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Suppose that X1, X2, X3 and X4 are independent Poisson where E[X1] = lab E[X2] = 11 – a)b E[X3] = da(1 – b) E[X2] = X(1 — a)(

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Answer #1

Solution iven that SUPRSE thot Poisgon wene E LX ab E LX2 =2 (1-ab E LX3) aC-b) E CX4) 1-al1-b) Let S= x1+ X2+X3+X4 CXX3 findeaxab (1-ab 201-a)b0-x 10-x) 2ab e Caabx e(-Abtab lab-abx 10-x dab 2abb PCR 1o lab cab-lab) PL1-6J 16, and R =12 S 16 and RE13 PC=6 X3+X4 4 PCX3+(4ニt) PCX1に6 ¥3こy4ニ4ーy) PCZ3 44- - 2ab 2ab = P£Lに63 R= Xt っ(2こ12 X2 ニ12- CニXt23こ lo X3 こ lo - →② SUb8tituL3+X4 2 m esuation O 3t scbetitute lo 144 L2424 ニ 14 10 Sub&titute euatio n arc 3+X42 メ2 -ズ3ニ2 →O ande の add の ae aet 2t3-2 22 2 Put 4 = 2 mesuation 3 242+0t2ニ14 X1+4こ14 メ」に14-4 01 Po8sible value of the lo cd) P(L= 1o nsこ14, Rこ(2,cこ10] P Lrt2t3t X4 1PCOL 2, X12 2) xab C2ablo (1-a01-b 2 rab 2ab2 e 21-a() 2Cta(Hb-2 C12-2 ! 2 (ac1-a1-b Caabo-2 lo-2 2 2 2! 2 12-2 10-2 Caablo-x

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