Mean of i.i.d Random Variables With No Expected Value 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.
 A: 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 $1/M$.  
