Show that if X ∼ Geom(p) then P(X = n + k|X > n) = P(X = k), for every n, k ≥ 1. This one of the ways to define the memoryless property of the geometric distribution. It states the following: given that there are no successes in the first n trials, the probability that the first success comes at trial n + k is the same as the probability that a freshly started sequence of trials yields the first success at trial k.

Accepted Solution

Since [tex]X\sim\mathrm{Geom}(p)[/tex], [tex]X[/tex] has PMF[tex]P(X=x)=\begin{cases}(1-p)^{x-1}p&\text{for }x\in\{1,2,3,\ldots\}\\0&\text{otherwise}\end{cases}[/tex]By definition of conditional probability,[tex]P(X=n+k\mid X>n)=\dfrac{P(X=n+k\text{ and }X>n)}{P(X>n)}[/tex][tex]X[/tex] has CDF[tex]P(X\le x)=\begin{cases}0&\text{for }x<1\\1-(1-p)^x&\text{for }x\ge1\end{cases}[/tex]which is useful for immediately computing the probability in the denominator:[tex]P(X>n)=1-P(X\le n)=(1-p)^n[/tex]Meanwhile, if [tex]X=n+k[/tex] and [tex]k\ge1[/tex], then it's always true that [tex]X>n[/tex], so[tex]P(X=n+k\text{ and }X>n)=P(X=n+k)=(1-p)^{n+k-1}p[/tex]Then[tex]P(X=n+k\mid X>n)=\dfrac{(1-p)^{n+k-1}p}{(1-p)^n}=(1-p)^{k-1}p[/tex]which is exactly [tex]P(X=k)[/tex] according to the PMF.