Statistical computing with SAS

1. A contin" rel="nofollow">inuous RV X has a pdf of the form f(x) = 2x/9 for 0 < x < 3 and zero otherwise (BE 2.18): (a) Fin" rel="nofollow">ind the CDF of X. (b) Fin" rel="nofollow">ind P(X ≤ 2). (c) Fin" rel="nofollow">ind P(−1 < X < 1.5) (d) Fin" rel="nofollow">ind a number m such that P(X ≤ m) = P(X ≥ m) (e) Fin" rel="nofollow">ind E[X]. Solution : (a) F(x) = ( 1 , x ≥ 3 x 2/9 , 0 < x < 3 0 , x ≤ 0 (b) 4/9, (c) 1/4, (d) 3√ 2/2, (e) 2, (f) 1/2 2. Suppose that X1 and X2 are discrete random variables with join" rel="nofollow">int pdf of the form f(x1, x2) = c(x1 + x2) x1 = 0, 1, 2; x2 = 0, 1, 2 and zero otherwise. Fin" rel="nofollow">ind the constant c. [BE4.Q7] Solution : 1/18 3. Let X and Y be contin" rel="nofollow">inuous random variables with a join" rel="nofollow">int pdf of the form f(x, y) = f(x, y) = k(x + y) , 0 ≤ x, y ≤ 1 and zero otherwise. [BE4.Q19] (a) Fin" rel="nofollow">ind k so that f(x, y) is a join" rel="nofollow">int pdf. (b) Fin" rel="nofollow">ind the margin" rel="nofollow">inal f(x) and f(y). (c) Fin" rel="nofollow">ind the conditional pdf f(y|x). (d) Fin" rel="nofollow">ind the conditional pdf f(x|y). Solution : (a) k = 1 (b) fX = 1/2 + x fY = 1/2 + y (c) f(y|x) = x + y 1/2 + x (d) f(x|y) = x + y 1/2 + y 4. Suppose X and Y are contin" rel="nofollow">inuous random variables with join" rel="nofollow">int pdf f(x, y) = 4(x − yx) if 0 < x, y < 1 and zero otherwise. [BE4.Q28] (a) Are X and Y in" rel="nofollow">independent? Why or why not? 1 (b) Fin" rel="nofollow">ind P(X ≤ Y ). Solution : (a) Yes. Because f(x, y) = f(x)f(y) for 0 < x < 1 and 0 < y < 1. (b) 1/6 5. A softball team has three pitchers, A, B, and C with win" rel="nofollow">innin" rel="nofollow">ing percentages of 0.4, 0.6 and 0.8 respectively. These pitchers pitch with frequency 2, 3, and 5 out of every 10 games, respectively. In other words, for a randomly selected game, P(A) = 0.2, P(B) = 0.3 and P(C) = 0.5. (BE 1.32) (a) P(team win" rel="nofollow">ins game) = P(W). (b) P(A pitched game|team won) = P(A|W). Solution : (a) 0.66, (b) 0.1212 6. A 2 × 1 vector X has mean zero and covariancce Σ = 1 0 0 2 . Let Y = X1 + X2 = AX. Fin" rel="nofollow">ind a lin" rel="nofollow">inear transformation matrix C then E[Y ] and Var(Y ). Solution : C = [1, 1] and E[Y ] = 0 and Var(Y ) = 3 7. Try the question 6 when Σ = 1 −0.3 −0.3 2 . Solution : E[Y ] = 0 and Var(Y ) = 2.4 8. Suppose that (x1, . . . , xd) are d-in" rel="nofollow">indepdent samples from a Bin" rel="nofollow">inomial distribution Bin" rel="nofollow">inom(N, p). Show that H = 1 d Pd i=1 xi/N is an unbiased esitimator for p. Solution : E[H] = 1 d X d i=1 E[xi ] N = p 2