# 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?

• Your approach wouldn't work because it doesn't account for the fact there was just one email sent during that interval. If you contemplate the continuation of the process, eventually there would have been a second email sent after time $1:$ you must also condition on this fact.
– whuber
Oct 23, 2020 at 18:52
• @whuber: As usual thank you for reply! The fact that I integrate over interval [0,1] and not [0, $\infty$] doesn't account for this? I mean I know that it happend in [0,1], I just need to get E for rv on this interval. I really can't understand how event A (information A) transform my rv from exp. into uniform... Oct 23, 2020 at 19:17

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.

• Thank you, that is what I more or less understand as explanation. Didn't you mistake in integral range, from $1-Y_1$ to $\infty$? $p(emails_1 = 1|Y_1)=P(Y_2 \ge 1-y_1)$, where $Y_2$ is also exp? Oct 24, 2020 at 11:45
• no I don't think so. $Y_2$ is the time of the second email being sent, and has an exponential distribution that starts at $Y_1$. That integral goes from 1 onwards because we need $Y_2$ to be greater than 1. Oct 24, 2020 at 12:28
• but given time $y_1$ between $[0,1]$ what we need is that $Y_2 \ge 1-y_1$ to be sure that second event doesn't happen in $[0,1]$. Oct 24, 2020 at 13:56
• @Sharov I call $Y_2$ what you call $Y_1+Y_2$ Oct 24, 2020 at 20:09