Question

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).

Answer 1

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)