Poisson process of sending e-mails

Question

Suppose a person receives and sends e-mails according to a Poisson Process with independent intensities mu and lambda respectively. He reads a received e-mail with probability p and deletes the mail unread with probability 1-p. Suppose that person receives just one e-mail in the last hour. Suppose further that that person checks his e-mail according to a Poisson Process with intensity r. What is the probability of him having deleted that one e-mail before the end of the hour?

Answer 1

E-mails arrive at a person's account according to a Poisson process $\{A(t),t \ge 0\}$ with rate $\mu$ per hour.He checks his email account at points in time that form a Poisson process $ \{C(t),t \ge 0\}$ with rate $ r.$ At each time that he checks, he processes (keeps or deletes) all accumulated e-mails (probability $1-p$ of deletion.).The way the problem is stated it seems to imply that if an e-mail is read it is not processed again at a subsequent check. Therefore, if we are interested in finding out if an e-mail is deleted, we only need consider the first checkpoint after the e-mail's arrival. (An extension of this model will be discussed later.)

We know exactly 1 email was received during time $(0,1].$ The conditional distribution of that e-mail's arrival time is Uniform$(0,1)$ because we can show its CDF is

$$P(A(t)=1 \text{ | }A(1)=1)= \frac{P(A(t)=1,A(1)=1)}{P(A(1)=1)} $$ $$ =\frac{P(A(t)=1,A(1)-A(t)=0)}{P(A(1)=1)} $$ $$=\frac{P(A(t)=1)P(A(1-t)=0)}{P(A(1)=1)} $$ $$=\frac{\mu t e^{-\mu t}e^{-\mu (1-t)}}{\mu e^{-\mu}} = t $$

for any $0<t<1.$ Let $U\sim $ Uniform$(0,1)$ be this e-mail's arrival time. Only those e-mail checks that occurred after $U$ have a chance to delete it. Let $D$ denote the event that the e-mail was deleted. We first want to evaluate $P(D\text { | }C(1)-C(U)=k).$ Note that for $k=0$ we have no chance of deletion and for if $k\ge 1$ we have probability $1-p.$

Then $$P[C(1)-C(U)\ge 1]=1-P[C(1)-C(U)=0]=1-P[C(1-U)=0]=1-P[C(U)=0]$$ since $U$ and $1-U$ have the same distribution. Evaluating this: $$1-\int_0^1 \frac{e^{-ru} (ru)^0}{0!}f_U(u)du=1-[1-e^{-r}]/r $$

So now we have $P(D)=$ $$P[D|C(1)-C(U)=0]P[C(1)-C(U)=0]+P[D|C(1)-C(U)\ge 1]P[C(1)-C(U)\ge 1] $$
$$ =(1-p)(1-\frac1r [1-e^{-r}] ) $$ That is our final answer.

Alternate interpretation: Suppose at each check we see all the e-mails that have $not$ been previously deleted. For each of these e-mails, we get another chance to now delete it with probability $1-p.$ We still assume that we know that only one e-mail arrived in $(0,1]$ and we did not have any undeleted e-mails present at $t=0.$ So now the difference is that additional checks are potentially beneficial since we get another chance to delete that one e-mail. (If an e-mail is ever deleted it stays deleted.) If there are $k$ checks made on the e-mail, the probability that the e-mail is deleted is $ 1-p^k .$

$$P(D\text{ | }C(1)-C(U)=k)=\int^1_0 P(D|C(1)-C(u)=k,U=u)du$$ $$P(D)= \int^1_0 \sum_{k=0}^\infty (1-p^k )\frac{e^{-r(1-u)}(r(1-u))^k}{k! }du $$ $$ = 1-\int_0^1 \sum_{k=0}^{\infty} p^k\frac{e^{-rv}(rv)^k}{k!}dv $$ $$ = 1-\frac{1-e^{-r(1-p)}}{r(1-p)}$$