Discussion between kamek and Alexander Vigodner

3:04 PM

A: Weighted least squares with angular data

EDITED THIS IS INCORRECT Tecnically this is straightforward. You specify the equations $$ z_{ij}=\theta_i-\theta_j + \sigma_{ij} \epsilon_{ij} $$ and try to minimize the error with respect to $\theta$s $$ \sum_{ij}\epsilon_{ij}^2=\sum_{ij}\frac{1}{\sigma_{ij}^2}(z_{ij}-\theta_i+\theta_j)^2 $$ wit...

Alexander Vigodner

@Rahul I don't get you comment and btw $z_{ij}\in [-\pi/2,\pi/2]$. So please give a contrexample that can not be solved by the LS above. There are constraints on the $\theta$ which eliminate any periodicity.

@kamek, please clarify $z_{ij}$ are exactly $\theta_i-\theta_j$ or they are defined up to $2\pi$ to fit into $[-\pi/2,\pi/2]$? If $\theta_i=-\theta_j=\pi$, what is $z_{ij}$?

kamek

@AlexanderVigodner I apologize, that was a mistake on my part. The correct bounds are now stated. I also added another measurement (a direct measurement of one of the unknown free variables) to make the problem observable.

Alexander Vigodner

I still did not understand. If $θ_i=π$ and $\theta_j=-\pi/2$, what is $z_{ij}$? $-\pi/2$?

kamek

@AlexanderVigodner Sorry about the confusion. I've edited the question to clarify $z_{ij}$ for all cases.

I'm not sure your equations get one $\theta$ on the right side. For example, $\tilde z_{23}^k = z_{23}^k-\tilde z_{12}^k$ is equal to $(\theta_2 - \theta_3) - [(\theta_1 - \theta_2) - \theta_1]$, which is equal to $2\theta_2 - \theta_3$.

Perhaps you meant $\tilde z_{1i}^k = z_{1i}^k+\tilde z_{01}^k$? (Also noting that $\tilde z_{01}^k = z_{01}^k = -\theta_1$.).

Alexander Vigodner

Yes, of course "+", not "-". Whatever, you can easily rearrange $z$. First you have equation for $\theta_1$ and derive equations for $\theta_2,.4$. Using them you can eliminate one of the $\theta$ in the rest of equations.

Since you finally have only 1 variable into each equation your editing of my answer generally is correct but IMHO too complicating. You just solve equations for each $\theta$ separately.

kamek

3:04 PM

Well solving for each $\theta$ separately isn't really that much more straightforward. In the example, there are four equations for $\theta_4$. That would mean you still need to take the weighted average of those four measurements. I've edited the answer again to reflect this.

Alexander Vigodner

Well, the weighted average is quite straightforward for me :). The main problem was modulo $2\pi$. Once we recalculate $y$s having only 1 $\theta$ - this problem is resolved. There is only one thing that bothers me. What is if $\sigma$ is too high, comparable to $\pi$.

kamek

3:18 PM

Alexander, thanks again for your insight into my question. I have thought of a couple remaining concerns with some special cases.

First special case: suppose you have both $z_{13}$ and $z_{35}$ in your measurements. By using the variable substitution formula, you'll end up with $y_{35} = z_{35} + z_{23} + z_{12} + z_{01}$, whose variance will be the sum of four variances. However, it would be better (if the variance is smaller) to use the substitution $y_{35} = z_{35} + z_{13} + z_{01}$, which combines the variance of only three terms. In other words, you should look for the shortest path …

3 hours later…

Alexander Vigodner

6:22 PM

Kamek, you probably would agree that these manipulations are necessary only for proper removing (or adding) extra $2\pi$ from (to) new $y$s. Without this all our manipulation are linear combinations of the original matrix of $\theta$s - they IMXO cannot change the solution. But... under one condition. I was wrong. The new system has already not independent measurements so in LS scheme $\sigmas$ are not diagonal it should be a covariance matrix.

if the correct covariance matrix is used - it does not matter what way your rearrange $z$s. Moreover... after removing/adding extra $2\pi$ you can revert your system back to your $z$s with 2 $\theta$s and solve standard weighted LS but with adjusted $z$.

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