I observe $\vec y$ and know $\vec x$. I assume that $\vec y$ mostly consists of $\vec x$, with some added residual $\vec r$.

This gives me the problem $\vec y = a\vec x + \vec r$, where $a \in [0, 1]$. Furthermore, I assume that we should maximize $a$. Otherwise, $a = 0, \vec r = \vec y$ would be a trivial solution.

Another way to express the problem for vectors of length 3 would be:

$\begin{pmatrix} y_1 \\ y_2 \\ y_3 \\ 1 \end{pmatrix} = \begin{bmatrix} a & 0 & 0 & r_1 \\ 0 & a & 0 & r_2 \\ 0 & 0 & a & r_3 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ 1 \end{pmatrix}$

In this version, the matrix would need to be calculated, but we need to do it in a way that maximizes $a$.

How to best approach this problem?

  • $\begingroup$ What exactly is your problem? What are you trying to find out based on your observations? Clearly you're not estimating $x$: you already know it. Consequently it would seem you're trying to find out the way in which "$y$ mostly consists of $x$", but that's not sufficiently precise or quantitative to permit any kind of answer. The "problem" ($y=ay+r$) you quote makes no mention of $x$ at all and therefore isn't soluble. Finally, the equation you write has an obvious trivial solution given the constraints: set $a=1$. That determines $r_i=y_i-x_i$ for all $i$. $\endgroup$
    – whuber
    Sep 10 '14 at 15:14
  • $\begingroup$ Thank you. I corrected the typo to $\vec y= a \vec x+ \vec r$. The idea is that $\vec y$ is a linear mixture of $\vec x$ and other influences, e.g. noise. If $a=1$ it means that $\vec y$ consists of $\vec x$ alone, $\vec r$ is $0$ then. However I see the problem with the unconstrained $\vec r$.. $\endgroup$
    – ypnos
    Sep 10 '14 at 15:51
  • $\begingroup$ Thanks--that helps clear up the apparent inconsistencies. But how do you address my last comment, which is that the solution is obviously $a=1$? I suspect you may have additional criteria in mind which you haven't yet mentioned. $\endgroup$
    – whuber
    Sep 10 '14 at 15:53
  • 1
    $\begingroup$ Yes the problem is that we would need to constraint $\vec r$ appropriately. Originally, the statement would be $\vec y = a\vec x + (1-a)\vec r$. But that does not help. I have to think about $\vec r$. Thank you for your comments so far! $\endgroup$
    – ypnos
    Sep 10 '14 at 15:56
  • $\begingroup$ I have the additional knowledge that all elements of y, x, and r are positive. $\endgroup$
    – ypnos
    Sep 11 '14 at 12:17

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