Question

X is a Poisson random variable of parameter 3 and Y an exponential random variable of parameter 3. Suppose X and Y are independent. Then A Var(2X + 9Y + 1) = 22 B Var(2X + 9Y + 1) = 7 CE[2X2 + 9Y2] = 19 D E[2X2 + 9Y2] = 26

Answer 1

IT. We have and by X~ Poisson (3) and y~ Exponential (3) • YN Exponential (3), we mean the pdf of Y is fy (y) = 8 303 goo. given by otherwise ECY) = y.seby ons let 3y=t dt 0 dy = ft x 7 1 3 I telat 6 t. 2 (where Th= fx et la convi 3 11 1 [n = (n-1)!] 3 13 ELY": "Jigaz-ely, let t=34 dt- dy o DO dt 0 e at 13 34 2. 3 ... Var (Y) = ECV') - (EY!) ()* 2 9 9

and X. Poisson (3) E (X) = 3 and Var (X) = 3 → ECX"). Var (x) +(EX))" 3 + 9 = 12 ... Var (2X+9Y+1) = Var (2x+94) [: variance remains unchanged with shift of origin] P = 4 Var (X) + 9 + Vas (Y) + 2x2x9x Cov (X,Y) - 4 Var (x) + 81 Var (Y)+ 0 [1' X med Yare independent & Cor (X,Y)=0 = 4x3 + 81 x L 12 + 9 =21 .. Both (A) and (B) are incorrect Now, E (2x+979) = 2 E(X) + 9 E(Yl) 2 x 12 + 9x 2 24+2 = 26 ..(c) is incorrect and (D) is correct.