Let X be a discrete random variable with values in N = {1,2,...}. Prove that if...

 

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Let X be a discrete random variable with values in N = {1,2,...}. Prove that if X is a Geometric random variable then the following property, known as the memoryless property, holds: P(X=n+m| X > n) = P(X = m) for all m, n € N. Conversely, show that if the memoryless property holds, then X is a Geo- metric random variable with parameter p = P(X = 1). Hint: To show that the memoryless property implies that X is Geomet- ric with parameter p = P(X = 1) you need to prove that the p.m.f. of X is P(X = m) = P(X = 1) (1 - P(X = 1))m-1 = p(1 - p)m-1 for all m € N. It indeed holds for m = 1. For m > 1 use the memoryless property to show that P(X = m|X 1) = P(X = m - 1). Use this to get a recursion for P(X= m) whose solution you then need to determine. Alternatively, it suffices to verify that² P(X > m) = P(X > 1)m = (1 − p)m for all m € Z+ := NU {0}. This indeed holds for me {0, 1} and for larger m use the memoryless property to show that P(X > m X > m − 1) = P(X > 1). Use this to get a recursion for P(X > m) whose solution you then need to determine. Let X be a discrete random variable with values in N = {1,2,...}. Prove that if X is a Geometric random variable then the following property, known as the memoryless property, holds: P(X=n+m| X > n) = P(X = m) for all m, n € N. Conversely, show that if the memoryless property holds, then X is a Geo- metric random variable with parameter p = P(X = 1). Hint: To show that the memoryless property implies that X is Geomet- ric with parameter p = P(X = 1) you need to prove that the p.m.f. of X is P(X = m) = P(X = 1) (1 - P(X = 1))m-1 = p(1 - p)m-1 for all m € N. It indeed holds for m = 1. For m > 1 use the memoryless property to show that P(X = m|X 1) = P(X = m - 1). Use this to get a recursion for P(X= m) whose solution you then need to determine. Alternatively, it suffices to verify that² P(X > m) = P(X > 1)m = (1 − p)m for all m € Z+ := NU {0}. This indeed holds for me {0, 1} and for larger m use the memoryless property to show that P(X > m X > m − 1) = P(X > 1). Use this to get a recursion for P(X > m) whose solution you then need to determine.