Question about solution: Poisson process & conditional expectation 
Given the following problem:
Alice shows up at an Athena cluster at time $0$ and spends her time exclusively in typing emails. The times that her emails are sent are a Poisson process with rate $\lambda_A$ per hour. Let $Y_1$ time at which Alice’s first  email was sent. You show up at time $1$ and you are told that Alice has sent exactly one email so far. What is the conditional expectation of $Y_1$ given this information?
Solution:
Let $A$ be the event $\{$exactly one arrival in the interval $[0,1]\}$. Given $A$, the times in this interval are equally likely for the arrival $Y_1$. Thus,
$E[Y_1 | A] = \frac{1}{2}$.

I am completely lost because $Y_1 \sim\exp(\lambda_A)$, so shouldn't I just calculate expected value for exp. rv on interval $[0,1]$? Why is my approach wrong?
 A: This is actually a bayesian problem. Time $Y_1$ if you don't know the number of mails at time 1, is exponentially distributed, you got that right. But when you get the additional information that at time 1 Alice only sent one e-mail, you have to update your distribution of $Y_1$. Applying Bayes rule:
$$p(Y_1|emails_1 = 1) \propto p(emails_1 = 1|Y_1)p(Y_1).$$
So you have the exponential distribution $p(Y_1)$ that we already understand, and the other stranger thing $p(emails_1 = 1|Y_1)$ which is the probability of not having any other mail sent after the first one, until time 1, given $Y_1$. It is equal to $\int_1^\infty p(Y_2|Y_1) dY_2$ and it raises as $Y_1$ gets closer to 1. Actually, as the integral of an exponential function is exponential as well, it turns out, with few passages, that $p(emails_1 = 1|Y_1) = exp(1-Y_1)$, so:
$$p(Y_1|emails_1 = 1) \propto exp(Y_1)exp(1-Y_1) \propto 1$$
where $exp$ is the esponential distribution density, whatever is the (equal) rate parameter. Last passage is easily verifiable from multiplying those exponential densities.
Your updated distribution of $Y_1$ is uniform, and expected value can be immediately derived.

If you'd like a much shorter, more intuitive explanation, mind that Poisson process is symmetrical, and that the distribution of $Y_1$ is the same looked from 0 and from 1, so of course its expected value must be 0.5. This explanation doesn't account for the rest of its distribution though.
