Krdytkyu

Heres an example of a sequence of r.vs $(X_n)$ that obey the Weak LLN but do not obey the Strong LLN,
$P(X_n=pm n)=1/(2nln n)$ for any $n=1,2dots$ and $P(X_n=0)=1-1/(nln n)$.

I am trying to show why this is so, but I am having trouble with it. We don't have any closed form expressions of the partial sums of $(X_n)$, only the probabilities that each $X_n$ takes at $n$. How do you even compute $sum_n^NX_n/N$?

With regards to the strong Law, it suffices to show that $E[X_1]$ is infinite, but we dont even have that they are identically distributed?

Any help would be greatly appreciated.

Assuming OP means to ask whether or not the random series $Z_N = sumlimits_{n = 2}^N dfrac{X_n}{N}$ converges almost surely. And that OP assumes $(X_n)_{n geq 2}$ is an independent sequence (we also notice $X_1$ makes no sense).

Abridged proof: we fundamentally apply Lindeberg-Feller Central Limit Theorem for triangular array with row-independent random variables (see 8.3 here https://www.ssc.wisc.edu/~xshi/econ715/Lecture_9_local_power.pdf). Our triangular array is $U_{N, k} = dfrac{X_k}{N}.$ Observe $mathbf{E}(U_{N, k}) = 0$ and $mathbf{V}mathrm{ar}(U_{N, k}) = dfrac{k}{N^2 log k}.$ Hence, for $Z_N = sumlimits_{k = 2}^N U_{N, k}$ we have $sigma_N^2 = mathbf{V}mathrm{ar}(Z_N) = sumlimits_{k = 2}^N dfrac{k}{N^2 log k}$ and this series clearly converges to some positive number $sigma^2.$ Finally, the LFCLT holds because
$$sum_{k = 2}^N mathbf{E} left(U_{N, k}^2 mathbf{1}_{left{|U_{n, k}| > frac{varepsilon}{2} sigma right}} right) leq sum_{frac{varepsilon}{2} sigma N leq k leq N} dfrac{k}{N^2 log k} leq dfrac{N(1 - frac{varepsilon}{2} sigma) times N}{N^2 log (frac{varepsilon}{2} sigma N)} to 0.$$ Therefore, $Z_N to mathrm{Norm}(0; sigma^2).$ Q.E.D.

Ammend. What the Lindeberg-Fellet CLT theorem gives is weak convergence. However, we know that convergence almost surely implies weak convergence; in other words, if the random variables $Z_N$ converges almost surely to any limit (which I did not prove), then said limit is normally distributed. Unfortunately,

$Z_N$ will not, in fact, converge almost surely to any limit.

My reasoning follows from the fact that $Z_{N + 1} = dfrac{N}{N+1}Z_N + dfrac{X_{N + 1}}{N + 1}$ so that, assuming $Z_N to Z$ almost surely, for some (finite-valued) limit $Z,$ then $dfrac{X_N}{N} to 0$ almost surely. However, the random variable $dfrac{X_N}{N}$ only has three values: $-1,$ $0$ and $1;$ whence, if the sequence $dfrac{X_N}{N}$ converges to zero, it has to be zero eventually. In other words, what I argumented was that
$$left{ lim_{N to infty} dfrac{X_N}{N} = 0 right} subset bigcup_{N = 1}^infty bigcap_{n = N}^infty {X_n = 0} mathop{=}^{mathrm{def}} bigcup_{N = 1}^infty mathrm{J}_N.$$
By independence, $mathrm{J}_N$ is a null event, hence, except on a null set, no point satisfy $dfrac{X_N}{N} to 0$ and therefore, $Z_N$ does not converge on any subset of positive probability. Q.E.D.