Proving the memoryless property of the exponential distribution

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To prove the memoryless property of the exponential distribution, start with the equation P(X ≤ a + b | X > a) = P(X ≤ b). Rewriting the left side as P((X ≤ a + b) ∩ (X > a)) / P(X > a) is a valid approach. From there, P((X ≤ a + b) ∩ (X > a)) can be expressed as P[X ≤ a + b] - P[X ≤ a]. This breakdown clarifies the relationship between the ranges of X, confirming the memoryless property holds true. The discussion emphasizes the importance of understanding the disjoint ranges of the exponential distribution to validate the proof.
DanielJackins
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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?
 
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P[(X ≤ a + b) ^ (X > a)] = P[X ≤ a + b] - P[X ≤ a] , right?
 
Thanks. Using that I was able to prove it. But why is what you said true?
 
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.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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