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Chapter 3: Problem 23

Let \(X\) have a geometric distribution. Show that $$ P(X \geq k+j \mid X \geq k)=P(X \geq j) $$ where \(k\) and \(j\) are nonnegative integers. Note that we sometimes say in this situation that \(X\) is memoryless.

Short Answer

Expert verified
The expression \(P(X \geq k+j | X \geq k) = P(X \geq j)\) shows the memoryless property of the geometric distribution, meaning that the remaining time until an event happens does not depend on how much time has already elapsed. The geometric distribution is the only discrete distribution that has this property.

Step by step solution

01

Understand the problem requirements

Given a random variable \(X\) which follows geometric distribution. We have to show that the conditional probability \(P(X \geq k+j | X \geq k)\) equals the marginal probability \(P(X \geq j)\). This characteristic is known as memorylessness.
02

Break down the left hand side

According to the definition of conditional probability, \(P(X >= k+j | X \geq k) = P(X \geq k+j , X \geq k) / P(X \geq k)\). Now because \(X\) can't be greater than or equal to \(k+j\) without also being greater than or equal to \(k\), this is same as \(P(X \geq k+j) / P(X \geq k)\).
03

Applying geometric distribution properties

In a geometric distribution, the probability that \(X\) is greater than or equal to some value \(x\) is given by \(1-(1-p)^{x}\) where \(p\) is the success probability. So, \(P(X \geq k+j) = 1-(1-p)^{k+j}\) and \(P(X \geq k) = 1-(1-p)^{k}\).
04

Substitute and Simplify

Substitute the expressions of \(P(X \geq k+j)\) and \(P(X \geq k)\) in the earlier equation: \(P(X \geq k+j | X \geq k) = [1-(1-p)^{k+j}] / [1-(1-p)^{k}]\). Simplify this to get \(P(X \geq k+j | X \geq k) = 1-(1-p)^{j}\), which equals \(P(X \geq j)\). Thus the memorylessness property of geometric distribution is proved.

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Most popular questions from this chapter

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