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Задача 4

Y=−1Y=-1Y=0Y=0Y=1Y=1
X=−1X=-10.20.10.2
X=1X=10.10.30.1

Подзадача D

Подзадача А

E[X]=(−1)×0.2+(−1)×0.1+(−1)×0.2+(1)×0.1+(1)×0.3+(1)×0.1=0\EE[X]=(-1)\times0.2+(-1)\times0.1+(-1)\times0.2+(1)\times0.1+(1)\times0.3+(1)\times0.1=0

Задача 5

Y=−1Y=-1Y=0Y=0Y=1Y=1
X=−1X=-10.20.10.2
X=1X=10.10.30.1

Подзадача А

D(X)=E[X2]−[EX]2=1−02=1D(X)=\EE[X^2]-[\EE X]^2=1-0^2=1
E[X2]=(−1)2×0.2+(−1)2×0.1+(−1)2×0.2+(1)2×0.1+(1)2×0.3+(1)2×0.1=1\begin{align*}\EE[X^2]&=(-1)^2\times0.2+(-1)^2\times0.1+(-1)^2\times0.2\\&+(1)^2\times0.1+(1)^2\times0.3+(1)^2\times0.1=1\end{align*}

Подзадача С

cov(X,Y)=E[(X−E[X])⋅(Y−E[Y])]=д/зE[XY]−E[X]E[Y]\cov(X,Y)=\EE[(X-\EE[X])\cdot(Y-\EE[Y])]\overset{\text{д/з}}{=}\EE[XY]-\EE[X]\EE[Y]

Подзадача D

corr(X,Y)=cov(X,Y)D(X)D(Y)\corr(X,Y)=\frac{\cov(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)}}
∣corr(X,Y)∣≤1|\corr(X,Y)|\leq1
∣cov(X,Y)∣≤D(X)D(Y)\boxed{|\cov(X,Y)|\leq\sqrt{D(X)}\sqrt{D(Y)}}
D(Y)=0  ⟹  Y=C  ⟹  cov(X,Y)=cov(X,C)=0D(Y)=0\implies Y=C\implies\cov(X,Y)=\cov(X,C)=0
D(X±Y)=D(X)+D(Y)±2cov(X,Y)D(X\plusmn Y)=D(X)+D(Y)\plusmn 2\cov(X,Y)
D(X+Y)=cov(X+Y,X+Y)=cov(X,X+Y)+cov(Y,X+Y)=cov(X,X)+cov(X,Y)+cov(Y,X)+cov(Y,Y)=D(X)+D(Y)+2cov(X,Y)\begin{align*}D(X+Y)&=\cov(X+Y,X+Y)=\cov(X,X+Y)+\cov(Y,X+Y) \\&=\cov(X,X)+\cov(X,Y)+\cov(Y,X)+\cov(Y,Y)\\&=D(X)+D(Y)+2\cov(X,Y) \end{align*}

t∈Rt\in\RR

g(t)=D(X+t⋅Y)=D(X)+D(tY)−2cov(X,tY)=D(X)+t2D(Y)−2tcov(X,Y)=D(Y)t2−2cov(X,Y)⋅t+D(X)≥0\begin{align*}g(t)&=D(X+t\cdot Y)=D(X)+D(tY)-2\cov(X,tY) \\&=D(X)+t^2D(Y)-2t\cov(X,Y) \\&=D(Y)t^2-2\cov(X,Y)\cdot t+D(X)\geq 0 \end{align*}
Диск=4cov2(X,Y)−4D(Y)D(X)≤0\text{Диск}=4\cov^2(X,Y)-4D(Y)D(X)\leq 0
cov2(X,Y)≤D(Y)D(X)\cov^2(X,Y)\leq D(Y)D(X)
∣cov(X,Y)∣≤D(X)D(Y)|\cov(X,Y)|\leq\sqrt{D(X)}\sqrt{D(Y)}
E[XY]=(−1)×(−1)×0.2+(−1)×0×0.1+(−1)×1×0.2+1×(−1)×0.1+1×0×0.3+1×1×0.1=0\EE[XY]=(-1)\times(-1)\times0.2+(-1)\times0\times0.1+(-1)\times1\times0.2+1\times(-1)\times0.1+1\times0\times0.3+1\times1\times0.1=0

Задача 6

Подзадача B

FX,Y(x,y)=P({X≤x}∩{Y≤y})F_{X,Y}(x,y)=\PP(\{X\leq x\}\cap\{Y\leq y\})
FX,Y(−1,−0.2)=P({X≤−1}∩{Y≤0.2})=0.2+0.1F_{X,Y}(-1,-0.2)=\PP(\{X\leq -1\}\cap\{Y\leq0.2\})=0.2+0.1

Подзадача F

FX,Y(2,3)=P({X≤2}∩{Y≤3})=1F_{X,Y}(2,3)=\PP(\{X\leq 2\}\cap\{Y\leq 3\})=1

Задача 7

Подзадача А

P({X=−1}∣{Y=0})=P({X=−1}∩{Y=0})P({Y=0})=0.10.1+0.3=14=0.25\PP(\{X=-1\}|\{Y=0\})=\frac{\PP(\{X=-1\}\cap \{Y=0\})}{\PP(\{Y=0\})}=\frac{0.1}{0.1+0.3}=\frac{1}{4}=0.25

Подзадача B

P({Y=0}∣{X=−1})=P({Y=0}∩{X=−1})P({X=−1})=0.10.2+0.1+0.2=15=0.2\PP(\{Y=0\}|\{X=-1\})=\frac{\PP(\{Y=0\}\cap\{X=-1\})}{\PP(\{X=-1\})}=\frac{0.1}{0.2+0.1+0.2}=\frac{1}{5}=0.2

Подзадача С

Таблица условного распределения случайной величины YY при условии {X=−1}\{X=-1\}

yy-101
$\PP({Y=y}{X=-1})$0.40.2

Подзадача D

E[Y∣{X=−1}]=(−1)×0.4+0×0.2+1×0.4=0\EE[Y|\{X=-1\}]=(-1)\times0.4+0\times0.2+1\times0.4=0

Подзадача E

D(Y∣X=−1)=E[Y2∣X=−1]⏟=0.8−(E[Y∣X=−1]⏟=0)2=0.8D(Y|X=-1)=\underbrace{\EE[Y^2|X=-1]}_{=0.8}-(\underbrace{\EE[Y|X=-1]}_{=0})^2=0.8

Листок 2, Задача 1

X∼Pois(λ),Y∼Pois(μ)X\sim\text{Pois}(\lambda),Y\sim\text{Pois}(\mu) — независимо

Z=X+Y∼  ?Z=X+Y\sim\ \, ?

k∈{0,1,2,…}k\in\{0,1,2,\ldots\}

P({Z=k})=P(⨆j=0k({X=j}∩{Y=k−j}))=∑j=0kP({X=j}∩{Y=k−j})=∑j=0kP({X=j})⋅P({Y=k−j})=∑j=0kλjj!e−λ⋅μk−j(k−j)!e−μ=1k!e−(λ+μ)∑j=0kk!j!(k−j)!(kj)λiμk−i=1k!e−(λ+μ)(λ+μ)k\begin{align*}\PP(\{Z=k\})&=\PP(\bigsqcup^k_{j=0}(\{X=j\}\cap\{Y=k-j\})) \\&=\sum^k_{j=0}\PP(\{X=j\}\cap\{Y=k-j\}) \\&=\sum^k_{j=0}\PP(\{X=j\})\cdot\PP(\{Y=k-j\}) \\&=\sum^k_{j=0}\frac{\lambda^j}{j!}e^{-\lambda}\cdot\frac{\mu^{k-j}}{(k-j)!}e^{-\mu}=\frac{1}{k!}e^{-(\lambda+\mu)}\sum^k_{j=0}\boxed{\frac{k!}{j!(k-j)!}}_{{k\choose j}}\lambda^i\mu^{k-i} \\&=\frac{1}{k!}e^{-(\lambda+\mu)}(\lambda+\mu)^k \end{align*}
ξ∼Pois(λ)  ⟺  defP({ξ=k})=λkk!e−λ,k∈{0,1,2,…}\boxed{\xi\sim\text{Pois}(\lambda)\overset{def}\iff\PP(\{\xi=k\})=\frac{\lambda^k}{k!}e^{-\lambda},\quad k\in\{0, 1, 2, \ldots\}}

Листок 2, Задача 2

X1∼Be(p),X2∼Be(p)X_1\sim\text{Be}(p),X_2\sim\text{Be}(p) — независимы

Y=X1+X2∼Bi(2,p)Y=X_1+X_2\sim\text{Bi}(2,p)

k=0,1,2k=0,1,2

P({Y=0})=P({X1=0∩{X2=0}})=P({X1=0})⋅P({X2=0})=(1−p)2\begin{align*}\PP(\{Y=0\})&=\PP(\{X_1=0\cap\{X_2=0\}\}) \\&=\PP(\{X_1=0\})\cdot\PP(\{X_2=0\})=(1-p)^2 \end{align*}
P({Y=1})=P({X1=0}∩{X2=1})+P({X1=1}∩{X2=0})=P({X1=0})⏟1−pP({X2=1})⏟p+P({X1=1})⏟p⋅P({X2=0})⏟1−p=2p(1−p)=(21)p1(1−p)1\begin{align*}\PP(\{Y=1\})&=\PP(\{X_1=0\}\cap\{X_2=1\})+\PP(\{X_1=1\}\cap\{X_2=0\}) \\&=\underbrace{\PP(\{X_1=0\})}_{1-p}\underbrace{\PP(\{X_2=1\})}_p+\underbrace{\PP(\{X_1=1\})}_p\cdot\underbrace{\PP(\{X_2=0\})}_{1-p} \\&=2p(1-p)={2\choose 1}p^1(1-p)^1 \end{align*}
P({Y=2})=P({X1=1}∩{X2=1})=…=p2=(22)p2(1−p)0\PP(\{Y=2\})=\PP(\{X_1=1\}\cap\{X_2=1\})=\ldots=p^2={2\choose 2}p^2(1-p)^0

Листок 2, Задача 2

X∼fX(x),Y∼Be(p)X\sim f_X(x), Y\sim \text{Be}(p) — независимая

Z=X+Y∼  ?Z=X+Y\sim\,\ ?

FZ(z)=P({Z≤z})=P({Z≤z}∩{Y=0})+P({Z≤z}∩{Y=1})=P({Z≤z}∩{Y=0})+P({Z≤z}∩{Y=1})=P({X≤z}∩{Y=0})+P({X≤z−1}∩{Y=1})=P({X≤z})⋅P({Y=0})+P({X≤z−1})⋅P({Y+1})=FX(z)(1−p)+FX(z−1)p\begin{align*}F_Z(z)&=\PP(\{Z\leq z\})=\PP(\{Z\leq z\}\cap\{Y=0\})+\PP(\{Z\leq z\}\cap\{Y=1\}) \\&=\PP(\{Z\leq z\}\cap\{Y=0\})+\PP(\{Z\leq z\}\cap\{Y=1\}) \\&=\PP(\{X\leq z\}\cap\{Y=0\})+\PP(\{X\leq z-1\}\cap\{Y=1\}) \\&=\PP(\{X\leq z\})\cdot\PP(\{Y=0\})+\PP(\{X\leq z-1\})\cdot\PP(\{Y+1\}) \\&=F_X(z)(1-p)+F_X(z-1)p \end{align*}

Дифференцируем:

fZ(z)=fX(z)(1−p)+fX(z−1)p\boxed{f_Z(z)=f_X(z)(1-p)+f_X(z-1)p}
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Семинар 8, 08.11.2024
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