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Let us suppose a scenario with two clients, $a$ and $b$, each one generating load at rate $\lambda_a$ and $\lambda_b$, respectively. The server receives the requests from both clients.

What will be the arrival rate at server?

(consider inter-arrival times independently generated by an exponential distribution)

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    $\begingroup$ The answer depends on whether the client request generation processes are independent processes or dependent processes. In the former case, the answer is $\lambda_a+\lambda_b$; in the latter, it is not. $\endgroup$
    – Dilip Sarwate
    Aug 31, 2015 at 14:25
  • $\begingroup$ @DilipSarwate, consider they are independent processes. Could you give an example of why the answer, then, is the sum of the rates? (thank you) $\endgroup$
    – Lourenco
    Aug 31, 2015 at 15:52

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Let $A$ and $B$ denote the number of requests in $(0,T]$ from clients $a$ and $b$ relatively. Then, $A$ and $B$ are Poisson random variables with parameters $\lambda_aT$ and $\lambda_bT$ respectively, and you have declared them to be independent random variables. Thus, $C = A+B$, the total number of requests received during $(0,T]$ is a Poisson random variable also with parameter $(\lambda_a+\lambda_b)T$. This is because for any $N \geq 0$, \begin{align} P\{C = N\} &= P\{A+B = N\}\\ &= \sum_{n=0}^N P\{A= n, B = N-n\}\\ &= \sum_{n=0}^N P\{A= n\}P\{B = N-n\}&{\scriptstyle{\text{because $A$ and $B$ are independent random variables}}}\\ &= \sum_{n=0}^N \exp(-\lambda_aT)\frac{(\lambda_aT)^n}{n!} \exp(-\lambda_bT)\frac{(\lambda_bT)^{N-n}}{(N-n)!}\\ &= \exp(-(\lambda_a+\lambda_b)T)\sum_{n=0}^N \frac{(\lambda_aT)^n}{n!}\frac{(\lambda_bT)^{N-n}}{(N-n)!}\\ &= \frac{\exp(-(\lambda_a+\lambda_b)T)}{N!} \sum_{n=0}^N \binom{N}{n}(\lambda_aT)^n(\lambda_bT)^{N-n}\\ &= \exp(-(\lambda_a+\lambda_b)T)\frac{((\lambda_a+\lambda_b)T)^N}{N!} \end{align} When $A$ and $B$ are not independent, this does not work because the step above where independence was used is no longer valid.

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  • $\begingroup$ Supposing $(0,T]$ is the time in which the samples happen, why let the Poisson random variables' parameters be $\lambda_a T$ and $\lambda_b T$ instead of $\lambda_a$ and $\lambda_b$? $\endgroup$
    – Lourenco
    Sep 1, 2015 at 2:42
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    $\begingroup$ $\lambda_a$ is the request rate, measured in number of requests per unit time. So, the average number of arrivals in $(0,T]$ is $\lambda_aT$, not just $\lambda_a$. $\endgroup$
    – Dilip Sarwate
    Sep 1, 2015 at 2:45
  • $\begingroup$ Yes, I got it. For instance, I also expand this question here. Could you comment there also? (thank you very much) $\endgroup$
    – Lourenco
    Sep 1, 2015 at 2:48
  • $\begingroup$ Considering the same proof you provided, the case of $\lambda_a = \lambda_b$ still valid, isn't it? (sorry for the naive confirmation) $\endgroup$
    – Lourenco
    Sep 1, 2015 at 18:08

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