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# 06_24_2 - U = X Y Y X 1 1 30 2 4 30 3 9 30 4 16 30 1 1 1 1...

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STAT 410 Examples for 06/24/2009 Summer 2009 ( part 2 ) If X 1 , X 2 , … , X n are n independent random variables and a 0 , a 1 , a 2 , … , a n are n + 1 constants, and U = a 0 + a 1 X 1 + a 2 X 2 + … + a n X n , then M U ( t ) = e a 0 t M X 1 ( a 1 t ) M X 2 ( a 2 t ) M X n ( a n t ) . If random variables X and Y are independent, then M X + Y ( t ) = M X ( t ) M Y ( t ) . 1. Suppose X and Y are two independent discrete random variables with the following probability distributions: p X ( 1 ) = 0.2, p X ( 2 ) = 0.4, p X ( 3 ) = 0.3, p X ( 4 ) = 0.1, p Y ( 1 ) = 0.3, p Y ( 3 ) = 0.5, p Y ( 5 ) = 0.2. Find the probability distribution of W = X + Y. M W ( t ) = M X ( t ) M Y ( t ) = + + + + + t t t t t t t e e e e e e e 5 3 4 3 2 2 . 0 5 . 0 3 . 0 1 . 0 3 . 0 4 . 0 2 . 0 = t t t t t t t t e e e e e e e e 9 8 7 6 5 4 3 2 02 . 0 06 . 0 13 . 0 19 . 0 23 . 0 19 . 0 12 . 0 06 . 0 + + + + + + + . p W ( 2 ) = 0.06, p W ( 3 ) = 0.12, p W ( 4 ) = 0.19, p W ( 5 ) = 0.23, p W ( 6 ) = 0.19, p W ( 7 ) = 0.13, p W ( 8 ) = 0.06, p W ( 9 ) = 0.02.

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2. Suppose we have two 4-sided dice. Suppose that for the first die ( X ) , p X ( 1 ) = 1 / 10 , p X ( 2 ) = 2 / 10 , p X ( 3 ) = 3 / 10 , p X ( 4 ) = 4 / 10 . Suppose also that for the second die ( Y ) , p Y ( 1 ) = 1 / 30 , p Y ( 2 ) = 4 / 30 , p Y ( 3 ) = 9 / 30 , p Y ( 4 ) = 16 / 30 . Find the probability distribution of U = X + Y.
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Unformatted text preview: U = X + Y. Y X 1 1 / 30 2 4 / 30 3 9 / 30 4 16 / 30 1 ( 1 , 1 ) ( 1 , 2 ) ( 1 , 3 ) ( 1 , 4 ) 1 / 10 2 1 / 300 3 4 / 300 4 9 / 300 5 16 / 300 2 ( 2 , 1 ) ( 2 , 2 ) ( 2 , 3 ) ( 2 , 4 ) 2 / 10 3 2 / 300 4 8 / 300 5 18 / 300 6 32 / 300 3 ( 3 , 1 ) ( 3 , 2 ) ( 3 , 3 ) ( 3 , 4 ) 3 / 10 4 3 / 300 5 12 / 300 6 27 / 300 7 48 / 300 4 ( 4 , 1 ) ( 4 , 2 ) ( 4 , 3 ) ( 4 , 4 ) 4 / 10 5 4 / 300 6 16 / 300 7 36 / 300 8 64 / 300 u p ( u ) 2 1 / 300 3 6 / 300 4 20 / 300 5 50 / 300 6 75 / 300 7 84 / 300 8 64 / 300 OR M U ( t ) = M X ( t ) ⋅ M Y ( t ) = + + + + + + ⋅ 30 16 30 9 30 4 30 1 10 4 10 3 10 2 10 1 4 3 2 4 3 2 t t t t t t t t e e e e e e e e = …...
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