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)
