Answer to Question #112321 in Statistics and Probability for S3

Question

Show that the Geometric distribution r.v satisfies the equation
P(X > i + j|X > i) = P(X > j)

Answer 1

$A \text{ is an event}, p=P(A).\\ \text{Trials are conducted until }A\text{ occurs}.\\ P\{X=k\}=(1-p)^{k-1}p, k=1,2,\ldots\\ P\{X>i+j|X>i\}=\frac{P\{X>i+j,X>i\}}{P\{X>i\}}=\frac{P\{X>i+j\}}{P\{X>i\}}=\\ =\frac{\sum_{k=i+j}^\infty (1-p)^kp}{\sum_{k=i}^\infty (1-p)^kp}=\frac{\frac{(1-p)^{i+j}p}{1-(1-p)}}{\frac{(1-p)^{i}p}{1-(1-p)}}=\frac{(1-p)^{i+j}}{(1-p)^i}=(1-p)^j.\\ \text{Here first we used the formula of conditional probability}\\ \text{and then the formula of sum of geometric progression}.\\ P\{X>j\}=\sum_{k=j}^{\infty} (1-p)^k p=\frac{(1-p)^jp}{1-(1-p)}=(1-p)^j\\ \text{(we used the formula of sum of geometric progression)}.$