CH9 - Calculate CDF from 1 variable and joint CDF

This class will continue the lesson from CH9 - Independence and Conditioning, where we will learn how to calculate the CDF of a single variable from the joint CDF of two variables.

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Example

We are given:

$$jointCDF(F_{x,y}(x,y))$$

	$y_0$	$y_1$
$x_0$	$\frac{3}{10}$	$\frac{6}{10}$
$x_1$	$\frac{5}{10}$	$\frac{10}{10}$

To find the $F_x(x)={\lim }_{y\to \infty }{F}_{x,y}(x,y)$:

note that x in this case can be either 0 or 1

$$ForX=0:F_x(o)=\underset{y\to \infty }{\lim }{F}_{x,y}(0,y)=\frac{6}{10}$$ $$ForX=1:F_x(1)=\underset{y\to \infty }{\lim }{F}_{x,y}(1,y)=\frac{10}{10}$$

Same thing for Y:

We modify the previous given equation

$$F_y(y)=\underset{x\to \infty }{\lim }{F}_{x,y}(x,y)$$ $$ForY=0:\underset{x\to \infty }{\lim }{F}_{x,y}(x,0)=\frac{5}{10}$$ $$ForY=1:\underset{x\to \infty }{\lim }{F}_{x,y}(x,1)=\frac{10}{10}$$

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Independent RV’s

If you have some independent random variables, then the joint is the product of the masses.

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

But something new to be aware of:

$$NEW\to F_{x,y}(x,y)=F_x(x)\cdot F_y(y)$$

And something we will look at later is applying this rule to given rules:

$$\begin{array}{rl}LATER\to P_{x\mid y}(x\mid y)& =P_x(x)\\ P_{y\mid x}(y\mid x)& =P_y(y)\end{array}$$