Find a and b so that f(x)=ax+b has a minimum sum of squared errors for the points (1,2),(3,4),(4,5) sum of squares of errors, s = [a(1)+b-2]^2 +[a(3)+b-4]^2 +[a(4)+b-5]^2 =(a+b-2)^2 ...
Find a and b so that f(x)=ax+b has a minimum sum of squared errors for the points (1,2),(3,4),(4,5)
sum of squares of errors,
s = [a(1)+b-2]^2 +[a(3)+b-4]^2 +[a(4)+b-5]^2 =(a+b-2)^2 +(3a+b-4)^2 +(4a+b-5)^2
Differentiating with respect to a
ds/da =2(a+b-2)(1)+2(3a+b-4)(3)+2(4a+b-5)(4)
=2a+2b-4+18a+6b-24+32a+8b-40
=52a + 16b - 68 => 52a + 16b = 68
Differentiating with respect to b
ds/db =2(a+b-2)+2(3a+b-4)+2(4a+b-5)
=2a+2b-4+6a+2b-8+8a+2b-10
=16a + 6b - 22 => 16a + 6b = 22
We have simultaneous equations
52a + 16b = 68
16a + 6b = 22
gives a = 1, b = 1
f(x)=1x+1
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We have a linear model `f(x) = ax+b`
and want to minimise the sum of squares of the errors `(y - (ax+b))^2` given data
`(x,y) =` (1,2),(3,4),(4,5)
Rewrite the model as
`f(x) = a(x-bar(x)) + b_1` where `b_1 = b + abar(x)`
Then the least-squares estimate of `b_1` is `bar(y)`
So `hat(b_1) = (2+4+5)/3 = 11/3`
The least-square estimate of `a` is given by
`hat(a) = (Sigma_1^n(x_i-bar(x))(y_i-bar(y)))/(Sigma_1^n((x_i-barx)^2)) = ((1-8/3)(2-11/3)+(3-8/3)(4-11/3)+(4-8/3)(5-11/3))/((1-8/3)^2+(3-8/3)^2+(4-8/3)^2) `
`= (14/3)"/"(14/3) = 1`
Therefore the least-square estimate of `b` is given by
`hat(b_1) -hat(a)bar(x) = 11/3-(1)8/3 = 3/3 = 1`
least squares estimates are a=1, b=1
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