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7.5.5 Random variables N and K have the joint PMF 100” e-100 Pn.K(n, k) = { (n+1)! n=0,1,...; k=0,1,...,n, otherwise. (a) Fin
(c) Express the random variable E[K|N] as a function of N and use the iterated ex- pectation to find E[K).
math   Statistics-And-Probability   
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Solution s roomē 100 neo, 1, 2,...; k=0, 1, 2,...n Prvik (mk) - (141)! otherwise öll PMF of N Par(n)= P(N=ng NiK(n,K) R=o 100EN: n 100 SO n! n |оо - 100 е all (n-1)) mal a100 гоо е EMO EMO where m=n-1 mso m! 1 mo m! - 100 (12 = 100Conditional PMF PRIN (MIN) = PKN (kin) PN (M) n 100 - 100 e (n+1)! | oo е K: 0, 1,4--,n ntl Paliw coins for a room 721 r(km)E[KIN] = ECK] - ELECKIN] ] = E(^] = EU (100) = 50 Argi

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