Question

independence Ex: 46 Let X, Y be independent Poisson r.v. with parameters x,, ta respectively. Compute P (2X=k 1X+ = P({X= k} n{X+Y=n} PL&X=k} ^{ Y = n-k}) v 3 PL&X+Y= n3) Pl{X+Y=n}) (EV-n-ks) 1 to. Ank. Kle ni la ik' (n-kel! Continen) Gent This is Binth, t)! n - V tylne - (), the) HMW: By using the interpretation of Poisson & Binomial random variables, could we have guessed this result!

Answer 1

LIL y be independent Poisson rov. Parameters ,,de respectively ic. xnp(a) l You P (A2) 1. Aga 12 t =e ni - Ply=y) = é 12 *! y ! ; y=0,1,2,- ; X=0,1,2,--- K /: X & Y are P(x=k|x+y=n) = plx=k nx+y=n] P(x+y=n) = p[x= kn Y= n -x] - P[x+y=n] - P[x=x) -P[Y=nu] 1. independent Note that, - if xnp (ai) and youP(22) then by additive property of Poisson distribution, Xry ~P(2, +12) =) P(x+y=n] = e utda) (title)" n!

eve thes -do k EKIX+y:n ndz nok ел, (n-k)! rence for K! - P(xemplarson) 8 (4t-B) (Antal)" = my mum! (m.)" en ni in-k | di+22 litda which is the pont of Binomial distribution with parameters ditaz Here, K - 0,1,2,--.n Hence, If X and Y are two independent Poisson variates, then the conditional distribution of X given x+y is binomial.