The geometric random variable is defined as the number of co

The geometric random variable is defined as the number of coin flips required to get the first \"Head.\" If heads occur with probability p. determine Pr {X(u) = k} and the cdf. Show that the Geometric random variable (see problem 2.3) has the memoryless property: Pr {x(u) m + k|X(u) > m} = Pr{X(u) k}, m, k 0 integers, FX(u)(m + k|X*(u) > m). = Fx_(u)(k).

Solution

X~Geometic(p),where X is the no of trials required to get the first head and p is the probability of getting head.

The pmf of X is given by,

f(x)=q^(x-1) *p ,q=1-p ,x=1,2,3,......................

The cdf of X is given by

F(x)=P(X<=x)=sum(t=1(1)x)(f(t))=p*(1-q^x)/(1-q)=(1-q^x) ,x=1,2,3,...............

P(X_(u)=k)=P(at least u-1 X_i\'s are less than or equal to k)=P(Z>=u-1)=sum(k=u-1(1)n)[nCk*{F(u-1)}^k *(1-F(u-1))^(n-k)

where Z follows binomial(n,F(x)) and Z is the no of Xi\'s which are less than or equal to x.

2.31.

Mean=E(X)=sum(x=1(1))(x*q^(x-1) *p)=p(1+2*q+3*q²+4*q³ +............... )=p*S

where S=(1+2*q+3*q²+4*q³ +............... )

S-q*S=1+q+q²+q³+...............=1/(1-q)

therefore,S=1/(1-q)² =1/p²

Therefore E(X)=1/p

E(X²)=sum(x=1(1))(x²*q^(x-1) *p)=2/p²

Therefore V(X)=1/p²