BIVARIATE RANDOM VARIABLES

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We want to look at sums when we add the observations from the spinning wheel We have been concerned with a single random variable, but rst note that the observations we want to add are those from different observations and so are different random variables So we turn our attention to different random variables and rst consider two different random variables Suppose we have a sample space and we have de ned two random variables, which we will call X and Y, on the sample points Now we must determine the probabilities that the random variables assume values together We need not show the sample space, but the probabilities with which the random variables take on their respective values together are shown in Table 81 This is called the joint probability distribution function The table is to be read this way: X can take on the values 1, 2, and 3 while the random variable Y can take on the values 1 and 2 The entries in the body of the table

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Table 81 X Y 1 2 1 1/12 1/3 2 1/12 1/12 3 1/3 1/12

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Continuous Probability Distributions

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are the probabilities that X and Y assume their values simultaneously For example, P(X = 1 and Y = 2) = 1/3 and P(X = 3 and Y = 2) = 1/12

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Note that the probabilities in the table add up to 1 as they should Now consider the random variable X + Y This random variable can take on the values 2, 3, 4, or 5 There is only one way for X + Y to be 2, namely, each of the variables must be 1 So the probability that X + Y = 2 is 1/12 and we write P(X + Y = 2) = 1/12 There are two mutually exclusive ways for X + Y to be 3, namely, X=1 and Y =2 or X=2 and Y = 1 So P(X + Y = 3) = P(X = 1 and Y = 2) + P(X = 2 and Y = 1) = 1/3 + 1/12 = 5/12 It is easy to check that there are two mutually exclusive ways for X + Y to be 4 and this has probability 5/12 Finally, the probability that X + Y = 5 is 1/12 These probabilities add up to 1 as they should This means that the random variable X + Y has the following probability distribution function: 1/12 if x + y = 2 5/12 if x + y = 3 f (x + y) = 5/12 if x + y = 4 1/12 if x + y = 5 where x and y denote values of the random variables X and Y This random variable then has a mean value We nd that 1 5 5 1 7 E(X + Y ) = 2 +3 +4 +5 = 12 12 12 12 2 How does this value relate to the expected values of the variables X and Y taken separately First, we must nd the probability distribution functions of the variables alone What, for example, is the probability that X = 1 We know that P(X = 1 and Y = 1) = 1/12 and P(X = 1 and Y = 2) = 1/3 These events are mutually exclusive and are the only events for which X = 1 So P(X = 1) = 1/12 + 1/3 = 5/12 Notice that this is the sum of the probabilities in the column in Table 81 for which X = 1 In a similar way, P(X = 2) = P(X = 2 and Y = 1) + P(X = 2 and Y = 2) = 1/12 + 1/12 = 1/6, which is the sum of the probabilities in the column of Table 81 for which X = 2 Finally, summing the probabilities in Table 81 for which X = 3 is 1/3 + 1/12 = 5/12 So the probability distribution for the random variable X alone can be found by adding up the entries in the columns; the probability distribution for the random variable Y alone can be found by adding up the probabilities in the rows of Table 81

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