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