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| + | [[Category:probabilities]] | ||
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X~Exp(1) | X~Exp(1) | ||
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What is Pr[ X>x+t | X>t ] | What is Pr[ X>x+t | X>t ] | ||
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We think it is Pr[ X>x ] | We think it is Pr[ X>x ] | ||
Pr[ X>x+t | X>t ] = Pr[ {X>t} <math> \cap </math> {X>x+t} ] / Pr[ X>t ] = Pr[ X>x+t ] / Pr[ X>t ] = e^(-k*(x+t)) / e^(-k*t) = e(-k*t) = Pr[ X>x ] | Pr[ X>x+t | X>t ] = Pr[ {X>t} <math> \cap </math> {X>x+t} ] / Pr[ X>t ] = Pr[ X>x+t ] / Pr[ X>t ] = e^(-k*(x+t)) / e^(-k*t) = e(-k*t) = Pr[ X>x ] | ||
| + | ---- | ||
| + | [[Main_Page_ECE302Fall2008sanghavi|Back to ECE302 Fall 2008 Prof. Sanghavi]] | ||
Latest revision as of 13:19, 22 November 2011
Question
X~Exp(1)
fX(x)= k*e^(-k*x)
Pr[ X>x ]= 1-FX (x)= e^(-k*x)
say we know X>t
What is Pr[ X>x+t | X>t ]
Answer
We think it is Pr[ X>x ]
Pr[ X>x+t | X>t ] = Pr[ {X>t} $ \cap $ {X>x+t} ] / Pr[ X>t ] = Pr[ X>x+t ] / Pr[ X>t ] = e^(-k*(x+t)) / e^(-k*t) = e(-k*t) = Pr[ X>x ]
