# Question

X~Exp(1)

f_{X}(x)= k*e^(-k*x)

Pr[ X>x ]= 1-F_{X} (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 ]