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Widgets of Type A arrive with Poisson Process with arrival rate $\lambda_A$, and, for Type B, with arrival rate $\lambda_B$ (independent).

During t, there have been b arrivals of Type B. What are the expected arrivals of Type A+B in time frame t?

Does one simply take the given value b, and add to that the expected arrivals for process A?:

$b+t \times \lambda_A$

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Assuming that both processes are independent, you can do like that. otherwise, it is inconclusive unless some auxiliary conditions are assumed. –  sos440 Oct 22 '12 at 6:56
    
thanks but why do you not post this as an answer, but as a comment? –  Wuschelbeutel Kartoffelhuhn Oct 22 '12 at 7:00
    
I usually post a comment when I think the answer lacks the required details to be an answer... –  sos440 Oct 22 '12 at 7:39
    
Wuschel: You could transform this into an acceptable question by adding the information @sos440 suggested. –  Did Oct 22 '12 at 13:58
    
@did Ok i added the indep. assumption –  Wuschelbeutel Kartoffelhuhn Oct 23 '12 at 4:16

1 Answer 1

up vote 1 down vote accepted

In general, $N_t=N^A_t+N^B_t$ implies $\mathbb E(N_t\mid N^B_t=b)=\mathbb E(N^A_t\mid N^B_t=b)+b$. If furthermore the processes $N^A$ and $N^B$ are independent, then $\mathbb E(N^A_t\mid N^B_t=b)=\mathbb E(N^A_t)=\lambda_At$.

Thus, in your setting, $\mathbb E(N_t\mid N^B_t=b)=b+\lambda_At$.

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