Question

If X is a geometric random variable, show analytically that P(X = n+k|X >n) = P(X = k) .Give a verbal argument using the interpretation of a geometricrandom variable as to why the preceding equation is true.A geometric random variable is a discrete random variable with aprobability mass function defined as follows:p(n)=(1-p)n-1p, n=1,2,....where 0 < p < 1.

Answer 1

P[(X=n+k)] ∩ X>n)] =P[X=K]

Explanation:

If X is a geometric random variable then

for success probability = p

so for failure q= 1-p

Now as given

`P_(x)(n)=(1-p)^(x-1)P_{}`

Now for the P parameter we have

x∈{1,2,3,....∞}

`P [X=K] = (1-P)^(K-1)P_{}`

`P [X=n+K] = (1-P)^(n+K-1)P_{}`

`p[(X=n+K)/(X>h)]=([(X=n+K)n,X>n])/(P(X>n))`

`P(X>n)=\sum_(i=n+1)^(\infty)(1-P)^(i-1)P^{}  \\`

`P(X>n)=P\sum_(i=n+1)^(\infty)(1-P)^(i-1)^{}`

`P(X>n)=P(1-P)/(1-(1-P)^(n) ) =(1-P)^(n)`

P[(X=n+k)] ∩ X>n)] = P(X=n+K)

P[(X=n+k)] ∩ X>n)] =
`((1-P)^(n+k-1)P )/((1-P)^(n) )`

P[(X=n+k)] ∩ X>n)] =
`(1-P)^(k-1) .P`

P[(X=n+k)] ∩ X>n)] =P[X=K]