Mathematical Statistics with Applications Chapter1 solution
233
Chapter 16: Differentiation
The home run hitters (probability p/4) score exactly k runs if and only if there are exactly k home runs before the third out. This has probability k +2 ak ( p/4)k (1 p/4)3 . The expectation is 2
16.67. If q is a polynomial, then k =0 q (k ) x k is the ratio of two polynomials in x (for x < 1). Proof 1 (differentiation). It suffices to prove that the statement holds k m , because the general case is a finite sum of multiples of when q(k ) ∞ k m x k pm ( x ) such answers. We prove by induction on m that for k =0 (1− x )m
=
∞
∞
∞ 1
ka k
k 0
k 0
=
=
(k
2
+ 2)(k + 1)k ( p4 ) (1 − p4 ) = 4 3− p p . 3
k
∞
(k
k 0
=
∞
k
+ 2)(k + 1)kx = x (
x k )
k 0
=
=
− −
= x ( 1 −1 x ) = 6 x (1 − x )− . 4
∞
+ −
∞
= 12 p (1 − p) 2
kb k
k 1
=
To sum the series, we use (k 1) 2(k 1) 6 to obtain
−
∞
+ −
+
(k 4)(k 3)k x k
+
k 1
=
∞
3
∞
(k
k 1
=
(k 4)(k 3)k x k
+
k 0
=
6
1
1
2
6
= (1 − x ) + (1 − x ) − (1 − x ) − 1 − x 3
2
p and inserting this into the formula for the expectation yields Setting x p 2 ( 1−3 p 1 (1 p) 3(1 p)2 ). As p approaches 1, the home run hitters score about .75 runs per inning, while the expectation for the singles hitters grows without bound. When p is very small, the home run hitters expect about 3 p/4 runs per inning, while the singles hitters expect only about 10 p 3 . When p > .279 (approximately), the singles hitters do better.
= + − − −
16.66.
n k k 0 k x
=
−
n
n
=
1
k
k x
= x =
k 1
k x −
k 0
=
= x d x (
d 1 x n+1 x ( ) d x 1 x
− −
d
=
x )
k 0
=
n +1 x (n x
n
− (n + 1) x + 1) . (1 − x ) 2
=
+1
=
x
d
pm −1 ( x )(1 d x
m
− x )
∞
=
+ (1 − x ) p − ( x ) = p ( x ) , (1 − x ) + (1 − x ) + − ( x ) + (1 − x ) p − ( x )]. Since p − is a polynomial, −
m
m 1
m 1
m 1
x [m pm 1 where pm ( x ) m 1 m 1 also pm is a polynomial. Proof 2 (generating functions). It suffices to prove that the statement k holds when q(k ) , because every polynomial is a finite sum of multim ples of such binomial coefficients (see Solution 5.29). By Theorem 12.35, ∞ k x k x m /(1 x )m+1 . k =0 m
=
−
16.68. The random variable X defined for nonnegative integer n by p(1 p)n , where 0 < p < 1. Prob ( X n) a) The probability generating function φ for X is given by φ (t ) p [1 ∞ Prob ( X n)t n . Using the geometric (1 p)t ]−1 . By definition, φ (t ) n =0 p(1 p)n t n p/(1 (1 p)t ). Since φ(1) series, we compute φ(t ) p/(1 (1 p)) 1, as desired. sums all the probability, we have φ (1) b) E ( X ) (1 p)/ p . We compute
The singles hitters score exactly k > 0 runs if and only if there are exactly k 2 singles before the third out. This has probability bk k +4 p k +2 (1 p)3 . The expectation is 2
some polynomial pm . When m 0, we have the geometric series k =0 x k 1/(1 x ). For the induction step, suppose that m > 0 and that the claim holds for m 1. We compute
