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Least squares estimation

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  1. Oct 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose that object 1 weighs [itex]\theta_1[/itex] and object two weights [itex]\theta_2[/itex]. Then each object is weighed once and then together getting three observations [itex] y_1,y_2,y_3[/itex]. The scale gives unbiased weights with normally distributed error (constant variance) Find the least square estimates for [itex]\theta_1[/itex] and [itex]\theta_2[/itex].
    2. Relevant equations
    The ls estimates for theta 1 and theta 2 is ## \hat{\theta}=(X^TX)^{-1}X^TY##

    3. The attempt at a solution
    I wrote the full mode as such
    [itex] y_1=\theta_1+\epsilon_1[/itex]
    [itex] y_2=\theta_2+\epsilon_2[/itex]
    [itex] y_3=\theta_1+\theta_2+\epsilon_3[/itex]
    Then it follows that in matrix form we get ## \begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \end{bmatrix} = \begin{bmatrix} \ \theta_1 \\ \theta_2 \\ \theta_1+\theta_2 \end{bmatrix}+\begin{bmatrix} \epsilon_1 \\ \epsilon_2 \\ \epsilon_3 \end{bmatrix}## ##=\begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0 & 1 \\1 & 1 \end{bmatrix} \begin{bmatrix} \theta_1 \\ \theta_2 \end{bmatrix}+\begin{bmatrix} \epsilon_1 \\ \epsilon_2 \\ \epsilon_3 \end{bmatrix}##.

    Then from here I got as my final answer
    ## \hat{\theta}=(X^TX)^{-1}X^TY=\begin{bmatrix} \hat{\theta_1} \\ \hat{\theta_2} \end{bmatrix} =\begin{bmatrix} y_1-y_2 \\ \frac{-1}{2}y_1+y_2+\frac{1}{2}y_3 \end{bmatrix}## The thing the I am not sure about is if I wrote my full model correctly. I was thinking that the observations are just the true weight of the object with some random error. But I am not sure.
     
    Last edited: Oct 30, 2015
  2. jcsd
  3. Oct 30, 2015 #2

    Ray Vickson

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    According to your model you are dealing with a very smart scale. It says to itself "I remember the errors ##\epsilon_1## and ##\epsilon_2## that I made when the guy weighed objects 1 and 2 separately. I see now that he is weighing the two objects together, so I had better add up the two errors I made before". As I said, a very smart scale indeed.
     
    Last edited: Nov 1, 2015
  4. Oct 30, 2015 #3

    mfb

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    Right, but the random error applies to the measurements, not to the weights.
     
  5. Oct 30, 2015 #4
    Oops now I fixed that. I don't know why I placed that in the first place. So I changed ##\epsilon_1+\epsilon_2## to ##\epsilon_3##. I think it looks correct now.
     
    Last edited: Oct 30, 2015
  6. Oct 30, 2015 #5

    Ray Vickson

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    Yes, I would say so.
     
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