数理統計後期演習プリント No.4
4.1(ex.4.3.1)
∫
x
FX (x) = P (X ≤ x) =
fX (x)dx
−∞
0 (x < 0)
= x2 (0 ≤ x < 1)
1 1 ≤ x
(i) X(4) は標本の中の一番大きな観測値なので
F(4) (x) = P (X(4) ≤ x)
= P (X1 ≤ x, X2 ≤ x, X3 ≤ x, X4 ≤ x)
= P (X1 ≤ x)P (X2 ≤ x)P (X3 ≤ x)P (X4 ≤ x)
= [FX (x)]4 = x8 (0 ≤ x < 1)
x = 0, 1 のとき微分不可なので
f(4) (x) =
d
F(4) (x) = 4{F (x)}3 fX (x) = 8x7 (0 < x < 1)
dx
fX (x) は 0 < x < 1 以外で 0 なので
8x7 (0 < x < 1)
f(4) (x) =
0 その他
(ii)X(1) は標本のなので一番小さい観測値なので
F(1) (x) = P (X(1) ≤ x)
= 1 − P (X(1) > x)
= 1 − P (X1 > x, X2 > x, X3 > x, X4 > x)
= 1 − P (X1 > x)P (X2 > x)P (X3 > x)P (X4 > x)
= 1 − {1 − F (x)}n
= 1 − (1 − x2 )4 (0 ≤ x < 1)
x = 0, 1 のとき微分不可なので
d
F(1) (x)
dx
= −4{1 − F (x)}3 (−f (x))
f(1) (x) =
= 4{1 − F (x)}3 f (x)
= 8x(1 − x2 )3
fX (x) は 0 < x < 1 以外で 0 なので
f(1) (x) =
8x(1 − x2 )3 (0 < x < 1)
0 その他
1
(iii)
F(3) (x) = P (X(3) ≤ x) =
( )
4
∑
4
j=3
=
4
∑
j=3
=
{F (x)}{1 − F (x)}4−j
j
4!
{F (x)}j {1 − F (x)}4−j
j!(4 − j)!
4!
{F (x)}3 {1 − F (x)}4−3 + {F (x)}4
3!(4 − 3)!
= 4x6 − 4x8 + x8
= 4x6 − 3x8
x = 0, 1 のとき微分不可なので
d
F(3) (x)
dx
= 24x5 (1 − x2 ) (0 < x < 1)
f(3) (x) =
fX (x) は 0 < x < 1 以外で 0 なので
f(3) (x) =
24x5 (1 − x2 ) (0 < x < 1)
0 その他
(iv)
∫
E(X(4) ) =
∞
−∞
∫ 1
xfX(4) (x)dx
x · 8x7 dx =
=
∫
0
1
E{(X(4) ) } =
x2 · 8x7 =
2
0
Var(X(4)
8
9
4
5
4
= E{(X(4) ) } − {E(X(4) )} = −
5
2
2
( )2
8
4
=
9
405
2
¯ ∼ N (5, 42 )
4.2(ex.4.4.1)
X
6
5−5
(a)P Z < √ = P (Z < 0) = Φ(0) = 0.5
16
36
6
−
5
(b) P Z < √ = P (Z > 23 ) = 1 − Φ(1.5) ≃ 0.0668
16
36
( ¯
)
(
)
¯ − 5| < 1 = P X − 5 < 1.5 = P (|Z| < 1.5) = 2Φ(1.5) − 1 ≃ 0.8664
(c)P |X
2 3
c−5
(d)P Z < √ = 0.95
c−5
2
3
16
36
= 1.65 c =
2
3
· 1.65 = 6.10
¯ − 5| ≥ 5) = 0.01
(e)P (|X
P (|Z| ≥ 32 c) = 0.01
1 − (2Φ( 32 c) − 1) = 0.01
2Φ( 23 c) = 1.99
Φ( 32 c) = 0.995
2
32 c = 2.58, c = 1.72
4.3
(1) カイ二乗分布
fX (x) =
(2) ガンマ分布
1
x
(n
2 )−1 e− 2 (x > 0)
n x
2 Γ( 2 )
n
2
0 (その他)
α
β xα−1 e−βx (x > 0)
fX (x) = Γ(α)
0 (その他)
カイ二乗分布はパラメータ α = n2 , β = 21 のガンマ分布,(ex.4.1.3) より
[
]α
β
ガンマ分布の m.g.f は
(t < β) より, カイ二乗分布の m.g.f は
β−t
mX (t) = E[etX ]
∫ ∞
=
etX fX (x)dx
−∞
∫ ∞
n
1
1
= n n
(x 2 −1 e−( 2 −t)x )dx
2
2 Γ( 2 ) −∞
Γ( n )
1
= n n · 1 2 n
2 2 Γ( 2 ) ( 2 − t) 2
(
) n2
1
1
=
(t < )
1 − 2t
2
[
] n2 −1
1
1
d
(3)E(X) = dx
mX (0) = n2
(−1)
(−2) = n
1−2·0
(1 − 2 · 0)2
( )2
[
]2
d
d
Var(X) =
mX (0) −
mX (0)
dx
dx
1
]2
)[
(n
1
1
+1
(−1)
(−2) − n2 = n2 + 2n − n2 = 2n
= n
2
1−2·0
(1 − 2 · 0)2
4.4(ex.4.4.4)
(4.4.4)
X1 , X2 , . . . , Xn ∼ N (µ, σ 2 )
∑n
¯ 2 とする.
標本分散 S 2 = n1 i=1 (Xi − X)
(ii)
P.144(3.9.2) E(X) = n,Var(X) = 2n
n 2
S ≪∼ χ2n−1
σ2
(n )
n
S 2 = 2 E(S 2 ) = n − 1
σ2
σ
n2 − 1
E(S 2 ) =
n
(
)
(n )
n2
Var 2 S 2 =
Var(S 2 ) = 2(n − 1)
σ
σ4
σ4
Var(S 2 ) = 2 2(n − 1)
n
E
3
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