Confusing about applying weighted least square for constant fitting

I'm now fitting a line with noise. My equation is to minimize corresponding to equation , then I have and with data. I want to caculate the best y. The WSL gives for the answer. But now my confusing is what is Y? Is this , which means my code is

(1) is the matrix with number 1. Is this right for me? or I should use other function such as fminsearch(I saw in the community, maybe it's still my missunderstanding)...Thanks

Answer 1

[Matt J]

I would recommend lscov

p=lscov(x(:).^[1,0],y,w/N);

yfit=polyval(p,x)

[WeiHao Xu]

Thanks sir! I use this get the result with dimension , so this right! And could you please point out my misstake above question? Thanks so much!

[Matt J]

@Torsten has corrected your formulation of the normal equations, but it would be inadvisable to use the normal equations directly. lscov handles the numerical issues involved in the inversion of (X^T*W*X) more reliably.

[WeiHao Xu]

Maybe I can get it, I will test it later . To prevent my missunderstanding, the code above x(:).^[1,0] is matrix with one? The standard solution of WSL confused me with vector X, for my case, y=y+noise, has no X. In this case I think it's ones(N,1), same as x above. Does my comprehension correct?

[Matt J]

You can execute some example code to see what x(:).^[1,0] gives you

x=1:5;
x(:).^[1,0]

ans = 5×2

1 1 2 1 3 1 4 1 5 1

This is appropriate if you are fitting a line with potentially non-zero slope. If you are fitting a line with zero slope, then you could substitute just x(:).^0

[WeiHao Xu]

Thanks for your helping, I will try it later!

[Matt J]

You're welcome, but if you find that one of the answers does what you want, please do Accept-click it.

Answer 2

[Torsten]

X = ones(N,1)

W = diag(w)

Y = y

where y is the (Nx1) column vector of the measurements and w is the (Nx1) column vector of weights.

The result of your formula is the coefficient a of the line y=a that best approximates the measurements.

[WeiHao Xu]

I have tried with ones(N,1)...According to my comprehension, my code is y = pinv( ones(N,1) * w * ones(N,1) )' *ones(N,1)' * w * y, which w and y is . Then matlab gives a dimension error. Then I change w to diag(w) which is , the result is a number. But I think the true result is , Could you please tech me some about it? Thanks so much!

[Torsten]

The result is a number since you do constant fitting (you want to fit your vector y to a horizontal line y = a).

It is this number a that you get from the formula.

[WeiHao Xu]

I may get it. Acording to y=a, I thought the result is [a a a a a ]. But the true answer is that I use N point to fit a data a, so it gives a. So even I use a sufficiently large number N, then I solve this equation, the result is a number, a, right? Thanks so much.

[Torsten]

Correct.

[WeiHao Xu]

Thanks so much!