We have shown that if two independent Poisson streams are merged, we stillget a Poisson

Chapter 6, Problem 1

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We have shown that if two independent Poisson streams are merged, we stillget a Poisson stream. Now consider two independent renewal counting processesN1(t) and N2(t) being merged into the process N(t) [SEVC 1977]. Let the under-lying distribution functions be FX1 , FX2 , and FX. First show that, conditionedon the fact that the last event in the output stream was contributed by the firststream, the time to next event in the output stream is the minimum of X1 andY2 = Y (X2), where Y (X2) is the residual time to next event in input stream 2.Next show that xFX(t) =E[X2]FZ1 (t) + E[X1]FZ2 (t)E[X1] + E[X2]whereFZ1 (t) = FX1 (t) +1 FX1 (t)E[X2] t0[1 FX2 (x)]dxandFZ2 (t) = FX2 (t) +1 FX2 (t)E[X1] t0[1 FX1 (x)]dx .

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