Kết luận thống kê về nhu cầu chăm sóc sức khỏe ở Hải Dương

a
a + c
b
b + d
a/(a + c)
b/(b + d)
a
c
b
d
a/c
b/d
Y
X
i
E(Y/X
i
)
E(Y/X
i
) X
i
E(Y/X

i
) = f(X
i
)
f(X
i
)
f(X
i
) = β
0
+ β
1
X
β
0
, β
1
f(X
i
)
β
0
, β
1
π(x)
E(Y/X) =
e
β
0

+β
1
X
1 + e
β
0
+β
1
X
β
0
β
1
π(x) = E(Y/X) =
e
β
0
+β
1
X
1 + e
β
0
+β
1
X
π(x).(1 + e
β
0
+β

1
X
) = e
β
0
+β
1
X
π(x) = (1 − π(x)).e
β
0
+β
1
X
π(x)
1 − π(x)
= e
β
0
+β
1
X
g(x) = ln



π(x)
1 − π(x)




g(x) = β
0
+ β
1
x x x ∈ (−∞, +∞)
0 ≤ π(x) ≤ 1
y = π(x)+ε
ε
ε
y = 1 ε = 1 − π(x) π(x)
y = 0 ε = −π(x) 1 − π(x)
ε E (ε) = 0 V ar (ε) = V ar (Y ) =
π (x) . [1 − π (x)]
X
1
, X
2
, , X
k
= a
1
X
1
+ a
2
X
2
+ · · · + a
k

X
k
+ b
log
 
= a
1
X
1
+ a
2
X
2
+ · · · + a
k
X
k
+ b
log

p
1 − p

= a
1
X
1
+ a
2
X

2
+ + a
k
X
k
+ b
= exp(a
1
X
1
+ a
2
X
2
+ · · · + a
k
X
k
+ b)
= exp(b). exp(a
1
X
1
). exp(a
2
X
2
) . . . exp(a
k
X

k
)
(x
i
, y
i
), i = 1, 2, . . . , n y
i
x
i
i
P (Y = 1 |x) = π (x) P (Y = 0 |x) = 1 − π (x)
Y 1 π(x
i
) 0
1 − π (x
i
) , i = 1, , n
(x
i
, y
i
), i = 1, 2, . . . , n
ξ(x
i
) = π(x
i
)
y
i

[1 − π(x
i
)]
1−y
i
(x
i
, y
i
), i = 1, 2, . . . , n
l(β) =
n

i=1
ξ(x
i
) =
n

i=1
π(x
i
)
y
i
[1 − π(x
i
)]
1−y
i

( )
L(β) =
n

i=1
(y
i
ln(π(x
i
)) + (1 − y
i
) ln(1 − π(x
i
)))
L(β) β
0
β
1







n

i=1
[y
i

− π(x
i
)] = 0
n

i=1
x
i
[y
i
− π(x
i
)] = 0
(1.3.5)
β = (β
0
, β
1
)
ˆ
β
β
D = −2 ln






ˆπ

i
= ˆπ(x
i
)
D = −2 ln
n

i=1



ˆπ
y
i
i
(1 − ˆπ
i
)
1−y
i
y
y
i
i
(1 − y
i
)
1−y
i




= −2
n

i=1



y
i
ln

ˆπ
i
y
i

+ (1 − y
i
) ln

1 − ˆπ
i
1 − y
i





G = −2 ln






β
1
= 0 β
1
= 0
β
1
= 0
ˆ
β
0
= ln

n

i=1
y
i

n

i=1
(1 − y

i
)

n
1
=
n

i=1
y
i
, n
0
=
n

i=1
(1 − y
i
) β
0
=
n
1
n
0
G = −2 ln





n
1
n

n
1

n
0
n

n
0


n

i=1
ˆπ
y
i
i
(1 − ˆπ
i
)
1−y
i





G = 2

n

i=1

y
i
ln(ˆπ
i
) + (1 − y
i
) ln(1 − ˆπ
i
)

−

n
1
ln(n
1
) + n
0
ln(n
0
) − n ln(n)



β
1
= 0
χ
2
β
1
= 0 −2l
χ
2
(1)
α = P [χ
2
(1) > −2l]
α α
0
α ≤ α
0
α
0
α > α
0
100(1 − α
0
)
W =
ˆ
β
1

SE(
ˆ
β
1
)
ˆ
β
1
β
1
SE(
ˆ
β
1
) β
1
β
1
= 0
β
1
= 0 β
1
= 0
α = P [|Z| > W ]
α α
0
α ≤ α
0
α

0
α > α
0
100(1 − α
0
)%
X
1
, X
2
, , X
p
X = (X
1
, X
2
, , X
p
)
P (Y = 1/x) = π(x)
P (Y = 0/x) = 1 − π(x)
g(x) = β
0
+ β
1
x
1
+ · · · + β
p
x

p
β
i
, i = 1, p β
0
π(x) =
e
g(x)
1 + e
g(x)
k − 1 D
1
, D
2
, . . . , D
(
k − 1)
D
1
=
D
2
=
. . . . . . . . .
D
k−1
=
k − 1
k − 1
D

i
= 0 i = 1, . . . , k − 1
k
j
k
j
− 1
D
ju
u = 1, k
j
− 1
β
ju
g(x) = β
0
+ β
1
x
1
+ · · · +
k
j
−1

u=1
β
ju
D
ju

+ · · · + β
p
x
p
(1.4.7)
β = (β
0
, β
1
, . . . , β
p
)
(x
i
, y
i
), x
i
=
(x
i1
, . . . , x
in
), i = 1, p
l(β) =
n

i=1
π(x
i

)
y
i
[1 − π(x
i
)]
1−y
i
L(β) =
n

i=1

y
i
ln π(x
i
) + (1 − y
i
) ln[1 − π(x
i
)]

β
0
, β
1
, . . . , β
p
























n

i=1
[y
i
− π(x
i
)] = 0

n

i=1
x
i1
[y
i
− π(x
i
)] = 0
. . .
n

i=1
x
ip
[y
i
− π(x
i
)] = 0
β = (β
0
, β
1
, . . . , β
p
)
ˆ
β = (

ˆ
β
0
,
ˆ
β
1
, . . . ,
ˆ
β
p
)
β
0
, β
1
, . . . , β
p
∂
2
L(β)
∂β
2
j
= −
n

i=1
x
2

ij
π
i
(1 − π
i
);
∂
2
L(β)
∂β
i
∂β
j
= −
n

i=1
x
ij
x
iu
π
i
(1 − π
i
)
u = 0.p
I(β) =
















∂
2
L(β)
∂β
2
1
= −
n

i=1
x
2
i1
π
i
(1 − π
i

) . . .
∂
2
L(β)
∂β
1
∂β
p
= −
n

i=1
x
i1
x
ip
π
i
(1 − π
i
)
∂
2
L(β)
∂β
1
∂β
2
= −
n


i=1
x
i1
x
i2
π
i
(1 − π
i
) . . .
∂
2
L(β)
∂β
2
∂β
p
= −
n

i=1
x
i2
x
ip
π
i
(1 − π
i

)
. . . . . . . . .
∂
2
L(β)
∂β
1
∂β
p
= −
n

i=1
x
i1
x
ip
π
i
(1 − π
i
) . . .
∂
2
L(β)
∂β
2
p
= −
n


i=1
x
2
ip
π
i
(1 − π
i
)
















(β) = I
−1
(β)


(β) =








σ
2
(β
1
) σ(β
1
, β
2
) . . . σ(β
1
, β
p
)
σ(β
1
, β
2
) σ
2
(β
2

) . . . σ(β
1
, β
p
)
. . . . . . . . . . . .
σ(β
1
, β
p
) σ(β
2
, β
p
) . . . σ
2
(β
p
)








σ
2
(β

j
)
ˆ
β
j
, j = 1, p, σ(β
j
, β
u
)
ˆ
β
j
ˆ
β
u
j, u = 1, p
ˆ

(
ˆ
β)

(β)
ˆ
β
ˆ

(
ˆ

β) =








σ
2
(
ˆ
β
1
) σ(
ˆ
β
1
,
ˆ
β
2
) . . . σ(
ˆ
β
1
,
ˆ
β

p
)
σ(
ˆ
β
1
,
ˆ
β
2
) σ
2
(
ˆ
β
2
) . . . σ(
ˆ
β
1
,
ˆ
β
p
)
. . . . . . . . . . . .
σ(
ˆ
β
1

,
ˆ
β
p
) σ(
ˆ
β
2
,
ˆ
β
p
) . . . σ
2
(
ˆ
β
p
)








ˆ
SE(
ˆ

β
i
) =

ˆσ
2
(
ˆ
β
j
)

1
2
β
1
= β
2
= · · · = β
p
= 0
β
1
, β
2
, . . . , β
p
l
H
(β) =

e
β
0
1 + e
β
0
l
K
(β) =
n

i=1
π(x
i
)
y
i
[1 − π(x
i
)]
1−y
i
ˆ
l
H
,
ˆ
l
K
G = −2 ln


ˆ
l
H
ˆ
l
K

χ
2
χ
2
p
α = P [χ
2
p
> −2l]
α α
0
α ≤ α
0
α
0
α > α
0
100(1 − α
0
)%
W =
ˆ

β


ˆ

(
ˆ
β)

−1
ˆ
β =
ˆ
β

(X

V X)
ˆ
β
X =








1 x

11
. . . x
1p
1 x
21
. . . x
2p
1 x
n1
. . . x
np








V = diag[ˆπ
1
(1 − ˆπ
1
), ˆπ
2
(1 − ˆπ
2
), . . . , ˆπ
n
(1 − ˆπ

n
)]
χ
2
χ
2
p+1
α = P [Z > W ]
α α
0
α ≤ α
0
α
0
α > α
0
100(1 − α
0
)%
g(x) = ln

π(x)

[1 − π(x)]

= β
0
+ β
1
x

β
1
= g(x + 1) − g(x)
X = 1 X = 0
Y = 1 π(1) =
e
β
0
+β
1
1 + e
β
0
+β
1
π(0) =
e
β
0
1 + e
β
0
Y = 0 π(1) =
1
1 + e
β
0
+β
1
π(0) =

1
1 + e
β
0
X = 1
π(1)
1 − π(1)
X = 0
π(0)
1 − π(0)
g(1) = ln

π(1)
1 − π(1)

= β
0
+ β
1
g (0) = ln

π (0)
1 − π (0)

= β
0
ψ
X = 1 X = 0
ψ =
π(1)/[1 − π(1)]

π(0)/[1 − π(0)]
ln(ψ) = ln

π(1)/[1 − π(1)]
π(0)/[1 − π(0)]

= g(1) − g(0)