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!

  • $\begingroup$ 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

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

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

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