AP Notes, Outlines, Study Guides, Vocabulary, Practice Exams and more!

Conditional Probability Distribution

Conditional Probability Distribution

The conditional probability distribution of a discrete random variable X, given a second discrete random variable Y, is the probability distribution of X conditioned on the fact that Y assumes a value y. The conditional probability distribution of X given Y is denoted by pX|Y(x|y) and is defined as

5rv021

Conditional probabilities, like simple probabilities, also sum up to 1, i.e.,

5rv022

If X and Y are independent random variables, then

5rv023

i.e., the probability distribution of X is not affected by the value of Y, and vice-versa.

EX. The random variables X and Y have the following joint probability distribution:

(x, y) pX,Y(x, y)
   
(0,0) 2/9
(0, 1) 1/3
(0, 2) 1/15
(1, 0) 2/9
(1, 1) 2/15
(2, 0) 1/45

The marginal distributions px(x) and py(y) can be calculated as follows:

5rv024

= px,y(x,0) + px,y(x,1) + px,y(x,2)

px(0) = pX,Y(0,0) + pX,Y(0,1) + pX,Y(0,2)

= 2/9 + 1/3 + 1/15
= 28/45

px(1) = pX,Y(1,0) + pX,Y(1,1) + pX,Y(1,2)

= 2/9 + 2/15
= 16/45

px(2) = pX,Y(2,0) + pX,Y(2,1) + pX,Y(2,2)

= 1/45 

 5rv025

py(0) = pX,Y(0,0) + pX,Y(1,0) + pX,Y(2,0)

= 2/9 + 2/9 + 1/45
= 7/15

py(1) = pX,Y(0,1) + pX,Y(1,1) + pX,Y(2,1)

= 1/3 + 2/15
= 7/15

py(2) = pX,Y(0,2) + pX,Y(1,2) + pX,Y(2,2)

= 1/15

The conditional probability of X given Y is calculated as follows:

5rv026

px(0|0) = (15/7) pX,Y(0,0) = (15/7) (2/9) = 10/21
px(1|0) = (15/7) pX,Y(1,0) = (15/7) (2/9) = 10/21
px(2|0) = (15/7) pX,Y(2,0) = (15/7) (1/45) = 1/21

5rv027

px(0|1)= (15/7) pX,Y(0,1) = (15/7) (1/3) = 5/7
px(1|1)= (15/7) pX,Y(1,1) = (15/7) (2/15) = 2/7
px(2|1)= (15/7) pX,Y(2,1) = (15/7) (0) = 0

5rv028

px(0|2) = 15 pX,Y(0,2) = 15 (1/15) = 1
px(1|2) = 15 pX,Y(1,2) = 15 (0) = 0
px(2|2) = 15 pX,Y(2,2) = 15 (0) = 0

The conditional probability of Y given X is calculated as follows:

5rv029

5rv030

py(0|0) = (45/28) pX,Y(0,0) = (45/28)(2/9) = 5/14
py(1|0) = (45/28) pX,Y(0,1) = (45/28)(1/3) = 15/28
py(2|0) = (45/28) pX,Y(0,2) = (45/28)(1/15) = 3/28

5rv031

py(0|1) = (45/16) pX,Y(1,0) = (45/16)(2/9) = 5/8
py(1|1) = (45/16) pX,Y(1,1) = (45/16)(2/15) = 3/8
py(2|1) = (45/16) pX,Y(1,2) = (45/16)(0) = 0

5rv032

py(0|2) = 45 pX,Y(2,0) = 45 (1/45) = 1
py(1|2) = 45 pX,Y(2,1) = 45 (0) = 0
py(2|2) = 45 pX,Y(2,2) = 45 (0) = 0

 

Subject: 
Statistics
Subject X2: 
Statistics

Need Help?

We hope your visit has been a productive one. If you're having any problems, or would like to give some feedback, we'd love to hear from you.

For general help, questions, and suggestions, try our dedicated support forums.

If you need to contact the Course-Notes.Org web experience team, please use our contact form.

Need Notes?

While we strive to provide the most comprehensive notes for as many high school textbooks as possible, there are certainly going to be some that we miss. Drop us a note and let us know which textbooks you need. Be sure to include which edition of the textbook you are using! If we see enough demand, we'll do whatever we can to get those notes up on the site for you!