A clinic is equally likely to have 2, 3, or 4 doctors volunteer for service on a

Question

A clinic is equally likely to have 2, 3, or 4 doctors volunteer for service on a given day. No matter how many volunteer doctors there are on a given day, the numbers of patients seen by these doctors are independent Poisson random variables with mean 30. Let X denote the number of patients seen in the clinic on a given day. Find the expected value of X and justify your reasoning for it. Be clear in your explanation.

Explanation / Answer

Sum of Poisson Processes: If X1(t) and X2(t) represent two independent Poisson processes, then their sum X1(t) + X2(t) is also a Poisson process with parameter Suppose that {N1(t), t = 0} and {N2(t), t = 0} are independent Poisson of rates ?1 and ?2 respectively. We want to look at the sum process where N(t) = N1(t) +N2(t) for all t = 0. In other words, {N(t), t = 0} is the process consisting of all arrivals to both process 1 and process 2. We shall show that {N(t), t = 0} is a Poisson c process of rate ? = ?1 + ?2. We show this in three different ways, first using Definition3 of a Poisson process (since that is most natural for this problem), then using Definition 2, and finally Definition 1. We then draw some conclusions about the way in which each approach is helpful. Since {N1(t); t = 0} and {N2(t); t = 0} are independent and both possess the stationary and independent increment properties, it follows from the definitions that {N(t); t = 0} also possesses the stationary and independent increment properties. Using the approximations in eqn below for the individual processes, we see that Definition 2 of a Poisson process: A Poisson counting process {N(t); t = 0} is a counting process that satisfies (2.15) (i.e., has the Poisson PMF) and has the independent and stationary increment properties. Pr( N(t, t + d)=0) = e-?d ˜ 1 - ?d + o(d) Pr(N(t, t + d)=1) = ?e-?d ˜ ?d + o(d) Pr( N(t, t + d) = 2)˜ o(d). Pr( N(t, t + d) = 0)= Pr( N1(t, t + d) = 0)*Pr( N2(t, t + d) = 0)= (1 - ?1d)(1 - ?2d) ˜1-?d. where ?1?2d2 has been dropped. In the same way, Pr( N(t, t+d) = 1)is approximated by ?d and Pr( N(t, t + d) = 2)is approximated by 0, both with errors proportional to d2. It follows that {N(t), t = 0} is a Poisson process. In the second approach, we have N(t) = N1(t) + N2(t). Since N(t), for any given t, is the sum of two independent Poisson rv’s , it is also a Poisson rv with mean ?t = ?1t + ?2t. If the reader is not aware that the sum of two independent Poisson rv’s is Poisson, it can be derived by discrete convolution of the two PMF’s . More elegantly, one can observe that we have already implicitly shown this fact. That is, if we break an interval I into disjoint subintervals, I1 and I2, the number of arrivals in I (which is Poisson) is the sum of the number of arrivals in I1 and in I2 (which are independent Poisson). Finally, since N(t) is Poisson for each t, and since the stationary and independent increment properties are satisfied, {N(t); t = 0} is a Poisson process. From the above conclusion addition of poisson gives poisson with ?=3*30=>?=90 which is the mean of the combined process.In poisson process mean and variance are same so variance=90

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