Let X have an exponential distribution with mean ? > 0.

Chapter 3, Problem 3E

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QUESTION:

Let \(X\)  have an exponential distribution with mean \(\theta>0\). Show that

                   \(P(X>x+y \mid X>x)=P(X>y) \).

     

Equation Transcription:

Text Transcription:

X  

Theta > 0  

P(X > x + y | X > x) = P(X > y)

Questions & Answers

QUESTION:

Let \(X\)  have an exponential distribution with mean \(\theta>0\). Show that

                   \(P(X>x+y \mid X>x)=P(X>y) \).

     

Equation Transcription:

Text Transcription:

X  

Theta > 0  

P(X > x + y | X > x) = P(X > y)

ANSWER:

Solution :

Step 1of 3:

X have an exponential distribution with mean > 0. We have to show that

P(X> x+y| X >x )  = P(X > y).( memoryless property of exponential distribution).


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