Memorylessness of exponential and expectation Suppose I have a teller who has servicing time that is exponential with mean of $2$ minutes. Say customer $A$ arrives at noon and begins being serviced by the teller. What is the expected length of time that $A$ will be in the system if $A$ is still in the system at 12:05?
Let $T_A$ denote the time customer $A$ is in the system (i.e. from arrival at noon until departure) in minutes. I have written the question as the expression:
$$E[T_A|T_A \geq 5]$$
Now the question is, can I apply the memorylessness property of the exponential distribution to simply write $E[T_A|T_A \geq 5] = E[T_A] = 2$? The other method I have is explicitly writing out the expectation:
$$E[T_A|T_A \geq 5] = \int_{0}^{\infty}xf(x|x\geq5)dx = \int_{5}^{\infty}xf(x|x\geq5)dx$$
and then apply memorylessness to get
$$\int_{5}^{\infty}xf(x|x\geq5)dx = \int_{5}^{\infty}xf(x)dx = 7e^{-5/2}\approx 0.57$$
My uneasiness in this question I suppose is the expression $E[X|X\geq5]$ since I am only used to working with things like $E[X|Y=y]$ that utilizes a different random variable.
 A: I think you're making a mistake in the application of memorylessness. The key idea is that if $X$ is memoryless then the probability of waiting $a+b$ units given that we've waited $a$ units is the same as just waiting $b$ units. So if we've waited $5$ minutes and want to know how long we're likely to wait from here, the expected value of the time remaining is just the original expected value plus the wait time.
For your more direct way of computing this, you're missing a piece in the conditional density. Working out the CDF of $X \mid X \geq t$ we have
$$
F_t(x) := P(X \leq x \mid X \geq t) = \frac{P(t \leq X \leq x)}{P(t \leq X)} = \frac{P(X \leq x) - P(X \leq t)}{P(t\leq X)}
$$
so differentiating w.r.t. $x$ to get the conditional PDF gives us
$$
f_t(x) = \frac{f(x)}{P(X \geq t)}\mathbf 1_{x \geq t}
$$
which is just the unconditional pdf normalized by the measure of the new support.
This means if we do this directly we'll have
$$
\text E[X \mid X \geq t] = \int_{t}^\infty x \frac{f(x)}{P(X \geq t)}\,\text dx \\
= e^{t/\lambda}\int_t^\infty x \cdot \frac 1\lambda e^{-x/\lambda}\,\text dx \\
= e^{t/\lambda} \cdot (\lambda + t)e^{-t/\lambda} \\
= \lambda + t.
$$
In your example you can see that you have the correct factor of $\lambda + t$ there, you just missed the division by $P(X \geq5)$ to cancel out the other term.

Contitioning on events with positive probability like $\{X \geq t\}$ is actually easier to reason about than things like $\text E[X \mid Y=y]$ even though we're more used to that, since we can just use the standard rules for probabilities that we learned in our first intro to probability class. Really it's $\text E[X\mid Y=y]$ that's the weird thing, since if $Y$ is continuous then we're conditioning on an event that has zero probability of happening (like in this example, the probability of a wait time of exactly 5 is zero). It takes a lot more work to be able to rigorously deal with this kind of thing (that's where the formalism of probability with measure theory becomes helpful).
