Timeline for why in Gaussian linear regression they put $y_i-\theta^Tx_i$ instead of $x$?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 4, 2020 at 10:52 | vote | accept | Raz | ||
| Jun 2, 2020 at 16:39 | comment | added | whuber♦ | The formula uses "$\mu,$" not $0,$ for the mean. The only thing going on here is that the formula you quote is being applied directly to the supposition that the mean of the random variable "$y_i$" is $\theta^\top x_i.$ This isn't even a matter of statistics or mathematics; it's purely a matter of notation: that is, applying the formula through substitution. | |
| Jun 2, 2020 at 14:48 | comment | added | Raz | you mean in $p(x)$, when we are dealing with Gaussian and regression, we consider x as $y_i$, and mean as $\mu$? Then, why the book write suppose mean is zero? | |
| Jun 2, 2020 at 14:46 | history | edited | Raz | CC BY-SA 4.0 |
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| Jun 2, 2020 at 14:43 | review | First posts | |||
| Jun 2, 2020 at 15:24 | |||||
| Jun 2, 2020 at 14:41 | comment | added | whuber♦ | Your edit doesn't characterize the use of the formula correctly. In the formula, which is a function of three variables $x,\mu,\sigma,$ you plug in "$y_i$" for "$x$" and "$\theta^\top x_i$" for "$\mu.$" (Note, also, that $\sigma^2,$ not $\sigma,$ should appear under the square root in the normalizing factor.) | |
| Jun 2, 2020 at 14:35 | history | edited | Raz | CC BY-SA 4.0 |
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| Jun 2, 2020 at 13:23 | answer | added | Aksakal | timeline score: 2 | |
| Jun 2, 2020 at 13:19 | history | edited | Xi'an | CC BY-SA 4.0 |
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| Jun 2, 2020 at 12:59 | comment | added | Do not reinstate Monica | The part that gets squared is the difference from the mean. In linear regression the conditional mean of $Y|X=x$ is $Y - \theta x$. | |
| Jun 2, 2020 at 12:47 | history | asked | Raz | CC BY-SA 4.0 |