2
$\begingroup$

Question: Alice shows up at time zero and spends her time exclusively in typing emails. The times that her emails are sent are a Poisson process with rate $λ_A$ per hour. And Bob just finished exercising (without email access) and sits next to Alice at time $1$. He starts typing emails at time $1$, and fires them according to an independent Poisson process with rate $λ_B$.

Given that a total of 10 emails were sent during the interval $[0, 2]$, what is the probability that exactly $4$ of them were sent by Alice?

The given answer is here: Given_Answer

I have a very different answer and I could not understand the answer.

My answer: $$ P( 4 \ sent\ by\ Alice | total\ 10\ is\ sent) = \sum_{k = 0}^{4} e^{-\lambda_A}\frac{\lambda_A^k}{k!} (\frac{\lambda_A}{\lambda_A + \lambda_B})^{4-k} (\frac{\lambda_B}{\lambda_A + \lambda_B})^6$$

explaination: There are 4 kind of probabilities for Alice to send email during which time interval: $0$ during $[0,1]$, $4$ during $[1,2]$; $1$ during $[0,1]$, $3$ during $[1,2]$; and $2$, $3$, $4$ by analogy. During$[0,1]$, the event is the Poisson process with $\lambda_A$; during$[1,2]$, the event is merged Posisson process with$\lambda_A + \lambda_B$, the probability that Alice sends emails is $\frac{\lambda_A}{\lambda_A + \lambda_B}$ and Bob $\frac{\lambda_B}{\lambda_A + \lambda_B}$ .

I don't know why I am wrong, and how $2\lambda_A + \lambda_B$ happens in the given answer.

Could you give an explanation to me? Thanks!

$\endgroup$
3
  • $\begingroup$ It is best to start by a clean use of the definition of conditional probability, as is done in the given answer (a minor typo in the given answer: the denominator should be "$e^{-(2\lambda_A + \lambda_B)}$" not "$e^{-2\lambda_A + \lambda_B}$"). Your answer is not considering the fact that, given there are a total of 10 sent, that skews the distribution on the number Alice sends. $\endgroup$ – Michael Jun 16 '16 at 23:49
  • $\begingroup$ Alice spends twice as long emailing as Bob; hence the combined rate of emails sent in time $(0;2]$ is $2\lambda_A+\lambda_B$ $\endgroup$ – Graham Kemp Jun 16 '16 at 23:49
  • $\begingroup$ @Michael Thanks! I know where I am wrong $\endgroup$ – stander Qiu Jun 17 '16 at 4:58
2
$\begingroup$

Another way of looking at it is :

Any email sent in the interval has an independent probability $p=\dfrac{2\lambda_A}{2\lambda_A+\lambda_B}$ of being one sent by Alice, because Alice sends emails at a constant average rate per unit-interval of $\lambda_A$ over two unit-intervals, and Bob sends them at rate per unit-interval of $\lambda_B$ over just the later one unit-interval.

Thus $\mathsf P(A_{(0;2]}=4\mid A_{(0;2]}+B_{(1;2]}=10) = \binom{10}{4} p^{4}(1-p)^6$ is the probability that four emails in the interval were sent by Alice given that 10 were sent by either Alice or Bob. (It is a Binomial Distribution.)


Your attempt to sum over the number of emails sent by Alice in the first minute appears to be:

$$\sum\limits_{k=0}^4 \mathsf P(A_{(0;1]}{=}k)\,\mathsf P(A_{(1;2]}{=}4{-}k\mid A_{(1;2]}{+}B_{(1;2]}{=}10{-}k)$$

This is not the required probability.   Rather $$\begin{align}\mathsf P(A_{(0;2]}{=}4\mid A_{(0;2]}{+}B_{(1;2]}{=}10) ~=&~ \sum\limits_{k=0}^4 \mathsf P(A_{(0;1]}{=}k\mid A_{(0;2]}{+}B_{(1;2]}{=}10)\,\mathsf P(A_{(1;2]}{=}4{-}k\mid A_{(0;2]}{+}B_{(1;2]}{=}10)\end{align}$$

$\endgroup$
1
  • $\begingroup$ Oh! I see ! Thank you!!!!! $\endgroup$ – stander Qiu Jun 17 '16 at 4:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.