Show that X enjoys a memoryless property.

P(X>x1+x2)|P(X>x1) = P(X>x2)

P(X>x1+x2)|P(X>x1) = P(X>x1+x2 and X>x1)/P(X>x1) = P(X>x1)/P(X>x1) = 1. Hence, P(X>x1+x2) and P(X>x2).

Is this correct?

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- September 25th 2008, 04:35 AMlord12memoryless proprety
Show that X enjoys a memoryless property.

P(X>x1+x2)|P(X>x1) = P(X>x2)

P(X>x1+x2)|P(X>x1) = P(X>x1+x2 and X>x1)/P(X>x1) = P(X>x1)/P(X>x1) = 1. Hence, P(X>x1+x2) and P(X>x2).

Is this correct? - September 25th 2008, 04:58 AMmr fantastic
No.

1. The notation for the conditional probability is wrong. It should be Pr(X > x1 + x2 | X > x1).

2. The calculation is wrong. The correct result is Pr(X > x1 + x2 | X > x1) = Pr(X > x1 + x2)/Pr(X > x1).

Nothing more can be done unless the distribution for X is given. Is X meant to be a geometric random variable? If so, see Q2 b) in this thread: http://www.mathhelpforum.com/math-he...ee-proofs.html