X_n=-n$ for infinitely many $n$ is almost sure), but the idea is all there. See my answer below if you want the details. – Brian Moehring Feb 08 '24 at 20:00

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Suppose $X_n$ is supported on $\{-n, a_n\}$ and that $\sum_{n=1}^\infty P(X_n = a_n) < \infty$. By the Borel-Cantelli lemma, we immediately have $$P\left(X_n = a_n \quad \text{infinitely often}\right) = 0,$$ which in particular means $$P(X_n=-n \quad \text{ for all but finitely many } n) = 1,$$ so that $P(\lim_n X_n = -\infty) = 1$. Finally, by the elementary real analysis of sequences, if $X_n \to a$, then $\frac1n(X_1+X_2+\ldots+X_n)\to a$, so the sample mean tends to $-\infty$ almost surely.

The above skipped over the details of whether it's possible to define $X_n$ in this manner, so I'll briefly deal with how to construct it.

First, choose some positive sequence $p_n := P(X_n=a_n)$ such that $\sum_n p_n < \infty$. This can be done arbitrarily, but for this example, I'll choose $p_n = 2^{-n}$. Since $P(X_n = -n) + P(X_n=a_n) = 1,$ we must have $P(X_n = -n) = 1-p_n = 1-2^{-n}.$

Then we'll solve for $a_n$: Since $E[X_n] = 0$, we have $$(-n)(1-p_n) + a_np_n = 0 \\ a_n = \frac{n(1-p_n)}{p_n} = n2^n(1-2^{-n})$$ and that's all. Specifically, if we let $X_n$ satisfy $$P(X_n = -n) = 1-2^{-n}, \qquad P(X_n = n2^n(1-2^{-n})) = 2^{-n},$$ then we may conclude, by the above argument, that the sample mean tends to $-\infty$ almost surely.

Note that we didn't need to use the independence of $X_n$ anywhere, as the result holds no matter the dependence/independence of the sequence $X_n$.

Brian Moehring
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  • Thank you! I am glad to see that the logic / order of elements of the construction is aligned with my above comment. I will now look at whether my example works as well and if not why. Thanks for your help! – JenniferK Feb 08 '24 at 20:01
  • One more note: There wasn't anything special about the $-n$, and you could replace it with any sequence that converges to $-\infty$. For that matter, there wasn't anything special about $-\infty$, as you could replace it with any extended real number you wish and end up showing the sample mean converges almost surely to that value. The construction and proof would proceed in the same manner. That said, while my solution is not guaranteed to be the simplest, changing both of those arbitrarily could end up with a nasty looking $a_n$ pretty quickly. – Brian Moehring Feb 08 '24 at 20:15