6

Let $\{X_i\}_{i=1}^{\infty}$ be i.i.d. random variables. Define $$ L_n = \frac{1}{n}\sum_{i=1}^n X_i \quad \forall n \in \{1, 2, 3, …\} $$ Using the central limit theorem, it can be shown that if $E[X_i]=0$ and $0<Var(X_i)<\infty$ then: $$ \lim_{n\rightarrow\infty} P[L_n\leq x] = \left\{ \begin{array}{ll} 1 &\mbox{ if $x > 0$} \\ c & \mbox{ if $x=0$}\\ 0 & \mbox{ if $x<0$} \end{array} \right.$$ where $c=1/2$. If the variance is infinite then the law of large numbers implies a similar structure for the cases $x>0$ and $x<0$, but the case $x=0$ is unclear to me.

Questions: For infinite variance, can we get different behavior for the case $x=0$, such as $c=1/3$? Can we get related step-function structure when the mean does not exist, but with different behavior for the case $x=0$?

Notes:

  • We can get such a limiting function with $c=1/3$ for random sequences with different structure, such as $L_n= A/n$ with $P[A=1]=2/3, P[A=-1]=1/3$.

  • I came up with this question while reflecting on the question here: Why does a C.D.F need to be right-continuous?

Michael
  • 23,905
  • I have added the clarification: "Can we get related [step-function] structure when the mean does not exist [but with different behavior for the case $x=0$]?" I originally intended this second part not so much as a "second question" but just to acknowledge that interesting examples when $L_n\rightarrow 0$ in distribution but with $c\neq 1/2$ might (for some reason) be easier to construct when the mean does not exist than when the mean exists but the variance is infinite. (Or, perhaps no such examples exist!) – Michael May 11 '19 at 18:48

2 Answers2

5

Yes it is possible for $c$ to take any value strictly between $0$ and $1$. The point is that there exist mean-zero stable distributions which are not symmetric about $0$ (of course, such a stable distribution cannot be Gaussian, and so it must have infinite variance). You may look at the Wikipedia page to see how some of these stable distributions look.

Specifically, if $\alpha \in (1,2)$ and $\beta \in [-1,1]$, then it turns out that there exists a random variable $X$ whose characteristic function will look like $$\phi_X(t) = e^{-|t|^{\alpha}\big(1-i\beta \tan(\frac{\pi\alpha}{2})\text{sign}(t)\big).}$$ As it turns out, this distribution will have mean zero, and moreover (by varying $\alpha$ and $\beta$), $P(X<0)$ can be any predefined number $c\in(0,1)$. Furthermore, for iid copies one may check directly from the characteristic function that $n^{-1/\alpha}(X_1+...+X_n)$ has the same distribution as $X_1$. From this we can easily conclude that $$P(L_n<0) =P(n^{1-1/\alpha}L_n<0)= P(X<0)=c \in (0,1),$$ for all $n$, as desired. I do not know if $c=0$ or $c=1$ is a possible limit for nonzero random variables $X_i$, though it'd be interesting to find out.

shalop
  • 13,703
  • Now, regarding the case of infinite mean, take $\alpha<1$ and one may obtain the same result, in fact in that case one has $P(L_n\le x) \to c$ for all $x\in \Bbb R$. – shalop May 12 '19 at 02:30
  • Thanks! I was able to verify the easy things such as $n^{-1/\alpha}(X_1+...+X_n)$ has the same characteristic function as $X_1$, and the $P[L_n<0]$ calculation. It is not obvious that such a characteristic function back-transforms to a valid PDF (i.e. nonnegative) or that $\beta$ controls $P[X<0]$ (though the linked pictures suggest that). But this was a nicely written description of a deep theory that I was not familiar with. Thanks again! – Michael May 12 '19 at 04:41
  • @Michael Those things are certainly not obvious at all. In fact, after closer inspection I am not even sure that $P(X<0)$ can take all values between $0$ and $1$: surely it is not true for a fixed value of $\alpha \in (1,2)$ as I write in my answer (note that the picture has $\alpha=1/2$ which is inadmissible if we want a finite mean). However, I strongly believe that by making $\alpha$ closer and closer to $2$ and setting $\beta=1$ it is possible to make $c$ approach $1$ (and similarly by setting $\beta=-1$ we can make $c$ approach $0$). Numerically this is clear but a proof might take work – shalop May 12 '19 at 04:57
  • I see. On your other comment: The $P[L_n\leq x]\rightarrow c$ for all $x \in \mathbb{R}$ is quite crazy! If $c \in (0,1)$, given the Kolmogorov 0/1 law we have $P[L_n\rightarrow \infty]=P[L_n\rightarrow-\infty]=0$. So it seems to me that we would need $L_n$ to take very large values for huge intervals of $n$, then "relatively quickly" change to taking very large-magnitude negative values for huge intervals of $n$, but infinitely oscillate in this way. – Michael May 12 '19 at 05:01
  • @Michael Yes this is true. If you search for (or simulate) pictures of $\alpha$-stable Levy processes for $\alpha<1$ then this is exactly the behavior you will see. In fact such processes have the property that for $c>0$, the distribution of $X_{c t}$ is the same as that of $c^{1/\alpha}X_t$. So the surprising thing is that (as opposed to the case of $\alpha>1$, e.g., Brownian motion) the spatial scaling as actually "bigger" than the temporal scaling. The Cauchy process ($\alpha=1$) is the case where the space and time scalings "equalize." – shalop May 12 '19 at 05:09
  • Also, in my second comment of this thread I of course meant "taking $\alpha$ closer and closer to $1$," rather than $2$. – shalop May 12 '19 at 05:17
  • I corrected the answer to be more precise. @sk – shalop May 13 '19 at 20:33
3

Partial answer: if $X_i$'s are non-negative with infinite mean than $L_n \to \infty$ a.s. an $P(L_n \leq x) \to 0$ for every $x$.

Kavi Rama Murthy
  • 311,013
  • Haha. Well yes, that is an easy side case (+1). There are certain situations where the mean does not exist but we still get convergence in distribution to 0, I was wondering if those cases can have weird things happening for $\lim_{n\rightarrow\infty} P[L_n\leq 0]$. Or, is $1/2$ fundamental? Obviously we get $1/2$ whenever $X_i$ has a continuous distribution and has the same distribution as $-X_i$. – Michael May 11 '19 at 14:30