About Posts which Tagged by 'Probability'
Let $\{E_n\}$ be arbitrary events in $\mathscr{F}$. If for each $m$, $\sum_{n>m}\mathscr{P}\{E_n\mid E_m^c\cap\cdots\cap E_{n-1}^c\}=\infty$, then $\mathscr{P}\{E_n\mbox{ i.o.}\}=1.$
$\bullet$ Proof.
2015年9月7日 星期一
2015年9月6日 星期日
Convergence of Moments (3)
About Posts which Tagged by 'Probability'
Let $\{X_n\}$ and $X$ be random variables. Let $0<r<\infty$, $X_n\in L^r$, and $X_n\rightarrow X$ in probability. Then the following three propositions are equivalent.
(1) $\{|X_n|^r\}$ is uniformly integrable;
(2) $X_n\rightarrow X$ in $L^r$;
(3) $\mathscr{E}|X_n|^r\rightarrow\mathscr{E}|X|^r<\infty$.
$\bullet$ Proof.
Let $\{X_n\}$ and $X$ be random variables. Let $0<r<\infty$, $X_n\in L^r$, and $X_n\rightarrow X$ in probability. Then the following three propositions are equivalent.
(1) $\{|X_n|^r\}$ is uniformly integrable;
(2) $X_n\rightarrow X$ in $L^r$;
(3) $\mathscr{E}|X_n|^r\rightarrow\mathscr{E}|X|^r<\infty$.
$\bullet$ Proof.
Convergence of Moments (2)
About Posts which Tagged by 'Probability'
Let $\{X_n\}$ and $X$ be random variables. If $X_n$ converges in distribution to $X$, and for some $p>0$, $\sup_n\mathscr{E}|X_n|^p=M<\infty$, then for each $r<p$, $$\underset{n\rightarrow\infty}{\lim}\mathscr{E}|X_n|^r=\mathscr{E}|X|^r<\infty.$$
$\bullet$ Proof.
Let $\{X_n\}$ and $X$ be random variables. If $X_n$ converges in distribution to $X$, and for some $p>0$, $\sup_n\mathscr{E}|X_n|^p=M<\infty$, then for each $r<p$, $$\underset{n\rightarrow\infty}{\lim}\mathscr{E}|X_n|^r=\mathscr{E}|X|^r<\infty.$$
$\bullet$ Proof.
Convergence of Moments (1)
About Posts which Tagged by 'Probability'
Let $\{X_n\}$ and $X$ be random variables. If $X_n\rightarrow X$ a.e., then for every $r>0$, $$\mathscr{E}|X|^r\leq\underset{n\rightarrow\infty}{\underline{\lim}}\mathscr{E}|X_n|^r.$$If $X_n\rightarrow X$ in $L^r$, and $X\in L^r$, then $\mathscr{E}|X_n|^r\rightarrow\mathscr{E}|X|^r$.
$\bullet$ Proof.
Let $\{X_n\}$ and $X$ be random variables. If $X_n\rightarrow X$ a.e., then for every $r>0$, $$\mathscr{E}|X|^r\leq\underset{n\rightarrow\infty}{\underline{\lim}}\mathscr{E}|X_n|^r.$$If $X_n\rightarrow X$ in $L^r$, and $X\in L^r$, then $\mathscr{E}|X_n|^r\rightarrow\mathscr{E}|X|^r$.
$\bullet$ Proof.
2015年9月4日 星期五
Characteristic Functions
About Posts which Tagged by 'Probability'
For any random variable $X$ with probability measure $\mu$ and distribution function $F$, the characteristic function (ch.f.) is a function $f$ on $\mathbb{R}$ defined as $$f(t)=\mathscr{E}\left(e^{itX}\right)=\int_{-\infty}^\infty e^{itx}\,dF(x)\mbox{ for all }t\in\mathbb{R}.$$There are some simple properties of ch.f.:
For any random variable $X$ with probability measure $\mu$ and distribution function $F$, the characteristic function (ch.f.) is a function $f$ on $\mathbb{R}$ defined as $$f(t)=\mathscr{E}\left(e^{itX}\right)=\int_{-\infty}^\infty e^{itx}\,dF(x)\mbox{ for all }t\in\mathbb{R}.$$There are some simple properties of ch.f.:
2015年9月3日 星期四
Cantelli's Law of Large Numbers
About Posts which Tagged by 'Probability'
If $\{X_n\}$ are independent random variables such that the fourth moments $\mathscr{E}(X_n^4)$ have a common bound and define $S_n=\sum_{j=1}^nX_j$, then $$\frac{S_n-\mathscr{E}(S_n)}{n}\rightarrow0\mbox{ a.e.}$$
$\bullet$ Proof.
WLOG, suppose $\mathscr{E}(X_n)=0$ for all $n$ and denote the common bound of $\mathscr{E}(X_n^4)$ to be $$\mathscr{E}(X_n^4)\leq M_4<\infty\mbox{ for all }n.$$Then by Lyapunov's inequality, we have the second moments $$\mathscr{E}|X_n|^2\leq\left[\mathscr{E}|X_n|^4\right]^\frac{2}{4}\leq \sqrt{M_4}<\infty.$$Consider the fourth moment of $S_n$, $$\begin{array}{rl}\mathscr{E}(S_n^4)
&=\mathscr{E}\left[\left(\sum_{j=1}^nX_j\right)^4\right]\\ &= \mathscr{E}\left[\sum_{j=1}^nX_j^4+{4\choose1}\sum_{i\neq j}X_iX_j^3+{4\choose2}\sum_{i\neq j}X_i^2X_j^2\right.\\ &\quad\left.+{4\choose1}{3\choose1}\sum_{i\neq j\neq k}X_iX_jX_k^2+{4\choose1}{3\choose1}{2\choose1}\sum_{i\neq j\neq k\neq l}X_iX_jX_kX_l\right]\\
&=\sum_{j=1}^n\mathscr{E}(X_j^4)+4\sum_{i\neq j}\mathscr{E}(X_i)\mathscr{E}(X_j^3)+6\sum_{i\neq j}\mathscr{E}(X_i^2)\mathscr{E}(X_j^2)\quad(\because\mbox{ indep.})\\ &\quad+12\sum_{i\neq j\neq k}\mathscr{E}(X_i)\mathscr{E}(X_j)\mathscr{E}(X_k^2)+24\sum_{i\neq j\neq k\neq l}\mathscr{E}(X_i)\mathscr{E}(X_j)\mathscr{E}(X_k)\mathscr{E}(X_l)\\ &=\sum_{j=1}^n\mathscr{E}(X_j^4)+6\sum_{i\neq j}\mathscr{E}(X_i^2)\mathscr{E}(X_j^2)\qquad\qquad(\because\mbox{ assuming }\mathscr{E}(X_n)=0.) \\ &\leq nM_4+3n(n-1)\sqrt{M_4}\sqrt{M_4}=n(3n-2)M_4.\end{array}$$By Markov's inequality, for $\varepsilon>0$, $$\mathscr{P}\{|S_n|>n\varepsilon\}\leq\frac{\mathscr{E}(S_n^4)}{n^4\varepsilon^4}\leq\frac{n(3n-2)M_4}{n^4\varepsilon^4}=\frac{3M_4}{n^2\varepsilon^4}+\frac{2M_4}{n^3\varepsilon^4}.$$Thus, $$\sum_n\mathscr{P}\{|S_n|>n\varepsilon\}\leq\sum_n\frac{3M_4}{n^2\varepsilon^4}+\frac{2M_4}{n^3\varepsilon^4}<\infty.$$By Borel-Cantelli Lemma I, we have $$\mathscr{P}\{|S_n|>n\varepsilon\mbox{ i.o.}\}=0\implies\frac{S_n}{n}\rightarrow0\mbox{ a.e.}$$
If $\{X_n\}$ are independent random variables such that the fourth moments $\mathscr{E}(X_n^4)$ have a common bound and define $S_n=\sum_{j=1}^nX_j$, then $$\frac{S_n-\mathscr{E}(S_n)}{n}\rightarrow0\mbox{ a.e.}$$
$\bullet$ Proof.
WLOG, suppose $\mathscr{E}(X_n)=0$ for all $n$ and denote the common bound of $\mathscr{E}(X_n^4)$ to be $$\mathscr{E}(X_n^4)\leq M_4<\infty\mbox{ for all }n.$$Then by Lyapunov's inequality, we have the second moments $$\mathscr{E}|X_n|^2\leq\left[\mathscr{E}|X_n|^4\right]^\frac{2}{4}\leq \sqrt{M_4}<\infty.$$Consider the fourth moment of $S_n$, $$\begin{array}{rl}\mathscr{E}(S_n^4)
&=\mathscr{E}\left[\left(\sum_{j=1}^nX_j\right)^4\right]\\ &= \mathscr{E}\left[\sum_{j=1}^nX_j^4+{4\choose1}\sum_{i\neq j}X_iX_j^3+{4\choose2}\sum_{i\neq j}X_i^2X_j^2\right.\\ &\quad\left.+{4\choose1}{3\choose1}\sum_{i\neq j\neq k}X_iX_jX_k^2+{4\choose1}{3\choose1}{2\choose1}\sum_{i\neq j\neq k\neq l}X_iX_jX_kX_l\right]\\
&=\sum_{j=1}^n\mathscr{E}(X_j^4)+4\sum_{i\neq j}\mathscr{E}(X_i)\mathscr{E}(X_j^3)+6\sum_{i\neq j}\mathscr{E}(X_i^2)\mathscr{E}(X_j^2)\quad(\because\mbox{ indep.})\\ &\quad+12\sum_{i\neq j\neq k}\mathscr{E}(X_i)\mathscr{E}(X_j)\mathscr{E}(X_k^2)+24\sum_{i\neq j\neq k\neq l}\mathscr{E}(X_i)\mathscr{E}(X_j)\mathscr{E}(X_k)\mathscr{E}(X_l)\\ &=\sum_{j=1}^n\mathscr{E}(X_j^4)+6\sum_{i\neq j}\mathscr{E}(X_i^2)\mathscr{E}(X_j^2)\qquad\qquad(\because\mbox{ assuming }\mathscr{E}(X_n)=0.) \\ &\leq nM_4+3n(n-1)\sqrt{M_4}\sqrt{M_4}=n(3n-2)M_4.\end{array}$$By Markov's inequality, for $\varepsilon>0$, $$\mathscr{P}\{|S_n|>n\varepsilon\}\leq\frac{\mathscr{E}(S_n^4)}{n^4\varepsilon^4}\leq\frac{n(3n-2)M_4}{n^4\varepsilon^4}=\frac{3M_4}{n^2\varepsilon^4}+\frac{2M_4}{n^3\varepsilon^4}.$$Thus, $$\sum_n\mathscr{P}\{|S_n|>n\varepsilon\}\leq\sum_n\frac{3M_4}{n^2\varepsilon^4}+\frac{2M_4}{n^3\varepsilon^4}<\infty.$$By Borel-Cantelli Lemma I, we have $$\mathscr{P}\{|S_n|>n\varepsilon\mbox{ i.o.}\}=0\implies\frac{S_n}{n}\rightarrow0\mbox{ a.e.}$$
$\Box$
2015年9月2日 星期三
The Converse of Strong Law of Number
About Posts which Tagged by 'Probability'
Let $\{X_n\}$ be a sequence of i.i.d. random variables, we have $$\frac{S_n}{n}\mbox{ converges a.e. }\implies\mathscr{E}|X_1|<\infty.$$
$\bullet$ Proof.
Let $\{X_n\}$ be a sequence of i.i.d. random variables, we have $$\frac{S_n}{n}\mbox{ converges a.e. }\implies\mathscr{E}|X_1|<\infty.$$
$\bullet$ Proof.
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