Geometric distribution

Question

Let y denote a geometric random variable with probability function

p(y)=p*(q)^(y-1) y=1,2,3,... o<=p<=1 , q = 1-p

Show that

1)for a positive integer a , P(Y>a)= q^a

And

2) for positive integers a,b

P(Y> (a+b) | Y>a) = q^b = P(Y>b)

Answer 1

Let y denote a geometric random variable with probability function

p(y)=p*(q)^(y-1) y=1,2,3,... o<=p<=1 , q = 1-p

Show that

1)for a positive integer a , P(Y>a)= q^a

The easiest way for me with a problem like this is first to find the complement of P(Y>a), which is P(Y ≤ a), and subtract it from 1.

P(Y ≤ a) = P(Y = 1) + P(Y = 2) + ... + P(Y = a)

P(Y ≤ a) = p · q⁰ + p · q¹ + ... p · q^{a - 1}

P(Y ≤ a) = p(1 + q + q² + q^{a - 1})

The part in the parenthesis is a geometric series, so it can be substituted as follows:

P(Y ≤ a) = p(1 - q^a) / (1 - q)

1 - q = 1 - (1 - p) = p

So the p's cancel, and P(Y ≤ a) = 1 - q^a

Then P(Y > a) = 1 - 1 - q^a = q^a

2) for positive integers a,b

P(Y> (a+b) | Y>a) = q^b = P(Y>b)

This is what is known as the "memoryless property" of geometric random variables, a property that it shares with exponential random variables.

The conditional probability general formula is P(A | B) = P(A ∩ B) / P(B) can be applied as follows:

P(Y> (a+b) | Y>a) = (P(Y> (a+b)) ∩ P(Y>a)) / P(Y>a)

The numerator is logically equivalent to P(Y > (a+b)) because being greater than a+b inevitably implies also being greater than a. According to the formula in part 1, P(Y > (a+b)) = q^{a + b}, and P(Y > a) = q^a.

So we can substitute as follows:

q^{a + b} / q^a = q^b