Because the definition fails for these two particular

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Because the definition fails for these two particular values, X and Y are not independent. Intuitively, this makes sense: if the sum X is large then we know that Y must be large, too. There outcomes are not “independent.” 23
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EXAMPLE ( Continuous ): In the example f X , Y x , y 6 xy 2 on the unit square, we showed f X x 2 x and f Y y 3 y 2 . It follows that f X x f Y y 2 x  3 y 2 f X , Y x , y for all 0 x , y 1, and so X and Y are independent. With most applications of independence we begin by assuming certain random variables are independent, and then we perform calculations under that assumption. 24
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EXAMPLE : Consider a group of 20 million adults. Suppose each person plays a lottery 50 times a year and the probability of winning the lottery on any attempt is 1/100,000 .00001. We first compute the probability that a particular individual wins the lottery at least twice. 25
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Let X j be the number of times person j wins the lottery. Then X j has the Binomial 50,.00001 distribution for each j . Further, P X j 2 1 P X j 0 P X j 1 1 .99999 50 50 .00001  49 .0000001225 or slightly better than a one in 10 million chance. 26
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We can get this value from Stata (version 10 or higher) by using the binomialtail function to compute P X j 2 : . di binomialtail(50,2,.00001) 1.225e-07 We get the same answer, of course, by computing 1 P X j 1 : . di 1 - binomial(50,1,.00001) 1.225e-07 27
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Next we compute the probability that at least on person in the group of 20,000,000 wins the lottery at least twice: P X 1 2or X 2 2 or . .. or X 20,000,000 2 1 P X 1 1 and X 2 1 and . .. and X 20,000,000 1 1 P X 1 1  P X 20,000,000 1 28
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Because of the many decimal places involved, we let Stata do the calculation: . di 1 - (binomial(50,1,.00001))^20000000 .91363876 So there is about a 91% chance that we see at least one person win the lottery twice, even though the chance that any particular person wins at least twice is very small. Statisticians have referred to this phenomenon as the “Law of Very Large Numbers.” 29
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An Important Fact about Independence Suppose that X 1 , X 2 ,..., X m are independent random variables. Let g 1 X 1 , g 2 X 2 , ..., g m X m be any (measurable) functions of X 1 X m . Then g 1 X 1 , g 2 X 2 g m X m are also independent. Proof : The general proof is easiest. If we let Y j g j X j then we have to show that for any Borel sets B 1 , B 2 B m , P Y 1 B 1 Y m B m P Y 1 B 1  P Y m B m But using the definition of g j 1 B j , 30
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P Y 1 B 1 ,..., Y m B m P g 1 X 1 B 1 g m X m B m P X 1 g 1 1 B 1 X m g m 1 B m  P X 1 g 1 1 B 1  P X m g m 1 B m  P g 1 X 1 B 1  P g 1 X 1 B 1
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Because the definition fails for these two particular...

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