Let X be from a geometric distribution with probability of success p. Given that P (X > y) = (1? p) y for any positive integer y. Show that for positive integers a and b, P (X > a + X > a) = P (X > b).

Full Question

Let X be from a geometric distribution with probability of success p.

Given that P (X > a + b | X > a) = q ^ { b } = P (X > b) for any positive integer x.

Show that for positive integers a and b, P (X > a + b | X > a) = P (X > b).

Answer:

P (X > a + b | X > a) = P (X > b)

Proved - - - See Explanation

Step-by-step explanation:

Given

P (X > a + b | X > a) = q ^ { b } = P (X > b)

From the above.

We can derive the following

P (X > a), P (X > b) and P (X > a + b)

P (X > a) = q^a

P (X > b) = q^b

P (X > a + b) = q^ (a + b)

Using the definition of conditional probability

P (X > a + b | X > a) can be represented by P (X > a + b and X > a) / P (X>a)

X>a+b and X>a is equivalent to X>a+b since a+b is larger than a

So,

P (X > a + b and X > a) / P (X>a) can be rewritten as

P (X>a + b) / P (X > a)

Bringing both sides together, we're left with

P (X > a + b | X > a) = P (X>a + b) / P (X > a)

By substituton

P (X > a + b | X > a) = q^ (a+b) / q^a

P (X > a + b | X > a) = q^ (a + b - a)

P (X > a + b | X > a) = q^ (a - a + b)

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

Since P (X > b) = q^b

So,

P (X > a + b | X > a) = P (X > b)