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Determine if the following sequences of independent rvs satisfy the SLLN:

  • $(a)$ $P(X_n=1) = \frac{1}{n^2},\, P(X_n=0) = 1-\frac{1}{n^2}$
  • $(b)$ $P(X_n=n) = \frac{1}{n^2},\, P(X_n=0) = 1-\frac{1}{n^2}$
  • $(c)$ $P(X_n=n) = \frac{1}{n}, \, P(X_n=0) = 1-\frac{1}{n}$

My attempt: Let $S_n = X_1 + X_2+\ldots + X_n$.

  • $(a)$: $E(X_n) = \frac{1}{n^2} = E(X_n^2)$. We need to verify: $\displaystyle \lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n}\rightarrow 0$ almost surely. Now, $E(S_n) = \sum_{k=1}^{n} \frac{1}{k^2}$, so as $n\rightarrow \infty$, $E(S_n)\rightarrow \frac{\pi^2}{6}$, which implies $\frac{E(S_n)}{n}\rightarrow 0$. However, $S_n\leq n$ for every $n$, thus as $n\rightarrow \infty$, $\frac{S_n}{n} = \frac{n-k}{n} \rightarrow 1$, unless all $X_i$'s $=0$ (which is an extreme case, since $X_n$ are random). Thus $X_n$ does not satisfy SLLN.

  • $(b)$: $E(X_n) = \frac{1}{n}$ and $E(X_n^2) = 1$. Thus $E(S_n)$ diverges as $n\rightarrow \infty$, while $S_n = kn$ for some nonnegative integers $k$. Thus, we could apply the L'Hospital rule (variable is $n$) to compute $\lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n} = \lim_{n\rightarrow \infty} (\sum_{k=1}^{n} \frac{1}{k^2}) = \frac{\pi^2}{6}\neq 0$, so $X_n$ does not satisfy SLLN.

  • $(c)$: $E(X_n) = 1$, so $E(S_n)= n$. Thus, $S_n - E(S_n) = (k-1)n$ for some nonnegative integers $k$. Thus, $\lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n} = k-1 = 0$ if and only if $k=1$. So, $\lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n}$ does not converge to $0$ almost surely (is this correct?). Thus, $X_n$ does not satisfy SLLN.

My question: I'm quite skeptical with my solution above since the way I tried to express $S_n$ seems to be weird. Could anyone please help me with these problems in case my solutions above are incorrect? Any thoughts would really be appreciated.

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  • $\begingroup$ Hint: In (a), what can you say about the random set of indices $n$ such that $X_n\ne0$? $\endgroup$ – Did Nov 5 '16 at 21:31
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    $\begingroup$ Is $P(\{ n : X_n=1 \} \text{ is infinite})$ positive in a? You may want to think about Borel-Cantelli... $\endgroup$ – Ian Nov 5 '16 at 21:43
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    $\begingroup$ See what my friend. In (a), where you say that $S_{n}= n-k$, you really should have said that $S_{n} = n-k(n)$. Here $k$ depends on $n$. And thus you can't trivially say that $\frac{S_{n}}{n} \to 1$. You should think over how $k$ depends on $n$. $\endgroup$ – kolobokish Nov 5 '16 at 23:04
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    $\begingroup$ I'm not a fan of probabilities , I'm sorry. Speaking just from the view of measure theory.)) About $2$. We should look for 0-s in the other random variables, as we know that the first one is almost surely non 0. (literally i mean that if we multiply from 1, we will for sure have 0.). Never mind about $\omega$. I mean we should have find the probability(summed up) of the cases, for which the $k=1$. That's only for the case where starting from the second random variable all the variables take value 0. $\endgroup$ – kolobokish Nov 5 '16 at 23:37
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    $\begingroup$ @user177196 The idea for (c) is that at each time $n$ such that $X_n=n$, the sequence $\frac{S_n}n$ "jumps" by nearly $+1$,and that these jumps prevent it to converge to any finite limit. More rigorously, assume that $$\frac{S_n}n\to L$$ almost surely, for some random almost surely finite $L$, then $$\frac{X_n}n=\frac{S_n}n-\frac{S_{n-1}}{n-1}\frac{n}{n-1}\to L-L=0$$ almost surely, which is impossible since $X_n=n$ infinitely many times hence $$P\left(\limsup \frac{X_n}n=1\right)=1$$ Note that this argument requires that $L-L=0$, that is, that $L$ is finite. $\endgroup$ – Did Nov 6 '16 at 11:03
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The SLLN by Kolmogorov for independent sequence $\left(X_n\right)_{n\in \mathbb N}$, states that if

  1. $Var(X_n)<\infty, \mathbb E[X_n]=μ_n$ for every $n$,
  2. $\sum_{n=1}^\infty\dfrac{Var(X_n)}{n^2}<\infty$

then $\frac1n(S_n-\mathbb ES_n)\overset{a.s.}\longrightarrow 0$. These are sufficient but not necessary meaning, that every sequence that satisfies these (actually the second is the one to check) satisfies the SLLN, but if a sequence does not satisfy these, then this does not mean that it cannot satisfy the SLLN.

So, in your case:

  • $(a)$: $\mathrm{Var}[X_n]\le \mathbb E[X_n^2]=\frac1{n^2}$ and so $$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}\le\sum_{n=1}^\infty\frac1{n^4}<\infty$$ and so, it does satisfy the SLLN.
  • $(b)$: $\mathrm{Var}[X_n]\le \mathbb E[X_n^2]=1$ and so, as in $(a)$ $$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}\le\sum_{n=1}^\infty\frac1{n^2}<\infty$$ and so, it does satisfy the SLLN.
  • $(c)$: Here $Var(X_n)=n-1$, hence $$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}=\sum_{n=1}^\infty\frac{n-1}{n^2}=\infty$$ and so, this way does not work! (But as mentioned above, it does not mean that we are done). However, if we consider the independent events $E_n:\{ω:X_n(ω)=n\}$, we have that $$\sum_{n=1}^\infty P(E_n)=\sum_{n=1}^\infty\frac1n=\infty$$ and hence by the second Borel-Cantelli lemma: $$P(\lim\sup_nE_n)=P(X_n=n \text{ occurs i.o.})=1$$ This, however, still does not allow you to show that \begin{align}P\left(\lim_{n}\frac{S_n-\mathbb E[S_n]}{n}\neq 0\right)&=P\left(\lim_{n}\frac{S_n}{n}\neq 1\right)\\[0.2cm]&\ge P\left(\lim_{n}\frac{S_n}{n}> 1\right)= P\left(\lim\sup E_n\right)=1\end{align} since there are unbounded sequences that are cesaro-summable.
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  • $\begingroup$ Re (c), why $$P\left(\lim_{n}\frac{S_n}{n}> 1\right)= P\left(\lim\sup \{X_n=n\}\right)\ ?$$ Note that there exists some sequences $(x_n)$ such that $x_n\in\{0,n\}$ for every $n$ and $x_n=n$ infinitely often and $\frac1n\sum\limits_{k=1}^nx_k\to1$, so you really need to explain what happens here. $\endgroup$ – Did Nov 6 '16 at 10:55
  • $\begingroup$ Re (a) and (b), you use a large hammer to solve elementary questions, as shown in the comments on main and as the OP themselves realized). $\endgroup$ – Did Nov 6 '16 at 10:55
  • $\begingroup$ @user177196 Thanks for accepting my answer, but what about c)? Did you solve it? $\endgroup$ – Jimmy R. Nov 7 '16 at 9:48
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    $\begingroup$ @JimmyR.: I used Did's help, as it's a bit slicker. $\endgroup$ – user177196 Nov 8 '16 at 4:01

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