Determine whether sequence of independent rvs satisfy SLLN 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.
The SLLN by Kolmogorov for independent sequence $\left(X_n\right)_{n\in \mathbb N}$, states that if


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


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 $$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}\le\sum_{n=1}^\infty\frac1{n^4}&lt;\infty$$
and so, it does satisfy the SLLN.
$(b)$: $\mathrm{Var}[X_n]\le \mathbb E[X_n^2]=1$ and so, as in $(a)$ 
$$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}\le\sum_{n=1}^\infty\frac1{n^2}&lt;\infty$$
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)&amp;=P\left(\lim_{n}\frac{S_n}{n}\neq 1\right)\\[0.2cm]&amp;\ge P\left(\lim_{n}\frac{S_n}{n}&gt; 1\right)= P\left(\lim\sup E_n\right)=1\end{align}
since there are unbounded sequences that are cesaro-summable.
The SLLN by Kolmogorov for independent sequence $\left(X_n\right)_{n\in \mathbb N}$, states that if


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


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 $$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}\le\sum_{n=1}^\infty\frac1{n^4}&lt;\infty$$
and so, it does satisfy the SLLN.
$(b)$: $\mathrm{Var}[X_n]\le \mathbb E[X_n^2]=1$ and so, as in $(a)$ 
$$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}\le\sum_{n=1}^\infty\frac1{n^2}&lt;\infty$$
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)&amp;=P\left(\lim_{n}\frac{S_n}{n}\neq 1\right)\\[0.2cm]&amp;\ge P\left(\lim_{n}\frac{S_n}{n}&gt; 1\right)= P\left(\lim\sup E_n\right)=1\end{align}
since there are unbounded sequences that are cesaro-summable.