John, a lawyer, receives a client according to a Poisson distribution on average once every 15 minutes. Between the hours of 10am and 12pm on Friday, John receives 10 clients. How many clients were expected to be received by John between 9am and 1pm on that day (Friday)?
My Solutions: even though the intervals are overlapping, can I assume that they occur according to the uniform distribution of that interval? Therefore,
Since $N(t) ~ \operatorname{Poisson}(\lambda = t/15)$, and $E[N(60)]=60/15=4$, then,
$$E[9\ am \text{ to }1\ pm] = E[9-10\ am]+E[10\ am-12\ pm]+E[12-1\ pm] = 4+10+4 = 18.$$
Is this correct?
Yes: expectations add, even when variables are dependent.