CH9 - Independence and Conditioning

Let’s say that we have 2 discrete random variables $A$ $B$.

Therefore, we say that:

$$A,Bevents$$

and:

$$P(A\cap B)=P(A)\cdot P(B)$$

---

Joint Probability Functions

A joint probability function is a function that gives the probability of 2 events happening at the same time.

$$P_{X,Y}(x,y)=P(X=x\cap Y=y)$$

This is a probability mass function.

---

Joint Cumulative Distribution Function

A joint cumulative distribution function is a function that gives the probability of 2 events happening at the same time, but in a cumulative way.

$$F_{X,Y}(x,y)=P(X\le x\cap Y\le y)$$

---

Conditional Probability Mass Function

A conditional probability mass function is a function that gives the probability of an event happening given that another event has happened.

$$P_{X\mid Y}(x\mid y)=P(X=x\mid Y=y)=\frac{P(X=x\cap Y=y)}{P(Y=y)}$$

And thus:

$$=\frac{P_{X,Y}(x,y)}{P_Y(y)}$$

---

Example

Example

Random Employee Hiring. Ten students apply for a job opening, but only 1 of the students will be selected. The employer chooses randomly; all ten outcomes are equally likely. If person 3,5,7, or 8 gets the job, let $X=1$; otherwise $X=0$. If person 1,2,3,4, or 5 gets the job, let $Y=1$; otherwise $Y=0$. Are $X$ and $Y$ independent random variables?

1. Find the mass of $X$ and the mass of $Y$

$x$	$P_x(X)$	$y$	$P_y(Y)$
$\{1,2,4,6,8,10\}\to 0$	$\frac{6}{10}$	$\{6,7,8,9,10\}\to 0$	$\frac{5}{10}$
$\{3,5,7,9\}\to 1$	$\frac{4}{10}$	$\{1,2,3,4,5\}\to 1$	$\frac{5}{10}$

1. Find the join of $X$ and $Y$

$$P_{X,Y}(x,y)=P(X=x\cap Y=y)$$ $$P_{x,y}(0,0)=P(X=0\cap Y=0)$$

for $X_0{Y}_0$

$$\begin{array}{r}P(X=0)\cdot P(Y=0\mid X=0)\\ \frac{6}{10}\cdot \frac{3}{6}=\frac{3}{10}\end{array}$$

for $X_0{Y}_1$

$$\begin{array}{rl}{P}_{x,y}(0,1)& =P(Y=0)\cdot P(Y=1\mid X=0)\\ \frac{5}{10}\cdot \frac{3}{5}& =\frac{3}{10}\end{array}$$

for $X_1{Y}_0$

$$\begin{array}{rl}{P}_{x,y}(1,0)& =P(X=1)\cdot P(Y=0\mid X=1)\\ \frac{4}{10}\cdot \frac{2}{4}& =\frac{2}{10}\end{array}$$

for $X_1{Y}_1$

$$\begin{array}{rl}{P}_{x,y}(1,1)& =P(X=1)\cdot P(Y=1\mid X=1)\\ \frac{4}{10}\cdot \frac{2}{4}& =\frac{2}{10}\end{array}$$

Therefore

	$Y_0$	$Y_1$
$X_0$	$\frac{3}{10}$	$\frac{3}{10}$
$X_1$	$\frac{2}{10}$	$\frac{2}{10}$

1. Use $P_{xy}(x,y)=P_x(x)\cdot P_y(y)$ to determine independence

Are $X,Y$ independent RV’s?

Let’s check $P(x,y)(x,y)=P_x(x)\cdot P_y(y)$

$$\begin{array}{rl}x=0,y=0:\frac{3}{10}& {=}^{?}\frac{6}{10}\cdot \frac{5}{10}\to YES\\ andsoon\dots & \end{array}$$

---

Given Joint Probability Mass Function…

Calculate mass of 1 variable

$$P_x(x)=\sum P_{x,y}(x,y)$$

	$Y_0$	$Y_1$
$X_0$	$\frac{3}{10}$	$\frac{3}{10}$
$X_1$	$\frac{2}{10}$	$\frac{2}{10}$

$$\begin{array}{rl}{P}_x(0)& =\sum _{all{Y}^{\prime }s}{P}_{x,y}(0,y)\\ & =P_{x,y}(0,0)+P_{x,y}(0,1)\\ & =\frac{3}{10}+\frac{3}{10}=\frac{6}{10}\end{array}$$ $$\begin{array}{rl}{P}_x(1)& =\sum _{all{Y}^{\prime }s}{P}_{x,y}(1,y)\\ & =\frac{2}{10}+\frac{2}{10}=\frac{4}{10}\end{array}$$

and so on for the Y variables…

---

Given Joint Mass, Caculate CDF…

$$F_{x,y}=P(X\le x\cap Y\le y)$$

Joint(given)

	$Y_0$	$Y_1$
$X_0$	$\frac{3}{10}$	$\frac{3}{10}$
$X_1$	$\frac{2}{10}$	$\frac{2}{10}$

CDF

$$\begin{array}{rl}{F}_{x,y}(0,0)& =P(X\le 0\cap Y\le 0)\\ & =\frac{3}{10}\end{array}$$ $$\begin{array}{rl}{F}_{x,y}(0,1)& =P(X\le 0\cap Y\le 1)\\ & =P((X=0\cap Y=0)\cup (X=0\cap Y=1))\\ & =\frac{3}{10}+\frac{3}{10}=\frac{6}{10}\end{array}$$

and so on for F(1,0), F(1,1)

	$Y_0$	$Y_1$
$X_0$	$\frac{3}{10}$	$\frac{6}{10}$
$X_1$	$\frac{5}{10}$	$\frac{10}{10}$