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19 Joint Dist II - Joint Distributions April to Click 12...

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Click to edit Master subtitle style 9/7/10 Joint Distributions April 12
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9/7/10 I ndependence of Random Variables Let’s compare conditional and marginal distributions for this example by finding: PX|Y(X|Y=1), PX| PX|Y(X|Y=2) P(X=1|Y=2) = 0.05/0.2 = 0.25 P(X=2|Y=2) = 0.1/0.2 = 0.5 P(X=3|Y=2) = Y 1 2 3 1 .1 .05 .1 .25 X 2 .2 .1 .2 .5 3 .1 .05 .1 .25 .4 .2 .4
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9/7/10 Independence Notice that in this example, all the conditional distributions and the marginal distribution of X are all the same! The reader can verify that the same is true for the conditional and marginal distributions of Y as well. What does this mean practically? It means that X is not effected by Y. No matter what the value of Y is, the
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9/7/10 Independence More formally, two random variables with joint distribution PX,Y(x,y) are said to be independent if any of the following 3 (equivalent) conditions hold: 1. PX,Y(x,y) = P(X=x)*P(Y=y) for all x and y 2. PX|Y(x|Y=y) = P(X=x) for all x and y such that P(Y=y) > 0
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9/7/10 Example Referring to the joint distribution given in the previous table, prove that X and Y are independent. The easiest way to prove this is to show that condition one in the definition of independence holds (i.e. the joint equals the product of the marginals). We can construct the product of the marginals as follows: Y 1 2 3 1 .1 .05 .1 .25 X 2 .2 .1 .2 .5 3 .1 .05 .1 .25 .4 .2 .4
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9/7/10 Review We have defined “Joint”, “Marginal” and “Conditional” distributions, and independence: 1.
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