0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ The problem is to show the least squares estimate $b$ estimates $\beta_{U,T}R_Z$. $\endgroup$ – lightfish Feb 8 '14 at 0:51
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.