Let $X$ be an integer-valued random variable and let $X\_n$ be the sum of $n$ independent realizations of $X$. I would like to understand the behavior of $X\_n/n$ for large $n$ in some cases where $X$ has no expected value and therefore the central limit theorem (along with Chebyshev's Theorem, etc.) does not apply. 


For a concrete example, let $X$ have the following distribution:
$$Prob(X=0)=1/3\qquad Prob(X=4^k)=1/4^{k+1}\qquad Prob(X=-4^{k})=1/4^{k+1}$$
with all other probabilities zero.


Note that $X$ has no expected value so we can't use the central limit theorem. Still, I'd like to have a good way to estimate $Prob(X\_n/n > M)$ for a given large $n$ and $M$. 


Note also that in this example $Prob(X\_n/n=M)$ can depend on number theoretic properties of $M$ (and in particular will not be monotonic in $M$), but I'm hoping that when $n$ is big and we consider $Prob(X\_n/n > M)$, this sort of thing will wash out.


I've done some relevant calculations, but I wonder whether I'm missing either some knowledge or some insight that would make this easier to understand.



Mean of i.i.d Random Variables With No Expected Value
We can attack this problem by using Fourier transforms (i.e. characteristic functions). I'll consider 
 the example in the problem where $X$ is a random variable taking the value $0$ with probability $1/3$, and $4^{k}$ and 
 $-4^{k}$ with probability $1/4^{k+1}$ (for $k=0$, $1$, $\ldots$). I'll show that the probability that $|X\_n|/n >M$ is 
 about a constant times $1/M$ (more precise result below). 


Fix a smooth function $\Phi$ compactly supported in $[-1,1]$ 
 and approximating the characteristic function of that interval. Concretely, suppose $\epsilon$ is small and 
 $\Phi(x)=1$ on $[-1+\epsilon, 1-\epsilon]$ and is between $0$ and $1$ on the rest of $[-1,1]$. Since $\Phi$ is 
 smooth, its Fourier transform ${\hat \Phi}(\xi) = \int\_{-\infty}^{\infty}\Phi(x) e^{-2\pi i x\xi} dx$ has rapid decay for $|\xi|$ large. 


Now let $n$ and $M$ be large and consider 
 $$
 {\Bbb E}\Big(\Phi\Big(\frac{X\_n}{nM}\Big)\Big). 
 $$  

 Note that 
 $$ 
 \text{Prob} (|X\_n| >nM) \le 1 -{\Bbb E}(\Phi(X\_n/(nM))) \le \text{Prob}(|X\_n| > (1-\epsilon)nM), 
 $$
 and so our problem is to understand the expectation above. By Fourier inversion, 
 $$ 
 {\Bbb E}(\Phi(X\_n/(nM))) = \int\_{-\infty}^{\infty} {\hat \Phi}(\xi) {\Bbb E}\Big( e^{2\pi i \xi X\_n/(nM)}\Big) d\xi 
 = \int\_{-\infty}^{\infty} {\hat \Phi}(\xi) \Big( {\Bbb E}\Big( e^{2\pi i \xi X/(nM)}\Big)\Big)^{n} d\xi. 
 $$


Now we compute that 
 $$ 
 {\Bbb E}\Big( e^{2\pi i \xi X/(nM)}\Big) = \frac{1}{3} + 2 \sum\_{k=0}^{\infty} \frac{1}{4^{k+1}} \cos \Big( \frac{2\pi \xi 4^k}{nM}\Big) 
 = 1 - 2\sum\_{k=0}^{\infty} \frac{1}{4^{k+1}}\Big (1-\cos \Big( \frac{2\pi \xi 4^k}{nM}\Big) \Big). 
 $$ 
 Now using that $(1-\cos(x)) = O(\min(x^2, 1))$ we see that 
 $$ 
 \sum\_{k=0}^{\infty} \frac{1}{4^{k+1}}\Big (1-\cos \Big( \frac{2\pi \xi 4^k}{nM}\Big) \Big) = O\Big( \frac{|\xi|}{nM}\Big). 
 $$ 
Therefore, using $(1-x)^n = 1-nx +O(n^2 x^2)$ for $0\le x\le 1$,  

$$
{\Bbb E}(\Phi(X\_n/(nM))) = \int\_{-\infty}^{\infty} {\hat \Phi}(\xi) \Big( 1- 2n \sum\_{k=0}^{\infty} \frac{1}{4^{k+1}}\Big (1-\cos \Big( \frac{2\pi \xi 4^k}{nM}\Big) \Big) 
+ O\Big(\frac{\xi^2}{M^2}\Big) \Big) d\xi.
$$ 
Since ${\hat \Phi}$ has rapid decay, the error term above is $O(1/M^2)$ (with the implied constant depending only on the 
fixed function $\Phi$). Using Fourier inversion, we conclude that 
$$ 
{\Bbb E}({\Phi }(X\_n/(nM))) = \Phi(0) - n \sum\_{k=0}^{\infty} \frac{1}{4^{k+1}} \Big(2 \Phi(0) - \Phi\Big(\frac{4^k }{nM} \Big) -\Phi\Big(-\frac{4^k}{nM}\Big)\Big) 
+ O\Big(\frac{1}{M^2}\Big). 
$$ 
Since $\Phi(0)=1$, we get 
$$ 
1- {\Bbb E}({\Phi }(X\_n/(nM))) = n \sum\_{k=0}^{\infty} \frac{1}{4^{k+1}} \Big(2 - \Phi\Big(\frac{4^k }{nM} \Big) -\Phi\Big(-\frac{4^k}{nM}\Big)\Big) 
+ O\Big(\frac{1}{M^2}\Big). 
$$ 
By our choice for $\Phi$, the main term above is 
$$ 
\ge 2n \sum\_{k, 4^{k} \ge nM} \frac{1}{4^{k+1}}, 
$$ 
and is 
$$ 
\le 2n \sum\_{k, 4^{k} \ge (1-\epsilon) nM} \frac{1}{4^{k+1}}.
$$ 


Thus we have obtained a good understanding of the probability that $|X\_n|/n$ is large. Note also that 
the precise answer will have discontinuities when $nM$ gets near a power of $4$, but in any case 
the probability is about a constant times $1/M$. 


