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7.2.7 Almost Sure Convergence
在某些问题中，直接证明almost sure收敛可能很困难。因此，需要知道一些almost sure收敛的充分条件。
Theorem 7.5
考虑序列$X_1, X_2, X_3, ⋯.$ 如果对于所有 $ϵ>0$，我们有
$$\sum_{n=1}^∞P(\abs{X_n-X}>ϵ)<∞,$$
则 $X_n\overset{a.s.}⟶X$.
Example 7.14
考虑序列$X_1, X_2, X_3, ⋯$使得\begin{cases}
-\frac{1}{n} & \quad \textrm{with probability } \frac{1}{2} \\
& \quad \\
\frac{1}{n} & \quad \textrm{with probability } \frac{1}{2}
\end{cases}求证 $X_n\overset{a.s.}⟶0$.
由上述定理，只需要证明 $\sum_{n=1}^{\infty} P\big(|X_n| > \epsilon \big) < \infty$.
注意$|X_n|=1/n$。因此，$|X_n|>ϵ$ 当且仅当 $n<1/ϵ$。我们得出\begin{align*}
\sum_{n=1}^{\infty} P\big(|X_n| > \epsilon \big) & \leq \sum_{n=1}^{{\large \lfloor\frac{1}{{\large \epsilon}}\rfloor}} P\big(|X_n| > \epsilon \big)\\
& = \left\lfloor\frac{1}{\epsilon}\right\rfloor< \infty.
\end{align*}
定理 7.5 为almost sure收敛提供了一个充分不必要条件。特别是，如果我们获得$$\sum_{n=1}^{\infty} P\big(|X_n-X| > \epsilon \big) = \infty,$$那么我们仍然不知道 $X_n$ 是否会almost sure收敛到 $X$。这里，我们提供一个充分必要条件。
Theorem 7.6
考虑序列 $X_1, X_2, X_3, ⋯.$ 对任何 $ϵ>0$, 定义事件
\[A_m=\{\abs{X_n−X}<ϵ,\text{ for all }n≥m\}.\]
则 $X_n\overset{a.s.}⟶X$ 当且仅当对任何 $ϵ>0$，我们有
\[\lim_{m→∞}P(A_m)=1.\]
Example 7.15
Let $X_1, X_2, X_3, ⋯$ be independent random variables, where $X_n∼\text{Bernoulli}(1/n)$ for $n=2,3,⋯$. The goal here is to check whether $X_n\overset{a.s.}⟶0$.
• Check that $∑^∞_{n=1}P(|X_n|>ϵ)=∞$
• Show that the sequence $X_1, X_2,⋯$ does not converge to 0 almost surely using Theorem 7.6
Solution
• We first note that for $0<ϵ<1$, we have\begin{align*}
\sum_{n=1}^{\infty} P\big(|X_n| > \epsilon \big)&= \sum_{n=1}^{\infty} P(X_n=1)\\
&= \sum_{n=1}^{\infty} \frac{1}{n} = \infty.
\end{align*}• To use Theorem 7.6, we define\[A_m=\{|X_n|< \epsilon, \text{ for all }n \geq m \}.\]Note that for $0<ϵ<1$, we have
\[A_m=\{X_n=0,\text{ for all }n≥m\}.\]
According to Theorem 7.6 , it suffices to show that
\[\lim_{m→∞}P(A_m)<1.\]
We can in fact show that $\lim_{m→∞}P(A_m)=0$. To show this, we will prove $P(A_m)=0$, for every $m≥2$. For $0<ϵ<1$, we have\begin{align*}
P(A_m) &=P \big(\{X_n=0,\text{ for all }n \geq m \}\big)\\
&\leq P \big(\{X_n=0,\text{ for } n=m, m+1, \cdots, N \}\big) \quad (\textrm{for every positive integer $N \geq m$})\\
&=P(X_m=0) P(X_{{\large m+1}}=0) \cdots P(X_{N}=0) \qquad (\textrm{since the $X_i$'s are independent})\\
&=\frac{m-1}{m} \cdot \frac{m}{m+1} \cdots \frac{N-1}{N}\\
&=\frac{m-1}{N}.
\end{align*}Thus, by choosing $N$ large enough, we can show that $P(A_m)$ is less than any positive number. Therefore, $P(A_m)=0$ for all $m≥2$. We conclude that $\lim_{m→∞}P(A_m)=0$. Thus, according to Theorem 7.6, the sequence $X_1, X_2, ...$ does not converge to 0 almost surely. |
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