Problem 9. Let Xi, Xj, Xa, Xa be independent random variables with common mean,

Question

Problem 9. Let Xi, Xj, Xa, Xa be independent random variables with common mean, variance, and transform, denoted by EX,var(X), and Mx (s), respectively. Let Y be a random variable that is independent of Xi, Xa, Xa, Xa, and is associated with the transform My () Each part of this problem introduces a new random variable either as a function of Xi, X, Xa. Xa, and Y, or as a transform defined in terms of Mx (s) and My (s). For each part, determine the mean and variance of the new random variable. (b) V-0.25(X1 + X2 + X3 + Xs). (d) R-4XY (e) Mo(s)Mx() h) Mp(s)-Mx(7s)

Explanation / Answer

(a)

E[W] = E[X1 + X2 + X3 + X4] = E[X1] + E[X2] + E[X3] + E[X4] = E[X] + E[X] + E[X] + E[X] = 4E[X]

So, E[W] = 4E[X]

Var[W] = Var[X1 + X2 + X3 + X4] = Var[X1] + Var[X2] + Var[X3] + Var[X4] = Var[X] + Var[X] + Var[X] + Var[X] = 4Var[X]

So,Var[W] = 4Var[X]

(b)

E[V] = E[0.25(X1 + X2 + X3 + X4)] = E[0.25X1 + 0.25X2 + 0.25X3 + 0.25X4)]

= 0.25E[X1] + 0.25E[X2] + 0.25E[X3] + 0.25E[X4] = 0.25E[X] + 0.25E[X] + 0.25E[X] + 0.25E[X] = E[X]

So, E[V] = E[X]

Var[V] = Var[0.25(X1 + X2 + X3 + X4)] = Var[0.25X1 + 0.25X2 + 0.25X3 + 0.25X4)]

= 0.252Var[X1] + 0.252Var[X2] + 0.252Var[X3] + 0.252Var[X4]

= 0.0625Var[X] + 0.0625Var[X] + 0.0625Var[X] + 0.0625Var[X] = 0.25Var[X]

So,Var[V] = 0.25Var[X]

(c)

E[U] = E[X1 + X2 + X3 + X4 + Y] = E[X1] + E[X2] + E[X3] + E[X4] + E[Y] = E[X] + E[X] + E[X] + E[X] + E[Y]

= 4E[X] + E[Y]

So, E[U] = 4E[X] + E[Y]

Var[U] = Var[X1 + X2 + X3 + X4 + Y] = Var[X1] + Var[X2] + Var[X3] + Var[X4] + Var[Y]

= Var[X] + Var[X] + Var[X] + Var[X] + Var[Y] = 4Var[X] + Var[Y]

So,Var[U] = 4Var[X] + Var[Y]

(d)

R = 4X - Y

E[R] = E[4X-Y] = E[4X] + E[-Y] = 4E[X] - E[Y]

So, E[R] = 4E[X] - E[Y]

Var[R] = Var[4X-Y] = Var[4X] + Var[-Y] = 42Var[X] + (-1)2 Var[Y] = 16Var[X] + Var[Y]

So,Var[R] = 16Var[X] + Var[Y]

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.