Is there a $p$-adic analogue of the Strong Law of Large Numbers? In particular, suppose that $f_i: \mathbb{Z}_p \longrightarrow \mathbb{Q}_p$ for $i = 1,2,\ldots$ is an sequence of random variables that is i.i.d. with respect to the $\mathbb{Q}_p$-valued normalized Haar measure $\mu$ on $\mathbb{Z}_p$. Must one have $$\int_{\mathbb{Z}_p} f_1 d \mu = \lim_{n \rightarrow \infty} \frac{f_1(x) + f_2(x) + \cdots + f_n(x)}{n}$$ for almost all $x \in \mathbb{Z}_p$? If not, is there a partial result in this direction?
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2$\begingroup$ Jesse, be more precise about what you mean by a $p$-adic random variable and integration wrt the $p$-adic-valued Haar "measure." Such integration is not well-defined in general, as a limit of Riemann sums can depend on the choice of sample points. See p. 39 of Koblitz's $p$-adics book (2nd ed.) for a problem with the simple example $f(x) = x$. If you want to define this integral using sample points $0,1,\ldots,p^n-1$ at "level $p^n$" then the integral exists for a restricted class of continuous functions. See Robert's $p$-adics book for "Volkenborn integral." $\endgroup$– KConradCommented Apr 23, 2015 at 9:17
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$\begingroup$ I'm using Chapter II Sections 3 and 5 of Koblitz as a reference. $\endgroup$– Jesse ElliottCommented Apr 23, 2015 at 9:26
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$\begingroup$ For i.i.d. of course you have to assume the integrals exist. Is continuity necessary? I don't know even if it's sufficient. $\endgroup$– Jesse ElliottCommented Apr 23, 2015 at 9:36
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2$\begingroup$ The $p$-adic Haar measure does not work well for the integration developed in his book, since it is not a bounded distribution. I've never heard of a robust $p$-adic version of the LLN. $\endgroup$– KConradCommented Apr 23, 2015 at 9:46
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$\begingroup$ Ok. The functions $f_i$ I was trying to apply it to are at least locally constant, so the question makes sense in that context at least. $\endgroup$– Jesse ElliottCommented Apr 23, 2015 at 18:08
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2 Answers
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I think this can't work. If you look at the simplest possible case, where the $f_i$ take values 0 and 1 with probabilities $1-1/p$ and $1/p$, say, then you are effectively looking at the average of the digit sums of $x$.
To make sense (since your $f_i(x)$ generally take values in $\mathbb Q_p$), you should look for a limit in $\mathbb Q_p$. But now, the numerator is an integer close to $n/p$. By standard random walk techniques, you know that the ratio will be exactly $1/p$ infinitely often (with $p$-valuation $-1$).
On the other hand, for $n$ not a multiple of $p$, the $p$-valuation of the ratio will be (at least) $0$. Hence there is no pointwise convergence.
answered Apr 23, 2015 at 15:38
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There seems to be an extensive oeuvre of one A. Khrennikov on the subject. Here is one paper. It seems one can get interesting results (his interest comes from mathematical physics).
answered Apr 23, 2015 at 20:31