Clase 7 - Independencia de variables aleatorias, cambio de variables y distribución Gamma

$Obs:$

$x$ e $y$ absolutamente continuas $\not\implies (x,y)$ absolutamente continuo.

Por ejemplo: $X \sim \mathcal{U}[0,1]$ $Y=X$ , $(x,y)=(x,x)$, $P(\underbrace{ (x,x) \in s }_{ =1 }) \neq \int_{s} f =0$

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Independencia

$x_{1},\dots,x_{n}$ son independientes.

$$P(x_{1} \in A_{1}, x_{2} \in A_{2},\dots,x_{n} \in A_{n})= \prod P(X_{i} \in A_{i}) \quad \text{Para todo }A_{1},\dots,A_{n}\text{ bolerianos}$$

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$Obs:$ Si $x_{1},\dots,x_{n}$son independientes cualquier subconjunto lo es. Si tengo infinitas variables, son independientes si toda subfamilia lo es.

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$Prop:$ Si $x_{1},\dots,x_{n}$ son independientes.

$$F_{x_{1},..,x_{n}}(t_{1},\dots,t_{n}) = F_{x_{1}}(t_{1}) \dots F_{x_{n}} ( t_{n})$$

Vale que Si $F_{X}(\overline{t})=F_{1}(t_{1})\cdot F_{2}(t_{2})\dots F_{n}(t_{n})$ con

$$\underset{ t\to \infty }{ \lim } F_{i}(t_{i})=1,$$

entonces son independientes.

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Criterio para discretas

$$x_{1},\dots,x_{n} \text{ son independientes} \iff P_{x_{1},x_{2},\dots,x_{n}}(x_{1}\dots x_{n})=P_{x_{1}}(x_{1})\dots P_{x_{n}}(x_{n})$$

$dem:$

para $n=2$ $\impliedby)$ $X\subseteq R_{X},Y\subseteq R_{Y}$

$$P(\{ x \in A \}\cap \{ Y \in B \})=\sum_{X \in A,Y \in B} P_{XY}(X,Y)=$$ $$=\sum \sum P_{X}(X)\cdot P_{Y}(Y)-\sum_{X \in A}P_{X}(A)\sum_{Y \in B}P_{Y}(Y)=$$ $$=\left( \sum_{X \in A}P_{X}(X) \right)\cdot\left( \sum_{Y \in B}P_{Y }(Y) \right)$$ $$=P(X \in A)\cdot P(Y \in B)$$

Tiro una moneda 3 veces. $Y=$ cantidad de caras

$$X=\begin{cases} 1 & \text{Cara la primera} \\ 0 & cc \end{cases}$$

Ejemplo:

Y \ X	0	1
1	$\frac{1}{9}$	$\frac{2}{9}$	$\frac{1}{3}$
2	$\frac{1}{9}$	$\frac{2}{9}$	$\frac{1}{3}$
3	$\frac{1}{9}$	$\frac{2}{9}$	$\frac{1}{3}$
	$\frac{1}{3}$	$\frac{2}{3}$

Caso continuo

$$x_{1},\dots x_{n} \text{ son independientes}\iff f_{x_{1},\dots,x_{n}} = f_{x_{1}}\cdot f_{x_{2}}\cdot\dots\cdot f_{x_{n}}$$

Más aún, si

$$f_{X}(X)=f_{1}(x_{1})\cdot f_{2}(x_{2})\dots f_{n}(x_{n}) \quad y\quad \int_{-\infty}^{+\infty} f_{1}(x) \, dx = 1$$

Entonces $f_{x_{1}}=f_{i}$ y son independientes.

Ejemplo: $(x,y)$ con densidad

$$f_{XY}(X,Y)= \frac{3}{4}\cdot X^{2}Y\cdot \mathbb{1}_{[-1,1]}(X) \cdot\mathbb{1}_{[0,2]}(Y)$$ $$=\frac{3}{4}\cdot X^{2}\cdot \mathbb{1}_{[-1,1]}(X) \cdot Y\cdot\mathbb{1}_{[0,2]}(Y)$$ $$\int_{0}^{2} Y \, dY = 2$$ $$f_{Y}(Y)= \frac{1}{2} \cdot Y \cdot\mathbb{1}_{[0,2]}(Y)\quad y\quad f_{X}(X)= \frac{3}{2} \cdot X \cdot\mathbb{1}_{[-1,1]}(X)$$

$X$ e $Y$ son independientes.

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Ejemplo 2:

$$f_{XY}(X,Y)= e^{-Y} \mathbb{1}_{(0,Y)}(X)$$ $$f_{X}(X)= e^{-X} \mathbb{1}_{(0,+\infty)}(X)$$ $$f_{Y}(Y)=Y \cdot e^{-Y} \mathbb{1}_{(0,+\infty)}(Y)$$ $$f_{XY}\neq F_{X}\cdot F_{Y}$$

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Ejemplo de vector aleatorio.

Uniforme, dado $G\subseteq \mathbb{R}^{n}$

$$X \sim \mathcal{U}\quad \text{si } f_{X}(X) = \frac{\mathbb{1}_{(G)(X)}}{vol(G)}$$

Donde $vol(G)=\displaystyle\int_{G} dx_{1}\dots dx_{n}$

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Ejemplo

$$D= \{ (x,y) \in \mathbb{R}^{2}\bigm| x^{2}+y^{2}\leq 1 \}$$ $$X \sim\mathcal{U}(D)$$ $$f_{X}(X)= \frac{1}{\pi}\cdot \mathbb{1}_{D}(X,Y)$$

$X$ e $Y$ no son independientes.

soporte rectangular no significa independencia.

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Si $(X,Y)$ es un vector discreto y $Z=g(X,Y)$

$$P_{Z}(k)=\sum_{X \in R_{X},Y \in R_{Y},g(X,Y)=k} P_{XY}(X,Y)$$

Ejemplo:

$$X \sim P(\lambda_{1}),Y\sim P(\lambda_{2}),\quad X,Y \text{ independientes}$$ $$Z=X+Y\sim P(\lambda_{1}+ \lambda_{2})$$ $$P_{Z}(k)=\sum_{s \in \mathbb{N}_{0},t \in \mathbb{N}_{0},s+t=k} P_{XY}(s,t)=\sum_{s=0}^{k} P_{X,Y}(s,k-s)$$ $$=\sum_{s=0}^{k} \frac{\lambda_{1}^{s} }{s!} \cdot e^{-\lambda_{1}} \cdot \frac{\lambda_{2}^{k-s} }{(k-s)!} \cdot e^{- \lambda_{2}}$$ $$=\frac{(\lambda_{1}+\lambda_{2})^{k} }{k!} \cdot e^{-(\lambda_{1}+\lambda_{2})}$$

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Ejemplo continuo

$$X,Y \sim\mathcal{D}[0,1] \quad \quad Z = \frac{X}{Y}$$

$X,Y$ independientes.

$$P(Z \leq t)= P \left( \frac{X}{Y}\leq t \right) = P(X\leq t \cdot Y)$$ $$f_{XY}= \mathbb{1}_{[0,1) \times[0,1)} ( X,Y)$$

Además $Y\geq \frac{X}{t}$

Si $t<1:$ Mirando el gráfico de la región vemos que

$$P(X\leq t \cdot Y)=\frac{t}{2}$$

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Teorema del cambio de variables

$U,V \in \mathbb{R}^{n}$ $g:U\longrightarrow V, C^{1}$ con inversa $C^{1}$
$A\subseteq V$

$$\int_{g ^{-1}(A)} f(X)\: dX = \int_{A} f(g ^{-1}(Y))\cdot |J_{g ^{-1}} ( Y)|\: dY$$

Si $X$ es un vector aleatorio y $f=f_{X}$

$$P(g(X) \in A) = P(X \in g ^{-1}(A)) = \int_{A} f_{X} (g ^{-1}(X)) \cdot |J_{g ^{-1}}(X)| \cdot \mathbb{1}_{V}(X) \: dY$$ $$f_{Y}(Y) = f_{X}(g ^{-1}(Y))\cdot |J_{g ^{-1}}(Y)| \cdot \mathbb{1}_{V}(Y) \cdot \mathbb{1}_{\mathrm{Im}(Y)}(Y)$$

Lo puedo usar para $X,Y \sim \mathcal{U}[0,1]$ independientes.

$$Z = \frac{X}{Y}$$

Agrego $W=Y$ así voy de $\mathbb{R}^{2}$ a $\mathbb{R}^{2}$

$$g(X,Y)= \left( \frac{X}{Y},Y \right)$$ $$g ^{-1}(Z,W) = (Z\cdot W,W)$$

$|J_{g ^{-1}}| = \begin{bmatrix}W & Z \\ 0 & 1\end{bmatrix}=W$

$$f_{ZW}(Z,W)= \mathbb{1}_{[0,1]}(Z\cdot W)\cdot \mathbb{1}_{[0,1]}(W) \cdot W$$

Para rescatar la marginal de $Z$ integro:

$$f_{Z}(Z)=\int_{-\infty}^{+\infty} f_{ZW}(Z,W) \, dW$$

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$$X,Y \sim N(0,1) \text{ independientes}$$ $$f_{X,Y}(X,Y)= \frac{1}{(\sqrt{ 2\pi })^{2}} \cdot e^{ - \frac{x^{2}}{2}} \cdot e^{- \frac{Y^{2}}{2}} = \frac{1}{2\pi} \cdot e^{\frac{-x^{2}-y^{2}}{2}} = \frac{1}{2\pi}\cdot e^{\frac{- || X || ^{2}}{2}}$$

$Obs:$ Si $A \subseteq \mathbb{R}^{2\times2}$ es ortogonal, $A\cdot A^{T}=I$

$$Z= A\begin{pmatrix} X \\ Y \end{pmatrix}$$ $$f_{Z}(Z)= \frac{1}{2\pi}\cdot e ^{- \frac{||A ^{-1} Z || ^{2}}{2}} \det(A ^{-1})= \frac{1}{\sqrt{ 2\pi }} \cdot \frac{1}{\sqrt{ 2\pi }} \cdot e^{- \frac{Z_{1}^{2}}{2}} \cdot e^{- \frac{Z_{2}^{2}}{2}}$$ $$Z_{1},Z_{2} \sim N(0,1) \quad \text{independientes.}$$

Si $|| (a,b) ||=1$ entonces $a\cdot X+b \cdot Y\sim N(0,1)$ tomo

$$A=\begin{pmatrix} a & b \\ b & -a \end{pmatrix}$$ $$Z_{1}= a\cdot X+b\cdot Y$$

Si $a,b \in \mathbb{R}$ $a \cdot X + b \cdot Y$ es normal.

Vale que si $X \sim N(\mu_X, \sigma_{X}^{2})$, $Y \sim N(\mu_{Y}, \sigma_{Y}^{2})$ . Independientes.

$$a \cdot X + b \cdot Y \text{ es normal.}$$

Escuchar audio, importante. 00:38:16 parte 2.

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Distribución $\Gamma$

$$X \sim \Gamma(\alpha,\lambda)$$ $$f_{X}(X)=\frac{\lambda ^{\alpha} }{\Gamma(\alpha)} \cdot X^{\alpha-1} \cdot e^{-\lambda X} \cdot \mathbb{1}_{(0,+\infty)}(X)$$ $$\Gamma(Y)=\int_{0}^{+\infty} X ^{Y-1} \cdot e^{-Y} \, dX$$ $$\begin{array}{c} \Gamma(Y+1)= Y \cdot P(Y) \\ \Gamma(n)=(n-1)! \\ \Gamma\left( \frac{1}{2} \right)=\sqrt{ \pi }\end{array}$$

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$$X \sim\Gamma(1,\lambda)$$ $$\begin{array}{c} X \sim\mathcal{E}(\lambda) \\ X \sim\mathcal{E}(\lambda) \\ Y \sim\mathcal{E}(\lambda) \end{array}$$

Entonces $X+Y \sim \Gamma(2,\lambda)$ , independientes $\frac{X}{X+Y} \sim \mathcal{U}[0,1]$