Joint Ditribution

Lecture_6 Joint Ditribution

1. Discrete Multivariate R.V.s
2. Continuous Multivariate R.V.s
3. Covariance and Correlation
4. Multinomial Distribution
5. Multivariate Normal
6. Change of Variables
7. Convolutions

Discrete Multivariate R.V.s

Foundation
$$
\begin{align}
\text{CDF: } &F_{_{X,Y}}(x,y)=P(X\leq x,Y\leq y)\\
\text{PMF: } &p_{_{X,Y}}(x,y)=P(X=x,Y=y)\\
&P(X=x)=\sum_y P(X=x,Y=y)
\end{align}
$$
Conditional PMF
For discrete r.v.s $X$ and $Y$, the conditional PMF of $X$ given $Y=y$ is

$$
P_{X|Y}(x|y)=P(X=x|Y=y)=\frac{P(X=x,Y=y)}{P(Y=y)}.
$$
E.G
Suppose a chicken lays a random number of eggs, $N$, where
and $N\sim\operatorname{Pois}(\lambda)$. Each egg independently hatches with probability $p$ and fails to hatch with probability $q=1-p$. Let $X$ be the number of eggs that hatch and Y the number eggs that hatch and Ythe number that do not hatch, so $X+Y=N.$ What is the joint PMF of $X$ and $Y?$

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Solution

Continuous Multivariate R.V.s

Conditional PDF Given an Event
- The conditional PDF $f_{X|A}$ of a continuous random variable $X$, given an event $A$ with $\mathbf{P}(A)>0$, satisfies
  $$
  \mathbf{P}(X\in B\mid A)=\int_{B}f_{X\mid A}(x)dx.
  $$
- If $A$ is a subset of the real line with $\mathbf{P} ( X\in A) > 0, $then
  $$
  f_{X|X\in A}(x)=\begin{cases}
  {\frac{f_{X}(x)}{\mathbf{P}(X\in A)}}&{\mathrm{if\ } x\in A,}\\{0,}&{\mathrm{otherwise}}.
  \end{cases}
  $$
- Let $A_1,A_2,\ldots,A_n$ be disjoint events that form a partition of the sample space, and assume that $\mathbf{P}(A_i)>0$ for all $i$. Then,
  $$
  f_{X}(x)=\sum_{i=1}^{n}\mathbf{P}(A_{i})f_{X|A_{i}}(x)
  $$
  (a version of the total probability theorem).
Joint PDF
$$
f_{X,Y}(x,y)=\frac{\partial^2}{\partial x\partial y}F_{X,Y}(x,y).
$$
Marginal PDF
$$
f_X\left(x\right)=\int_{-\infty}^\infty f_{X,Y}\left(x,y\right)dy
$$

E.Gs

The joint PDF of X and Yis given by
$$
f(x,y)=\begin{cases}\frac{12x(2-x-y)}{5}&\mathrm{if~}0<x<1,0<y<1\\
0&\mathrm{otherwise}&&\end{cases}
$$
Compute the conditional PDF of X given that $Y=y$, where $0<y<1.$

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Solution

Suppose that the joint PDF of $X$ and Yis given by

$$
f(x,y)=\begin{cases}\frac{e^{-x/y-y}}{y}&\text{if }0<x<\infty,0<y<\infty\\
0&\text{otherwise}\end{cases}
$$

Find $P \{ X>1|Y=y \} .$

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Solution

Continuous form of Bayes’ Rule and LOTP
$$
\begin{gathered}
f_{Y|X}\left(y|x\right)=\frac{f_{X|Y}\left(x|y\right)f_Y\left(y\right)}{f_X\left(x\right)}\\
f_X\left(x\right)=\int_{-\infty}^\infty f_{X|Y}\left(x|y\right)f_Y\left(y\right)dy
\end{gathered}
$$

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proof

$$
\begin{gathered}
\small\bullet \quad f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)} = \frac{f_{X|Y}(x|y)f_Y(y)}{f_X(x)}\\
\small\bullet \quad f_X(x) = \int_{-\infty}^\infty f_{X,Y}(x,y)dy = \int_{-\infty}^\infty f_{X|Y}\left(x|y\right)f_Y\left(y\right)dy
\end{gathered}
$$

E.G
A light bulb produced by the GE company is known to have an exponential distributed lifetime $Y$. However, the company has been experiencing quality control problems. On any given day, the parameter $\lambda$ of the PDF of $Y$is actually a random variable. uniformly distributed in the interval[1,3/2]. We test a ligut $\omega$ Jrd its lifetime. What we can say about the underlying parameter $\lambda?$

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Solution