Timeline for Conditional expectation of $X$ given $Z = X + Y$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 26, 2014 at 20:32 | vote | accept | David Melkuev | ||
| Aug 26, 2014 at 11:20 | comment | added | Dilip Sarwate | @DavidMelkuev Yes, the expressions are based on the essential idea behind OLS: we seek coefficients $a$ and $b$ that minimize the mean-square error $E[(X - (aZ+b))^2]$. This is standard stuff usually covered in undergraduate probability textbooks in the chapters on expectation which usually also point out that for the special case when $X$ and $Z$ have a bivariate normal distribution, the general minimum-mean-square-error (MMSE) estimator $E[X\mid Z]$ coincides with the linear minimum-mean-square error (LMMSE) estimator $aZ+b$. | |
| Aug 26, 2014 at 11:12 | comment | added | Dilip Sarwate | @StéphaneLaurent Carrying out the replacements you suggest gives $$\begin{align}\mu_X&=a(\mu_X+\mu_Y)+b\\\mu_Y&=(1-a)(\mu_X+\mu_Y)-b\\\end{align}$$ whose sum is $$\mu_X+\mu_Y = \mu_X+\mu_Y$$ and so, yes, the system is not invertible. | |
| Aug 26, 2014 at 10:31 | comment | added | Stéphane Laurent | @DilipSarwate After replacing $\alpha$ with $1-a$ and $\beta$ with $-b$, there are two equations and the solutions $a$ and $b$ depends on $\mu_X$ and $\mu_Y$ only. I think my approach is wrong: the system is not invertible. | |
| Aug 25, 2014 at 23:11 | comment | added | David Melkuev | Thank you both for your answers. @DilipSarwate I recognize your expression for $a$ in terms of covariance/variance as the coefficient in an OLS linear regression. Is that what it is based on or is there a more general idea that both are linked to? | |
| Aug 25, 2014 at 14:40 | history | edited | Dilip Sarwate | CC BY-SA 3.0 |
added more information
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| Aug 25, 2014 at 12:38 | comment | added | Dilip Sarwate | @StéphaneLaurent Your approach results in the equations $$\begin{align}\mu_X&=a(\mu_X+\mu_Y)+b\\\mu_Y&=\alpha(\mu_X+\mu_Y)+\beta\\a+\alpha &=1\\b+\beta &=0\end{align}$$ but do these four linear equations give $a$ and $b$ in terms of the means alone? or do they involve $\alpha$ and/or $\beta$ as well? | |
| Aug 25, 2014 at 12:25 | comment | added | Stéphane Laurent | There's something strange when comparing our two approaches: mine yields a result that doesn't involve any variance. | |
| Aug 25, 2014 at 12:17 | history | answered | Dilip Sarwate | CC BY-SA 3.0 |