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. 1862 . Sir Francis Galton Regression towards Mediocrity hereditary stature.
, , .
i | yi | xi |
y1 | x1 | |
y2 | x2 | |
y3 | x3 | |
n | yn | xn |
y x.
Y , .
X , , , , .
:
yi = β0 + β1 * xi + εi ( ).
[, x , y ] .
.
x y .
, y, x .
, , .. .
yi = β0 + β1 * xi + εi
|________|
yi = β0 + β1 * xi , .. .
|
.
. .
yi = E (y|xi)
yi = yi + ei
ei , , ( ).
β0 β1, .
1821 1822. .
yi = β0 + β1 * (xi x .) + εi
x (0;0).
, .
: , = |x|.
. .
Ψ = ∑ ei2 → min
Ψ = ∑ (yi β0 β1 (xi x .))2 → min
β0, β1
, .. .
:
Ψβ0 = ∑ ( 2 (yi β0 β1 (xi x .))) = 0
Ψβ1 = ∑ ( 2 (yi β0 β1 (xi x .))) (xi x .) = 0
. .
Ψβ0 = ∑ ( 2 (yi β0 β1(xi x .))) = 0
.
∑ yi nβ0 β1∑ (xi x .) = 0
, β1∑ (xi x .) = 0,
.. x . = ∑ xi /n,
∑ xi = nx .
∑ (xi x .) = ∑ xi x .n = x .n x .n = 0
∑ yi nβ0 = 0
β0 = ∑ yi / n = . |
- β0 β0 , - .
|
|
- β1
.
Ψβ1 = ∑ ( 2 (yi β0 β1 (xi x .))) (xi x .) = 0
.
∑ ((yi y .) *(xi x .) β1 (xi x .)2) = 0
∑ ((yi y .) *(xi x .)) β1 ∑ (xi x .)2 = 0
β1 = ∑ ((yi y .) *(xi x .)) / ∑ (xi x .)2 |
β1 = R * (∑ (yi y .)2)0,5 / (∑ (xi x .)2)0,5,
R .
.
,
β0 = 25
β1 = 0,7
yi = 25 + 0,7xi + exi
yi = 25 + 0,7xi
-
β1 .
x y 0,7.
β0 .
x = 0, y = 25.
. , x.
xi , , yi .
? ei.
ei.
- ei , E(ei) = 0.
yi = β0 + β1xi + ei,
β0 + β1xi . . ei 0!
- ei , D(ei) = ς12 = ς22
.
() x, ().
x, . , , .
.
- Corr (ei; ej) = 0 i≠j. .
- Corr (ei; xi) = 0.
(.. x ).
( -):
5*. Ei ~ N (0; ς2). , .
.
, - .
β0 β1 , .
β0 ~ N (β0; ς2/n)
β1 ~ N (β1; ς2/∑ (xi x .)2)
β0 β1 .
?
.
[, x ? .. , - ].
R2 = x y.
F .
H0: ( ).
VS H1: .
. = 0,05.
. .
F = R2 / [(1 R2) / (n 2)] ~ F (1; n-2),
n .
( ).
.
.
F: H0 H1 0,05, F .
β1 ?
yi = β0 + β1 * xi + εi
.
H0: β1 = 0.
VS H1: β1 ≠ 0.
. = 0,05.
, :
t = [β1 / .. (β1)] ~ t (n 2) ,
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( ).
.
.
: H0 H1 0,05, F .
. :
. y x . y x.
E (Y| X=x) = ∑ yi P (X=yi|X=x) |
E (Y| X) = β0 + β1xi
, , . .
yi = E (Y| X) + ei
(yi y .) = β0* + β1*(xi x .) + ei
- , .
.
, y . x x ., .. , y, x, .
y y.
1/n ∑ (yi y .)2 = 1/n ∑ (yi yi )2 + 1/n ∑ (yi y .)2
. . . . . . .
TSS RSS ESS
TSS = RSS + ESS
(. ).
. . . . .
R2 = ESS / TSS = (TSS RSS) / TSS = 1 RSS/TSS
RSS = ∑ (ei)2
F- ( ).
H0: R2 = 0
VS H1: R2 > 0
= 0,05.
:
F = (ESS/1) / [RSS/(n 2)], (ESS/k) / [RSS/(n k 1)].
F ~ F (1, n 2) .