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Unformatted text preview: ma3160a2 ma4535a2 ma5160a2 ma5610a2 Random Variables 1. A and B play a series of games with A winning each game with probability 3 5 . The process continues until a player has won two more games than the other. The overall winner is the first player to have won two more games than the other. (a) Find the probability that A gets the overall winner within the first 4 games. (b) Find the probability that A is the overall winner. (c) Let X denote the number of games played until a player gets the overall winner. Find ( 29 E X . Solution: (a) Let AA denote the event that A wins the first and the second games, etc. Observe that gets the overall winner within the first 4 games A AA ABAA BAAA = ∪ ∪ and , , AA ABAA BAAA are disjoint. Then ( 29 ( 29 ( 29 ( 29 ( 29 gets the overall winner within the first 4 games or or 3 3 3 2 3 3 2 3 3 3 225 54 54 333 5 5 5 5 5 5 5 5 5 5 625 625 P A P AA ABAA BAAA P AA P ABAA P BAAA = + + = + + = × + × × × + × × × = = (b) Let E denote the event that A is the overall winner. Observe that E AA AB BA ⊂ ∪ ∪ and , , AA AB BA are disjoint. ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 3 3 3 2 2 3 1 5 5 5 5 5 5 6 6 9 13 9 9 1 25 25 25 25 25 13 P E P AA P E AA P AB P E AB P BA P E BA P E P E P E P E P E = + + = × × + × × + × × ⇒ = ⇒ = ⇒ = (c) Observe that , , , AA AB BA BB is a partition of events. ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 E X P AA E X AA P AB E X AB P BA E X BA P BB E X BB = + + + . See that ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2, 2 2, 2 2, 2 E X AA E X AB E X E X E X BA E X E X E X BB = = + = + = + = + = , Thus, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 9 6 6 4 2 2 2 2 25 25 25 25 6 6 18 12 12 8 13 50 1 2 25 25 25 25 13 E X E X E X E X E X E X = × + × + + × + + × + + + ⇒ = ⇒ = ⇒ = 2. A random variable X has cdf X F . Find, in terms of the cdf X F , the cdf Y G of b aX Y + = , where a and b are real constants. Solution: For a , ( 29 ( 29 ( 29 ( 29 Y X y b y b G y P Y y P aX b y P aX y b P X F a a = ≤ = + ≤ = ≤ = ≤ = . For < a , ( 29 ( 29 ( 29 ( 29 ( 29 1 1 lim Y X y b x a y b y b G y P Y y P aX b y P aX y b P X P X F x a a →  = ≤ = + ≤ = ≤ = ≥ =  < =  For = a , we have b Y = . Then ( 29 ( 29 Y G y P Y y = ≤ = if b y < and ( 29 ( 29 1 Y G y P Y y = ≤ = if b y ≥ . 3. Let X be a random variable with pdf ( 29 1 1 0 otherwise X x x f x  < < = and 2 Y X = . (a) Determine the cdf of Y . 1 (b) Determine the pdf of Y . Solution: First we have 0 1 Y ≤ < (a) For 1 y ≥ , ( 29 1 Y F y = . For y < , ( 29 Y F y = ....
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 Spring '07
 CWWoo
 Probability, dx, lim FX, cdf FX, lim Fc

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