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I have $\epsilon_0 \sim N(0, \sigma_0)$ and $\epsilon_1 \sim N(0, \sigma_1)$ are independent and $\epsilon_1 - \epsilon_0 = \nu$

How do you go from $E\left(\epsilon_0 | \nu \right) = \frac{\sigma_{0\nu}}{\sigma_\nu^2}\nu$

to $E\left(\frac{\epsilon_0}{\sigma_0} | \frac{\nu}{\sigma_\nu} \right) = \frac{\sigma_{0\nu}}{\sigma_\nu^2} \frac{1}{\sigma_\nu^{-2}} \frac{1}{\sigma_0 \sigma_\nu} \frac{\nu}{\sigma_\nu}$

What rules are being used to make this step?

Edit: Noting that $\sigma_{0v}=Cov\left(\epsilon_0,v\right)=\sigma_0^2$

Rami
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    What is $\sigma_{0\nu}$? A covariance? Why do you have $\frac{1}{\sigma_\nu^{-2}}$ rather than $\sigma_\nu^{2}$ or $\sigma_\nu^{-2}$? Is $E\left(\frac{\epsilon_0}{\sigma_0} \mid \frac{\nu}{\sigma_\nu} \right) = E\left(\frac{\epsilon_0}{\sigma_0} \mid \nu \right)$ or some multiple of it? – Henry Nov 29 '21 at 09:33
  • I would have thought $E\left(\frac{\epsilon_0}{\sigma_0} \mid \nu \right) = \frac{1}{\sigma_0} E\left(\epsilon_0 \mid \nu \right)$ and you can cancel a lot of the $\sigma_\nu$ – Henry Nov 29 '21 at 09:38
  • I think $\sigma_{0v}=Cov\left(\epsilon_0,v\right)=\sigma_0^2$.

    I am not sure what is the relationship between $E\left(\frac{\epsilon_0}{\sigma_0} \mid \frac{\nu}{\sigma_\nu} \right)$ and $E\left(\frac{\epsilon_0}{\sigma_0} \mid \nu \right)$

     – Rami Nov 30 '21 at 09:45

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