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Emails arrive according to a Poisson process with rate $λ=2/hour$. You check your inbox (instantly reading all new emails) at time $t=5$ hours and also at some uniformly distributed random time between 0 and 5.

Suppose that at time $t=0$, you have no unread emails. What is the probability that between $t=0$ and $t=5$, at least 4 emails were unread for more than an hour?

(Example for clarification: suppose you check your inbox at $t=2$ and $t=5$. Four emails arrived, at $t=0.5$, $t=3$, $t=3.5$, and $t=4.5$. The first three were unread for more than an hour, the last one was not.)

My thinking is to break this down into cases based on the value of the uniform random variable $u$.

Case 1: $u \in$ (0, 1], then an email is unread for more than an hour iff it arrives in $(u, 4)$, total interval length $4-u$.

Case 2: $u \in$ (1, 4), then an email is unread for more than an hour iff it arrives in $(0, u-1)$ or $(u, 4)$, total interval length 3.

Case 3: $u \in$ [4, 5), then an email is unread for more than an hour iff it arrives in $(0, u-1)$, total interval length $u-1$.

Due to stationary and independent increments, the splitting of the interval in case 2 doesn't affect the distribution, right? Then the desired probability is $1-P($0, 1, 2, or 3 emails arrive in the desired interval(s)), weighted by the probability of each case. $N(t)$, the number of events in an interval of length $t$, has a Poisson ($λt = 10$) distribution.

For the first case and third cases, since the interval length is dependent on $u$, I will find the average by integrating the sum $P(X=0)+P(X=1)+P(X=2)+P(X=3)$ over the appropriate values of $u$ (0 to 1 in the first case, 4 to 5 in the third case). For example, in the first case I used $λt = 8-2u$ and plugged this into the Poisson mass function with $k=$0, 1, 2, and 3.

The second case is simpler because the interval has constant length 3.

The desired probability should then be $1-({1\over 5}$(average of case 1) + ${3\over5}$(case 2) + ${1\over5}$(average of case 3)$)$.

Any flaws in my work here? I'm really not sure how else to approach this. Any help would be appreciated. Thank you!

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  • $\begingroup$ I don’t understand your example: if you checked at time $t=2$ (this is the " uniformly distributed random time" right?) the first e-mail, received at time $t=0.5$, is no longer unread when you check at $t=5$. $\endgroup$
    – Elvis
    Mar 2, 2015 at 9:13
  • $\begingroup$ If you assume that the value of the intermediate checking time is known, the question becomes more easy, right? You just have to integrate the answer of this easy question for a uniform intermediate time... $\endgroup$
    – Elvis
    Mar 2, 2015 at 9:15
  • $\begingroup$ The only thing that matters is whether an email was ever unread for more than an hour at any point. In the example, the message received at $t=0.5$ was not read until the inbox was checked at $t=2$, so it counts. Also, the value of the first check time isn't known. It's random, uniformly distributed between 0 and 5. $\endgroup$
    – Shalikram L.
    Mar 2, 2015 at 9:27
  • $\begingroup$ "The only thing that matters is whether an email was ever unread for more than an hour at any point". OK, my bad, I get now. "the value of the first check time isn't known. It's random, uniformly distributed between 0 and 5." Yes. Assume it's known, call it $u$, and compute your answer as a function of $u$. Then integrate it over the density of $u$. $\endgroup$
    – Elvis
    Mar 2, 2015 at 9:44
  • $\begingroup$ Your approach sounds reasonable. If the purpose is self study, I would think about how the solution would look if the fixed time is arbitrary (instead of t=5, t=T), the fixed time is a random variable itself, if you randomly check your inbox n times (instead of once) between t and T, etc ... $\endgroup$
    – Cacti
    Mar 22, 2018 at 23:11

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