Homework #14 solutions
1. #6.80 f (y) = 6y(1 y) on (0, 1), so F (y) = 3y 2 2y 3 on (0, 1).
F(n) (y) = P (Y(n) y) = P (all Yi y) = F (y)n = [3y 2 2y 3 ]n .
(b) Then
f(n) (y) = nF (y)n1 f (y) = 6ny(3y 2 2y 3 )n1 (1 y).
Below are some examples for n = 2 and
Homework #13 solutions
2 /2
1. #6.40 For a standard normal Y , mY (t) = et
2
E(etY ) =
2
ety ey
2 /2
, and so the mgf for Y 2 is
1
dy =
.
1 + 2t
Then the sum of Y12 and Y22 has mgf 1/(1 + 2t). We recognize this as the
mgf for chi-squared with = 2.
2. #6.
Homework #15 solutions
1. #7.11 Let Y1 , . . . , y9 be the measurements of the n = 9 trees. The required
probability is P (|Y | 2). Were not given , but we know that
Z=
Y
N (0, 1),
4/ n
so
P (|Y | 2) = P
Y
2
4/ 9
4/ 9
= P (|Z| 1.5) = .8664.
2. #7.12