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Unformatted text preview: APPM 4/5520 Solutions to Problem Set One 1. The binomial random variable X counts the number of successes in n trials where p is the probability of success on any single trial. So, Y = n X represents the number of failures in n trials. If we rename “failure” as “success”, we see that we still have a binomial distribution, but now the “success” (failure) probability is 1 p . Let’s check this “analytically”: f Y ( y ) = P ( Y = y ) = P ( n X = y ) = P ( X = n y ) = parenleftBigg n n y parenrightBigg p n y (1 p ) y I { , 1 , 2 ,...,n } ( n y ) Now since parenleftBigg n n y parenrightBigg = n ! ( n y )![ n ( n y )]! = n ! ( n y )! y ! = n ! y !( n y )! = parenleftBigg n y parenrightBigg , and I { , 1 , 2 ,...,n } ( n y ) = I { , 1 , 2 ,...,n } ( y ) , (since n y = 0 ⇒ y = n , n y = 1 ⇒ y = n 1, ... n y = n ⇒ y = 0) we have that f Y ( y ) = parenleftBigg n y parenrightBigg (1 p ) y p n y which is the form of a binomial pdf with parameters n and 1 p . Y ∼ bin ( n, 1 p ) 2. Y = g ( x ) = ln x implies that x = g 1 ( y ) = e y which has derivative e y . The pdf for X ∼ unif (0 , 1) is f X (...
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 Fall '11
 Manuel
 Normal Distribution, Probability theory, CDF

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