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 ]
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 ]
