Question

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

The memoryless property of \(X\) having a geometric distribution has been confirmed by finding that \(P(X \ge k+j | X \ge k) = P(X \ge j)\), for all nonnegative integers \(k\) and \(j\). This was done by using the definition of conditional probability and the properties of a geometrically distributed variable.

Step-by-step solution

Know the definitions

Recall that a variable \(X\) has the geometric distribution if it models the number of trials needed to get the first success in independent Bernoulli trials. The variable is memoryless if \(P(X \ge k+j | X \ge k) = P(X \ge j)\), where \(k\) and \(j\) are nonnegative integers - this is what needs to be proved. Also recall the definition of conditional probability: \(P(A|B) = P(A \cap B) / P(B)\) whenever \(P(B) \ne 0\).

Set up the problem

We start with the left side of the equation, \(P(X \ge k+j | X \ge k)\). By definition of conditional probability, this is \(P(X \ge k+j \, \text{and} \, X \ge k) / P(X \ge k)\). However, \(X \ge k+j\) implies \(X \ge k\), so the event \(X \ge k+j \, \text{and} \, X \ge k\) is actually just \(X \ge k+j\). Thus, \(P(X \ge k+j | X \ge k) = P(X \ge k+j) / P(X \ge k)\).

Use the geometric distribution properties

Next, we need to use properties of the geometric distribution. Recall that if \(X\) is a geometrically distributed variable with success probability \(p\), then \(P(X \ge n) = (1-p)^{n-1}\), for \(n \ge 1\). Therefore, \(P(X \ge k+j | X \ge k) = (1-p)^{k+j-1} / (1-p)^{k-1}\). This simplifies to \(P(X \ge k+j | X \ge k) = (1-p)^j = P(X \ge j)\). This is the desired result, confirming the memoryless property of the geometric distribution.