# Making Sense of the weighted Least Squares method from ordinary Least Squares

It is stated that the equation for the best $$\hat{x}$$ uses the covariance matrix $$V$$ such that $$A^TV^{-1}A\hat{x}=A^TV^{-1}b$$ and the error vector is $$\vec{e}=(b-Ax)^TV^{-1}(b-Ax)$$ and we need to minimize $$e=\sum_{i=1}^m\dfrac{(b-Ax)_i^2}{\sigma_i^2}$$

How do I make sense of this idea of weighted least squares in comparison with ordinary least square method ?

And how does the inverse of variances become the respective weights ?

My Understanding

Least square

We have a system $$Ax=b$$ and $$b$$ need not have to be in the column space of $$A$$.

We need the vector $$A\hat{x}$$ in $$C(A)$$ thats best fit to $$b$$, ie., closest to $$b$$.

So we need to minimize the error vector $$\vec{e}=b-Ax$$, ie., we need to minimize $$e=e_1^2+\cdots+e_m^2=(b-Ax)^T(b-Ax)=||b-Ax||^2=\sum_{i=1}^m(b-Ax)_i^2$$

Weighted Least square

I think here we are assigning weights for each errors such that each error is not equally likely, say $$w_i$$

Then we need to minimize the error $$e=\sum_{i=1}^m w_ie_i=(b-Ax)^TW(b-Ax)=\sum_{i=1}^m w_i(b-Ax)_i^2$$.

But how does the matrix for the weights $$W$$ is the inverse of the covariance matrix $$V^{-1}$$ ?

Reference: Page 557, Introduction to Linear Algebra, Gilbert Strang

Note: I am actually looking for an exaplanation based on linear algebra, by not going deep into statistics

I don't think we can understand where $$V^{-1}$$ comes from without a bit of statistics, so here's a statistics viewpoint (but using notation similar to Strang's notation). Suppose there is an $$m \times n$$ matrix $$A$$ and a vector $$\bar x \in \mathbb R^n$$ such that $$Y = A \bar x + \epsilon.$$ Here $$A$$ and $$\bar x$$ are non-random but $$\epsilon$$ is a normally distributed random vector with mean $$0$$ and covariance matrix $$V$$. So $$Y$$ is a normally distributed random vector with mean $$A \bar x$$ and covariance matrix $$V$$. You can think of $$Y$$ as being a noisy measurement of the value of $$A \bar x$$.

Let $$b \in \mathbb R^m$$ be the observed value of $$Y$$. Our goal is to estimate $$\bar x$$, given $$A$$ and $$b$$. Let $$f_Y$$ be the probability density function for a normally distributed random vector with mean $$Ax$$ and covariance matrix $$V$$. Notice that $$f_Y$$ depends on $$x$$. A natural estimate of $$\bar x$$ is the maximum likelihood estimate \begin{align} \hat x &= \arg \max_x f_Y(b) \\ &= \arg \max_x \frac{1}{\sqrt{(2 \pi)^m |V|}}e^{-\frac12 (b - Ax)^T V^{-1}(b - Ax)} \end{align} Maximizing $$f_Y(b)$$ is equivalent to minimizing $$-\ln(f_Y(b))$$, which is equivalent to minimizing $$E(x) = (b - Ax)^T V^{-1} (b - Ax).$$ That is how we arrive at our weighted least squares problem. Setting the gradient of $$E$$ equal to $$0$$ yields Strang's equation (10).

If the components of the random variable $$\epsilon$$ are independent, then $$V$$ is diagonal, so the formula for $$E(x)$$ simplifies to Strang's equation (11).

Here is a derivation of the formula (in terms of the weight matrix $$W$$).

Note that the matrix $$W$$ is necessarily positive definite. Let $$M$$ be a matrix such that $$W = M^TM$$ (such a matrix can be found using the Cholesky decomposition, for instance). Note that for any vector $$x$$, the weighted norm satisfies $$x^TWx = (Mx)^T(Mx).$$ With that in mind, we can reframe our original problem as follows: $$\min_{x \in \Bbb R^n} (Ax - b)^TW(Ax - b) = \\ \min_{x \in \Bbb R^n} [M(Ax - b)]^T[M(Ax - b)] = \\ \min_{x \in \Bbb R^n} [(MA)x - (Mb)]^T[(MA)x - (Mb)].$$ In other words, the solution that we're looking for is the least squares solution to the equation $$(MA) x = Mb$$. Using the ordinary least-squares, the solution satisfies $$(MA)^T(MA)\hat x = (MA)^T(Mb) \implies\\ A^T(M^TM) A \hat x = A^T(M^TM)b \implies\\ A^TWA\hat x = A^TWb.$$

Now, a question remains: why is $$W = V^{-1}$$ the correct choice of weight matrix? Once answer is that even if entries of $$b$$ are not uncorrelated with identical variance, then entries of $$Mb$$ will be, where $$M$$ is chosen such that $$W = V^{-1} = M^TM$$.

Indeed: we note that the covariance matrix of $$b$$ is given by $$\Bbb E(bb^T) - \Bbb E(b)\Bbb E(b)^T.$$ We note that the covariance of the new variable $$Mb$$ is given by \begin{align} \Bbb E((Mb)(Mb)^T) - \Bbb E(Mb) \Bbb E(Mb)^T &= \Bbb E(Mbb^TM^T) - \Bbb E(Mb)\Bbb E(b^TM^T) \\ & = M\Bbb E(bb^T)M^T - M\Bbb E(b)\Bbb E(b)^T M^T \\ & = M[\Bbb E(bb^T) - \Bbb E(b)\Bbb E(b)^T]M^T \\ & = MVM^T = M(M^TM)^{-1}M^T \\ & = MM^{-1}M^{-T}M^T = I. \end{align} In other words, the covariance of each pair of components of $$Mb$$ is zero and the variance of each component is $$1$$.

Because the equation $$(MA)x = (Mb)$$ is such that the right side has entries that are uncorrelated with identical variance, it is more appropriate to compute its least squares solution (in the usual sense). As I explain in the first part of this answer, finding this least squares solution is equivalent to finding the weighted least squares solution for the original problem.

• I have added the statements in original reference. Still confused about the apprance of covariance matrix here. – Sooraj S Mar 8 at 6:41
• How do you say that $W$ s positive definite ? I was actually thinking the weight matrix should be diagonal ! – Sooraj S Mar 9 at 12:31
• @ss1729 A diagonal matrix with positive diagonal entries will be positive definite. More generally, though, covariance matrices are positive semidefinite. – Ben Grossmann Mar 9 at 14:26
• @ss1729 When you say that the covariance matrix is diagonal, the corresponding assumption about the data is that the errors in the components of $b$ are uncorrelated (which holds, for instance, if the errors are independent). – Ben Grossmann Mar 9 at 14:29
• @ss1729 See my latest edit. – Ben Grossmann Mar 9 at 14:50