xid-7779027_1

xid-7779027_1 - Bivariate Discrete Distributions Let X and...

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Unformatted text preview: Bivariate Discrete Distributions Let X and Y be two discrete random variables defined on a sample space S of an experiment. The joint probability mass function p(x, y) is defined for each pair of numbers (x, y) by In this class the pairs of numbers can be created as a product set and show as the margins in a two way table. The table cells can contain the joint probabilities. p(x,y) x 1 2 2 .05 .25 y 4 .15 .05 6 .20 .30 Such joint probabilities must satisfy twp conditions: Check sum of p(x, y) in table above: .05 + .15 + .20 + .25 + .05 + .30 = 1 Probabilities can be calculated for any set A of the (x, y) pairs. Example A = { (x,y) : x + y ≤ 5} = { (1,2), (1,4), ( 2,2) } p(x,y) x 1 2 y 4 .15 .05 2 .05 .25 6 .20 .30 P(A) = .05 + .15 + .25 = .45 Probabilities can be calculated for individual random variables X (or Y) Example p(x,y) x 1 2 2 .05 .25 y 4 .15 .05 6 .20 .30 p(x) .4 .6 Doing this for all possible values of X (1 and 2 below) yields the marginal probability mass function for X. p(x,y) x 1 2 y 4 .15 .05 2 .05 .25 6 .20 .30 p(x) .4 .6 The label marginal simply indicates the context of a joint distribution. A marginal pmf is like any other probability mass function. We can compute its expected value, variance and standard deviation. 1 2 .4 .6 .4 1.2 1.6 1 4 = .4 .6 .4 2.4 = 2.8 We obtain the marginal pmf for Y similarly. p(x,y) x 1 2 2 .05 .25 .30 y 4 .15 .05 .20 p(y) 6 .20 .30 .50 p(x) .4 .6 1.00 We can then compute the expected value, variance and standard deviation for y. The calculations can be done just as easily in a row format but have been put in a column format show the similarity to the x calculations above. 2 4 6 .3 .2 .5 .6 .8 3.0 4.4 4 16 36 = .3 .2 .5 1.2 3.2 18.0 = 22.4 Conditional Probability Mass Functions Knowing the outcome for X or Y limits the possible pairs of outcomes. Suppose we know that y=4. P(x,y) X 1 2 2 .05 .25 .30 y 4 .15 .05 .20 6 .20 .30 .50 x .4 .6 1.00 The conditional distribution of X|y=4 is another probability mass function. The probabilities for the possible values of X must sum to 1, so we divide the joint probabilities in the column by the marginal probability P(Y=4) = .20. 1 2 .15/.20 =.75 .05/.20 =.25 1.00 We could compute the expected value, variance and standard deviation of this probability mass function. However this time the task is to compare marginal pmf and conditional pmf for X. 1 2 .4 .6 1.0 . 75 .25 1.00 The pmfs are different. Thus knowing what Y is tells us something about pmf of X and hence X and Y are not independent. Similarly if the expected values for the marginal and conditional distributions of X differ then X and Y are not independent. 1 2 .4 .6 .4 1.2 1.6 1 2 .75 .25 .75 .50 = 1.25 Independence requires that The joint probabilities are the product of the margin probabilities. An example is below. The top left probability is .12 = .4 x .3. All the bivariate cell probabilities are the product of the respect margin probabilities. p(x,y) X 1 2 2 .12 .18 .3 y 4 .08 .12 .2 p(y) 6 .20 .30 .5 p(x) .4 .6 1.00 Expected Value of a Function of Random Variables, X and Y, For Bivariate Probability Mass Function. Let = XY. Find E(XY) using the first bivariate pmf from above. p(x,y) x 1 2 xy x 1 2 xy p(x,y) x 1 2 2 .05 .25 y 4 .15 .05 6 .20 .30 2 1 2=2 2 2=4 y 4 4 8 6 6 12 2 2 .05 = .1 4 .25 = 1.0 y 4 4 .15 = .6 8 .05 = .4 6 6 .20 = 1.2 12 .30 = 3.6 E(XY) = .1 + .6 + 1.2 + 1.0 + .4 + 3.6 = 6.9 Covariance of X and Y Cov(X,Y) = = So The pmf being used is still: p(x,y) x 1 2 2 .05 .25 Y 4 .15 .05 Previous results were E(XY) = 6.9 , E(X) = So – 6 .20 .30 = 1.6 and E(Y) = = 4.4. Correlation of X and Y This is denoted as Cor(X,Y), Previous results were and defined by 0 . ...
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This note was uploaded on 12/20/2011 for the course STAT 344 taught by Professor Staff during the Spring '08 term at George Mason.

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