Question

7. Suppose that waiting time, Y, at a particular restaurant follows an Exponential distribution with mean X, where X is a Geometric random variable with mean 1/ p. Find the unconditional mean and variance of Y.

Answer 1

mean and variance of general exponential

mean and variance of geometric

Note: Zn Exp.CA) : E(za)=52225ce3dzo Sz*de-42dz = . Je-dz.z* dz = x ge-k (kyr da azak 3 x dz=dk - dr. Sekse holder seg forsi (78e-1ms Tu, Tu =(r=)!) -- E(2) - - E(z2). LE I Van (2) -

So,

In Generat, lets say X~ Geometric (p). PMF of * : P(x+2) = (-1276, 2=1,2,.. ,O<pa! . Meandf*: E6): Z z P(x-2) = { 21-23.pl [Note that, sem of infinite Gp is. E x2:(1-)'.ocmas + { xxx!= (1-27? [ Differentiating both side wat o] (2-1)/27=241-23° [ Differentiating again untx]. E(8) = p.Ž 2(1-1,2! =P(1-(1-P): E(«(x-1)) - (-) P(x-x) = { 2(2-1)(1-43.!p. 36(1-4) Ž 2(2-3)/1342 = *(1+) 2 (1-C1-))* = 2(1-1) E(xt); E(XCX-1) + x) = E(x(x-1) +E(x) = 2(14) + ¢ = 5 Var (x) = E(x+)- É(x) = TE 5 Þ2 þz

using the concepts from above,

Law of iterated expectation : E(Y)- E (E(VIX)) here, E(/(x)= x {: Ylx~ Exch (mean= X) ] : fx xx("\x) = x E(X)= $ 73 x lyeometrie (mean =] [:-P(x=x)=4(-)] E(X)=E(E(41x)) = E(X) == -0 Law of iterated variance : Var (4) = Var (EC4\x)) + E(Var (elx)) Noo, Ylx~ Exp (*= 1/4) Var (YX)= Bert and Var (x)= 1-1 - Vaz (8)= Var (x) + E(x2) = Var (x)+ Var(x) +E*(x) = 2 (1+2+3= 3-2.1- [i: Var (x) = E(X2)- E(x) 7 Þ2