Expectation of $Y = [X-2|X>2]$ where $X$ is an exponential random variable?

Question

Random variable $X$ follows the exponential distribution with mean $2$.Define $Y = [X-2|X > 2]$.

Then value of $E(Y) = ?$

I was thinking of this $E(Y) = E(X-2)=E(X) - 2 = 2-2 = 0$,Why is this wrong?

Also how can "memorylessness " property of Exponential random variable help me in this case?

I saw that from memoryless property of exponential r.v - $P(X>s+t) = P(X>s) P(X>t)$.How can i use this here?

Answer 1

With "memoryless property" we find (third equality):

$$P(X-2>x\mid X>2)=\frac{P(X-x>2\wedge X>2)}{P(X>2)}$$$$=\frac{P(X>2+x)}{P(X>2)}=\frac{P(X>x)P(X>2)}{P(X>2)}=P(X>x)$$