Proving the memoryless property of the exponential distribution

[DanielJackins]

Given that a random variable X follows an Exponential Distribution with paramater β, how would you prove the memoryless property?

That is, that P(X ≤ a + b|X > a) = P(X ≤ b)

The only step I can really think of doing is rewriting the left side as [P((X ≤ a + b) ^ (X > a))]/P(X > a). Where can I go from there?

 

[haruspex]

P[(X ≤ a + b) ^ (X > a)] = P[X ≤ a + b] - P[X ≤ a] , right?

 

[DanielJackins]

Thanks. Using that I was able to prove it. But why is what you said true?

 

[haruspex]

X has three disjoint ranges, <a, (a,a+b), >a+b.
P[(X ≤ a + b) ^ (X > a)] is the probability X is in the middle range.
P[(X ≤ a + b)] is the probability X is in the first or middle range.
P[X ≤ a] is the probability X is in the first range.

 

[blue_raver22]

To prove the memoryless property of the exponential distribution, we can use the definition of conditional probability and the properties of the exponential distribution.

First, let's rewrite the left side of the equation as follows:

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

Next, we can use the definition of conditional probability to rewrite the numerator:

P(X ≤ a + b and X > a) = P(X ≤ a + b | X > a) * P(X > a)

Then, using the definition of the exponential distribution, we can substitute in the probability density function for X:

P(X ≤ a + b | X > a) = (1 - e^-(a+b)/β) * e^-a/β

And for the second term, we can use the complementary cumulative distribution function:

P(X > a) = 1 - P(X ≤ a) = 1 - (1 - e^-a/β) = e^-a/β

Substituting these values back into the original equation, we get:

P(X ≤ a + b|X > a) = [(1 - e^-(a+b)/β) * e^-a/β] / e^-a/β

Simplifying, we get:

P(X ≤ a + b|X > a) = 1 - e^-b/β

Which is equivalent to P(X ≤ b), proving the memoryless property of the exponential distribution.

In summary, we have used the definition of conditional probability, the properties of the exponential distribution, and basic algebraic manipulations to prove the memoryless property of the exponential distribution.