Question

Xj are Random Variables which are continous and independent, means of Xj are 4 (E[ Xj ]=4), PDFs(Probability density function) of Xj are exponantial and they have the same PDFs (Probability density function). Let N is a RV(random variable) with Poisson PMF(probability mass function), whose mean is 7.(E[N]=7) Y = X1 + … + XN. If you find a function of Y which produces moment from table or by using a software like Matlab, Find PDF(Probability density function), Mean[E] and Variance(Var) of function of Y without process method.

Answer 1

g=lun sedaj XN Given X; ~ Exponential (mean = 4) indeperdita The pay of xj is $(*) where I >o .4 Hiyo otherwise It is a RV with Poisson PMF with meant The ole N=0,1,2 == O ping of wis an ut otherwise Y = x;+ xy + + XN N I Xi LEI Р[Ү42] P[Y<y | N=n] P[N-n] PL I xi syy Nun] P[N=n] - Hy LE 1-0 to y n An n-XZ nt-u e Z e udz Mn) n n=0 O whore Z=3x; w Gamuda wa Pdf of Yis. $(3) = Иә d P[YS] dy dy rin) - AZ ht te An e Z e le dz nEO ni nyo

Mean of Y +X en Enx E(Y) = ELĖ (Y/N=n)] Now E(Y/Ar=n) = E(x,+X,+ + Xalatan = E(x, +X ₂ + . nEXI) [Since Xi's are independents and identically distribulos .: E = ELME(x)} ELE (Y N:o)] E [ N E(X)] = E(N) ELX:) . 1 7X4 = 28 - Nar Y = Narixit n var (x Var (EC^N=r)) + E[ Vor (YNun) Now Vor (YN=n) = Var (x + x2 + = Var (%;+x,+ + XN|N=n xi+ .. +Xn) + Xo) due to indipendence ( 0) due to id var (Y) = Var (NE(X)) + E [ N Var (x:]] vas (N) + Elar) Var (x) Vedebey 16% 7 224, :. EX + 7x16 =