# OLS with the lowest MSE question

I am struggling to prove that the an estimator $\beta_{ls}=\frac{\sum X_iY_i}{\sum X_i^2}$ has a lower MSE than $\beta_{ols}=\frac{\sum (X_i-\bar{X})(Y_i-\bar{Y})}{\sum (X_i-\bar{X})^2}$. The least square $\beta_{ls}$ has no constant. Does the MSE for $\beta_{ls}$ has a lower MSE because of one less degree of freedom than $\beta_{ols}$?

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If this is homework it should have the homework tag. –  Peter Flom Oct 7 '12 at 12:28
actually it is a past exam practice question –  CharlesM Oct 7 '12 at 17:27
Assuming "$\bar{X_n}$" is a typo for the mean $\bar{X}$, the assertion is false in general: you need more assumptions. E.g., take $n=2$, $X_1$=$0$, $X_2$=1, let $Y_1$ have any distribution with a positive variance, and suppose $Y_2=Y_1$ (i.e., the responses are not independent). Then the MSE of $\beta_{ols}$ is $0$ while the MSE of $\beta_{ls}$ is strictly positive. –  whuber Oct 7 '12 at 19:14
@whuber yes it was a typo sorry –  CharlesM Oct 7 '12 at 20:32
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