 # Let a sample ( x , y ) &#x2208;<!-- ∈ --> <mi mathvariant="double-stru Reginald Delacruz 2022-06-24 Answered
Let a sample $\left(x,y\right)\in {\mathbb{R}}^{2n}$ be given, where $y$ only attains the values $0$ and $1$. We can try to model this data set by either linear regression
${y}_{i}={\alpha }_{0}+{\beta }_{0}{x}_{i}$
with the coefficients determined by the method of least squares or by logistic regression
${\pi }_{i}=\frac{\mathrm{exp}\left({\alpha }_{1}+{\beta }_{1}{x}_{i}\right)}{1+\mathrm{exp}\left({\alpha }_{1}+{\beta }_{1}{x}_{i}\right)},$
where ${\pi }_{i}$ denotes the probability that ${y}_{i}=1$ under the given value ${x}_{i}$ and the coefficients are determined by the Maximum-Likelihood method. My question is whether the following statement holds true.
Claim: If ${\beta }_{0}>0$ (${\beta }_{0}<0$), then ${\beta }_{1}>0$ (${\beta }_{1}>0$).
I figure this could be due to the sign of the correlation coefficient.
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If ${\beta }_{0}>0$, i.e., $\frac{\mathrm{\partial }}{\mathrm{\partial }x}E\left[Y|X=x\right]={\beta }_{0}>0$, is suggests that increase in $x$ will increase the probability that $Y=1$, since $E\left[Y|X=x\right]=P\left(Y=1|X=x\right)=p$. Now, the logistic model is equivalent to
$\mathrm{ln}\left(\frac{p}{1-p}\right)={\alpha }_{1}+{\beta }_{1}x.$
The right hand side can be viewed as linear approximation of $\mathrm{ln}\left(p/\left(1-p\right)\right)$. Given that the original linear model is ok, and ${\beta }_{0}>0$, it translates into $\mathrm{\partial }/\mathrm{\partial }x\mathrm{ln}\left(p/\left(1-p\right)\right)=\left(p\left(1-p\right){\right)}^{-1}{\beta }_{0}>0$ for the $\mathrm{ln}$ odds model, namely, ${\beta }_{1}$ must be positive as well (as the slope of the linear approximation).

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