Question

Let X be the random variable with the geometric distribution with parameter 0 < p < 1.

(1) For any integer n ≥ 0, find P(X > n).

(2) Show that for any integers m ≥ 0 and n ≥ 0, P(X > n + m|X > m) = P(X > n) (This is called memoryless property since this conditional probability does not depend on m.)

Answer 1

we have XNGCP), ocP<1 we have to find i For any integer nyo P(x>n) Now PCX >n) - 1-PCXC6) Now Placn) - I- P(x>n) - 1-P (x>nt) - 1 - 5 Px (6) k=nti = 1- [(1-Php + (1-Bntle + C1-pjht2p+ -] - 1- (1-pj belit (1-P) + (1-p2+ ---] - 1-C1-pin & ci-Pok k=u - 1-(1-P) P (= (1-P) - 1-(-php ( ) P(XCD) = 1 (1-P) P(x>n) = 1- P(X ch. = 1-1 + (1-P)0 P(X>n) = (1-P)

③ we have to show that for my, of nxo P (x>ntml x m ) = P(x>n) The probability mass function for roux is. f(x) - P(1-P) / The prob that X is greater than or equal to a is P(x ,x) = (1-P)* . P(X), nemlx >m) = P ( x, htm, x>, im) P(x>, m) - P(x>, ntm) P(x , m) - = ... P(x>htm 1x, m) - (-p) htm (1-P)m (1-P)h P(x>n)