Assume $\beta_{U,T}$ is the underlying slope of straight line associating $U$ with $T$. We know that $X=U+f$ and $Z=T+e$ are measurable instead of $U$ and $T$, where $e$ and $f$ are uncorrelated random errors. $R_Z=\frac{\sigma_T^2}{\sigma_T^2+\sigma_e^2}$ and $R_X=\frac{\sigma_U^2}{\sigma_U^2+\sigma_f^2}$ are reliabilities of $Z$ and $X$.

I don't know how to convert the words into equations. Does this mean that $U=\alpha+\beta T$? Would the observed variables be $X=\hat{\alpha}+\hat{\beta}Z$ and then we have

$(observed-expected)^2=(\hat{\alpha}+\hat{\beta}Z-\alpha-\beta T)^2$?

Thank you

  • $\begingroup$ The problem is to show the least squares estimate $b$ estimates $\beta_{U,T}R_Z$. $\endgroup$ – lightfish Feb 8 '14 at 0:51

yes you're right and everything you said is true, just wonder why you asked, lol. It's like fitting a line to two observations $X$ and $Z$ that are subject to noise. I'd recommend you to take a look at this excelent paper from Jayenes. In the second part, you're calculating what is called residuals or square error. In statistical setting, you're having one than one observation, say $N$, and the sum of residuals is written as:

$\mathcal{X}^2 = \sum_{i=1}^{N}(x - \hat{x})^2 $

where, $\hat{x} = a + b. y$. And the task of model fitting is basically to minimize $\mathcal{X}^2$ with respect to the parameters $a$ and $b$.

was that your answer? or I'm missing something.

  • $\begingroup$ yes, thanks! I wasn't sure if I was approaching it correctly. The problem is to prove that the least square estimate $b$ estimates $\beta_{U,T}R_Z$. Does this mean I need to show $E[b]=\beta_{U,T}R_Z$? Thanks again $\endgroup$ – lightfish Feb 8 '14 at 0:34
  • $\begingroup$ In this case, would it be $\chi^2=\sum_{i=1}^N (X-U)^2=\sum(a+b(T+e)-(\alpha+\beta_{T,U}T))^2$? I found the minimum which is $\displaystyle b=\frac{\alpha-a+\beta_{T,U}T}{T+e}$ $\endgroup$ – lightfish Feb 8 '14 at 0:49

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