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Section 5.3

Section 5.3.

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Section 5.3

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  1. Section 5.3 Suppose X1 , X2 , … , Xn are independent random variables (which may be either of the discrete type or of the continuous type) each from the same distribution. Then we say that the collection of independent and identically distributed random variables X1 , X2 , … , Xn is a random sample of size n from the common distribution. Important Theorems in the Text: Theorem 5.3-1 (See Class Exercises #1(a)(b) in Section 5.2 and #1(e) in Section 5.3.) Theorem 5.3-2 (See Class Exercise #10 in Sections 4.1, 4.2, 4.3.) Theorem 5.3-3 (See Class Exercises #13 & 15 in Sections 4.1, 4.2, 4.3.)

  2. 1. (a) An urn contains six chips, one $3 chip, two $2 chips, and three $1 chips. Two chips are selected at random and with replacement. The following random variables are defined: X1 = dollar value of the first chip selected, X2 = dollar value of the second chip selected. Explain why X1 , X2 can be treated as a random sample of size 2 from a common distribution, and find the p.m.f. for that common distribution. Since the selection is done with replacement, X1 and X2 are independent and identically distributed. The p.m.f. for that common distribution is 4 –x —— 6 f(x) = if x = 1, 2, 3.

  3. 1. - continued (b) (c) Find the joint p.m.f. of (X1 , X2). The joint p.m.f. is (4 –x1)(4 –x2) —————— 36 f(x1) f(x2) = if x1 = 1, 2, 3 and x2 = 1, 2, 3 . Find the mean and variance for each of X1 and X2 . 3 — 6 2 — 6 1 — 6 5 — 3 E(X1) = E(X2) = E(X) = (1) + (2) + (3) = 10 — 3 3 — 6 2 — 6 1 — 6 E(X12) = E(X22) = E(X2) = (1) + (4) + (9) = (10/3) – (5/3)2 = 5/9 Var(X1) = Var(X2) = Var(X) =

  4. (d) (e) Find P(X1 + X2 < 4) two ways: from the p.m.f. of the common distribution and also from the joint p.m.f. P(X1 = 1 , X2 = 1) + P(X1 = 1 , X2 = 2) + P(X1 = 2 , X2 = 1) = (3/6)(3/6) + (3/6)(2/6) + (2/6)(3/6) = 21/36 = 7/12 or (3)(3)/36 + (3)(2)/36 + (2)(3)/36 = 21/36 = 7/12 Find E(X1 + X2) two ways: from the p.m.f. of the common distribution and also from the joint p.m.f. 5 — + 3 5 — = 3 10 — 3 E(X1 + X2) = E(X1) + E(X2) = or 9 — 36 6 — 36 3 — 36 6 — 36 (1+1) + (1+2) + (1+3) + (2+1) + E(X1 + X2) = 4 — 36 2 — 36 3 — 36 2 — 36 1 — 36 10 — 3 (2+2) + (2+3) + (3+1) + (3+2) + (3+3) =

  5. 1. - continued (f) Find E[1 / (X1 + X2)]. 1 ——— X1 + X2 1 9 —— — + 1 + 1 36 1 6 —— — + 1 + 2 36 1 3 —— — + 1 + 3 36 E = 1 6 —— — + 2 + 1 36 1 4 —— — + 2 + 2 36 1 2 —— — + 2 + 3 36 1 3 —— — + 3 + 1 36 1 2 —— — + 3 + 2 36 1 1 —— — = 3 + 3 36 73 —— 270

  6. (g) Let Y = X1 + X2 . Find the m.g.f. of Y, and find the p.m.f. of Y. t(X1 + X2 ) tX1 tX2 MY(t) = E(etY) = E[e ] = E[e e ] = tX1 tX2 E[e ] E[e ] = [(1/2)et + (1/3)e2t + (1/6)e3t ] 2 = (1/4)e2t + (1/3)e3t + (5/18)e4t + (1/9)e5t + (1/36)e6t 1/4 if y = 2 1/3 if y = 3 5/18 if y = 4 1/9 if y = 5 1/36 if y = 6 The p.m.f. of Y is g(y) =

  7. 2. (a) (b) Suppose W, X, and Y are mutually independent random variables each having an exponential distribution with E(W) = 1, E(X) = 5, and E(Y) = 10. Explain why W, X, and Y cannot be treated as a random sample of size 3. Since W, X, and Y, are not identically distributed (i.e., they do not share a common distribution), these three random variables cannot be treated as a random sample. Find the joint p.d.f. of (W, X, Y). e– x / 5 —— 5 e– y / 10 —— = 10 The joint p.d.f. is f(w,x,y) = e– w – w – x / 5 – y / 10 e —————— if w > 0, x > 0, y > 0 50

  8. (c) Find P[max(W, X, Y) < 2]. P[max(W,X,Y) < 2] = P(W < 2  X < 2 Y < 2 ) = 2 2 2 e– x / 5 —— dx 5 e– y / 10 —— dy = 10 P(W < 2) P(X < 2) P(Y < 2) = e– wdw 0 0 0 (1 – e– 2) (1 – e– 2 / 5) (1 – e– 2 / 10)

  9. 2. - continued (d) Find the mean and variance of 4X + 3Y − 5W . E(4X + 3Y − 5W) = 4E(X) + 3E(Y) − 5E(W) = 4(5) + 3(10) − 5(1) = 45 Var(4X + 3Y − 5W) = 42Var(X) + 32Var(Y) + 52Var(W) = 42(25) + 32(100) + 52(1) = 1325

  10. (e) Find E(XY– W2). E(XY − W2) = E(X)E(Y) − E(W2) = E(X)E(Y) − {Var(W) + [E(W)]2} = (5)(10) − {1 + 12} = 48

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