Solution Found!
Let X have an exponential distribution with mean ? > 0.
Chapter 3, Problem 3E(choose chapter or problem)
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).