# hw1sol - APPM 4/5520 Solutions to Problem Set One 1 The...

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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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hw1sol - APPM 4/5520 Solutions to Problem Set One 1 The...

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