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]
