Семинар 8, 08.11.2024

Задача 1¶

\Omega=\{\heartsuit,\diamondsuit,\spadesuit,\clubsuit\},\quad \FF=2^\Omega

\PP(\{\heartsuit\})=\ldots=\PP(\{\clubsuit\})=\frac{1}{4}

$\omega$	$\heartsuit$	$\diamondsuit$	$\spadesuit$	$\clubsuit$
$X(\omega)$	1	0	-1	0
$Y(\omega)$	1	1	-1	-1

\begin{align*}\PP(\{X=-1\}\cap\{Y=-1\})&=\PP(\{\omega\in\Omega\colon X(\omega)=-1\}\cap\{\omega\in\Omega\colon Y(\omega)=-1\})\\&=\PP(\{\spadesuit\}\cap\{\spadesuit,\clubsuit\})=\PP(\{\spadesuit\})=\frac{1}{4}\end{align*}

\begin{align*} \PP(\{X=-1\}\cap\{Y=1\})&=\PP(\{\omega\in\Omega\colon X(\omega)=-1\}\cap\{\omega\in\Omega\colon Y(\omega)=1\})\\&=\PP(\{\spadesuit\cap\{\heartsuit,\diamondsuit\}\})=\PP(\varnothing)=0 \end{align*}

\PP(\{X=0\}\cap\{Y=-1\})=\PP(\{\clubsuit\})=\frac{1}{4}

\PP(\{X=0\}\cap\{Y=1\})=\PP(\{\diamondsuit\})=\frac{1}{4}

\PP(\{X=1\}\cap\{Y=-1\})=\PP(\varnothing)=0

\PP(\{X=1\}\cap\{Y=1\})=\PP(\{\heartsuit\})=\frac{1}{4}

	$Y=-1$	$Y=0$	$Y=1$
$X=-1$	0.2	0.1	0.2
$X=1$	0.1	0.3	0.1

\PP(\{X=1\}\cap\{Y=-1\})=0.1

\begin{align*}\PP(\{X=1\})&=\underbrace{\PP(\{X=1\}\cap\{Y=-1\})}_{=0.1}+\underbrace{\PP(\{X=1\}\cap\{Y=0\})}_{=0.3}\\&+\underbrace{\PP(\{X=1\}\cap\{Y=1\})}_{=0.1}=0.5 \end{align*}

\PP(\{Y=-1\})=\underbrace{\PP(\{X=-1\}\cap\{Y=-1\})}_{=0.2}+\underbrace{\PP(\{X=1\}\cap\{Y=-1\})}_{=0.1}=0.3

$X$ и $Y$ зависимы?

Proof 1

$(\Rightarrow)$ $B_1=\{a_i\}$ , $B_2=\{b_j\}$

\underbrace{\PP(\{X\in\{a_i\}\}\cap\{Y\in\{b_j\}\})}_{=\PP(\{X=a_i\}\cap\{Y=b_j\})}=\underbrace{\PP(\{X\in\{a_i\}\})}_{=\PP(\{X=a_i\})}\cdot\underbrace{\PP(\{Y\in\{b_j\}\})}_{\PP(\{Y=b_j\})}

B_1=(B_1\cap\{a_1,a_2,\ldots, a_m,\ldots\})\sqcup(B_1\setminus\{a_1,a_2,\ldots,a_m,\ldots\})

\forall B_1, B_2\in\BB(\RR)

B_1=(B_1\cap\{a_1,a_2,\ldots, a_m,\ldots\})\sqcup \underbrace{(B_1\setminus\{a_1,a_2,\ldots,a_m,\ldots\})}_{=B_{1,0}}

B_2=(B_2\cap\{a_1,a_2,\ldots, a_m,\ldots\})\sqcup \underbrace{(B_2\setminus\{a_1,a_2,\ldots,a_m,\ldots\})}_{=B_{2,0}}

\begin{align*}\PP(\{X\in B_1\}\cap\{Y\in B_2\})&=\PP([\{X\in B_{1,1}\}\sqcup\{X\in B_{1,0}\}]\cap[\{Y\in B_{2,1}\}\sqcup\{Y\in B_{2,2}\}])\\&= \PP(\{X\in B_{1,1}\}\cap\{Y\in B_{2,1}\})+\PP(\overbrace{\{X\in B_{1,0}\}}^{=\varnothing}\cap\{Y\in B_{2,1}\})\\&+\PP(\{X\in B_{1,1}\}\cap\overbrace{\{Y\in B_{2,0}\}}^{=\varnothing})+\PP(\overbrace{\{X\in B_{1,0}\}}^{\varnothing}\cap\overbrace{\{Y\in B_{2,0}\}}^{=\varnothing}) \\&=\PP\left(\bigsqcup^\infty_{\underset{\scriptstyle i\colon a_i\in B_1}{i=1}}\{X=a_i\}\right)\cap\left(\bigsqcup^\infty_{\underset{\scriptstyle j\colon b_j\in B_2}{j=1}}\{Y= b_j\}\right) \\&=\PP\left(\bigsqcup^\infty_{\underset{\scriptstyle i\colon a_i\in B_1}{i=1}}\bigsqcup^\infty_{\underset{\scriptstyle j\colon b_j\in B_2}{j=1}}(\{X=a_i\}\cap\{Y=b_j\})\right) \\&=\sum^\infty_{\underset{\scriptstyle i\colon a_i\in B_1}{i=1}}\sum^\infty_{\underset{\scriptstyle j\colon b_j\in B_2}{j=1}}\underbrace{\PP(\{X=a_i\}\cap\{Y=b_j\})}_{=\PP(\{X=a_i\})\cdot\PP(\{Y=b_j\})} \\&=\left(\sum^\infty_{\underset{\scriptstyle i\colon a_i\in B_1}{i=1}}\PP(\{X=a_i\})\right)\left(\sum^\infty_{\underset{\scriptstyle j\colon b_j\in B_2}{j=1}}\PP(\{Y=b_j\})\right) \\&=\PP(\{X\in B_{1,1}\})\cdot\PP(\{Y\in B_{2,1}\}) \\&=\PP(\{X\in B_1\})\cdot\PP(\{Y\in B_2\}) \end{align*}

B_{1,1}=\bigsqcup^\infty_{\underset{\scriptstyle i\colon a_i\in B_1}{i=1}}\{a_i\},\quad B_{2,1}=\bigsqcup^\infty_{\underset{\scriptstyle j\colon b_j\in B_2}{j=1}}\{b_j\}

\underbrace{\PP(\{X=1\}\cap\{Y=-1\})}_{=0.1}\neq \underbrace{\PP(\{X=1\})\cdot\PP(\{Y=-1\})}_{=0.15}

$X$ и $Y$ являются зависимыми.

	$Y=-1$	$Y=0$	$Y=1$
$X=-1$	0.2	0.1	0.2
$X=1$	0.1	0.3	0.1

$X$	-1	1
$\PP_X$	0.5	0.5

\EE[X]=(-1)\times-0.5+1\times0.5=0

$X$	-1	0	1
$\PP_Y$	0.3	0.4	0.3

\EE[Y]=(-1)\times 0.3+0\times0.4+1\times0.3=0

| $X\cdot Y$ | -1 | 0 | 1 | | $\PP_{X\cdot Y}$ | $0.2+0.1=0.3$ | $0.1+0.3=0.4$ | $0.2+0.1=0.3$ |

\EE[XY]=(-1)\times 0.3+0\times0.4+1\times0.3=0

| $\max(X,Y)$ | -1 | 0 | 1 | | $\PP_{X\cdot Y}$ | $0.2+0.1=0.3$ | $0.1+0.3=0.4$ | $0.2+0.1=0.3$ |

| $\max(X, Y)$ | -1 | 0 | 1 | | :-: | :-: | :-: | | $\PP(X, Y)$ | 0.2 | 0.1 | $0.2+0.1+0.3+0.1=0.7$ |

\EE[\max(X, Y)]=(-1)\times0.2+0\times0.1+1\times0.7=0.5

\EE[\min(X, Y)]=-0.5

X+Y=(X\land Y)+(X\lor Y)

0+0=\min(X,Y)+0.5