Probability Theory proof help?

[slaux89]

Homework Statement

Let X be a geometric(p) random variable for some p ∈ (0, 1). For n, k ∈ N, show that
P(X = n + k |X > n) = P(X = k).
Now, explain this property using the interpretation of X as the first successful trial
among independent trials each of probability p.

Homework Equations

The Attempt at a Solution

Here's my attempt at the proof. Is this a rigorous enough proof?

Since X is a geometric (p) random variable and since we are given that x>n, we know that the first n trials are failures. Since the first n trials are failed trials and an assumption of the geometric distribution is that X is independent, then we can treat x=0 without changing the probability of P(X = n + k |X > n) = P(X = k). Thus P(X = 0 + k |X > 0) = P(X = k). Since X>0 is always true, we can eliminate this condition and thus. P(X = n + k |X > n) = P(X = k)

 

[lippyka]

is true. To explain the property using the interpretation of X as the first successful trial among independent trials, we can say that the probability of a successful trial on n+k trials is equal to the probability of a successful trial on k trials. This is because the first n trials are failures and thus do not affect the probability of a successful trial on the remaining k trials.