Paraphrase-this-paragraph-Put-it-in-your-own-words-

https://stats.stackexchange.com/questions/46185/qu…

Multiple regression can be obtained by sequential matching

Returning to the setting of the question, we have one target

y

y

and two matchers

x1

x

1

and

x2

x

2

. We seek numbers

b1

b

1

and

b2

b

2

for which

y

y

is approximated as closely as possible by

b1x1+b2x2

b

1

x

1

+

b

2

x

2

, again in the least-distance sense. Arbitrarily beginning with

x1

x

1

, Mosteller & Tukey match the remaining variables

x2

x

2

and

y

y

to

x1

x

1

. Write the residuals for these matches as

x2â‹…1

x

2

â‹…

1

and

yâ‹…1

y

â‹…

1

, respectively: the

â‹…1

â‹…

1

indicates that

x1

x

1

has been “taken out of” the variable.

We can write

y=λ1x1+y⋅1 and x2=λ2x1+x2⋅1.

y

=

λ

1

x

1

+

y

â‹…

1

and

x

2

=

λ

2

x

1

+

x

2

â‹…

1

.

Having taken

x1

x

1

out of

x2

x

2

and

y

y

, we proceed to match the target residuals

yâ‹…1

y

â‹…

1

to the matcher residuals

x2â‹…1

x

2

â‹…

1

. The final residuals are

yâ‹…12

y

â‹…

12

. Algebraically, we have written

y⋅1y=λ3x2⋅1+y⋅12; whence=λ1x1+y⋅1=λ1x1+λ3x2⋅1+y⋅12=λ1x1+λ3(x2−λ2x1)+y⋅12=(λ1−λ3λ2)x1+λ3x2+y⋅12.

y

â‹…

1

=

λ

3

x

2

â‹…

1

+

y

â‹…

12

;

whence

y

=

λ

1

x

1

+

y

â‹…

1

=

λ

1

x

1

+

λ

3

x

2

â‹…

1

+

y

â‹…

12

=

λ

1

x

1

+

λ

3

(

x

2

−

λ

2

x

1

)

+

y

â‹…

12

=

(

λ

1

−

λ

3

λ

2

)

x

1

+

λ

3

x

2

+

y

â‹…

12

.

This shows that the

λ3

λ

3

in the last step is the coefficient of

x2

x

2

in a matching of

x1

x

1

and

x2

x

2

to

y

y

.

We could just as well have proceeded by first taking

x2

x

2

out of

x1

x

1

and

y

y

, producing

x1â‹…2

x

1

â‹…

2

and

yâ‹…2

y

â‹…

2

, and then taking

x1â‹…2

x

1

â‹…

2

out of

yâ‹…2

y

â‹…

2

, yielding a different set of residuals

yâ‹…21

y

â‹…

21

. This time, the coefficient of

x1

x

1

found in the last step–let’s call it

μ3

μ

3

–is the coefficient of

x1

x

1

in a matching of

x1

x

1

and

x2

x

2

to

y

y

.

Finally, for comparison, we might run a multiple (ordinary least squares regression) of

y

y

against

x1

x

1

and

x2

x

2

. Let those residuals be

yâ‹…lm

y

â‹…

l

m

. It turns out that the coefficients in this multiple regression are precisely the coefficients

μ3

μ

3

and

λ3

λ

3

found previously and that all three sets of residuals,

yâ‹…12

y

â‹…

12

,

yâ‹…21

y

â‹…

21

, and

yâ‹…lm

y

â‹…

l

m

, are identical.

 

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