= Gaussian regression toward the mean
{c}
{title2=$Y\mid X=x\sim N(0.6x,0.64)$}
Take the original synthetic model $T\sim N(0,0.6)$ and $\varepsilon_1,\varepsilon_2\sim N(0,0.4)$, mutually <independent random variables>, with the second argument denoting <variance>. The pair $(X,Y)$ has a <multivariate normal distribution> with <mean> zero, unit marginal <variances>, and <Pearson correlation coefficient> $0.6$. Its <conditional multivariate normal distribution> is
$$
Y\mid X=x\sim N(0.6x,1-0.6^2)=N(0.6x,0.64).
$$
For a first reading $x=2$, the second reading has <conditional expectation> $1.2$ and <conditional variance> $0.64$, hence <standard deviation> $0.8$. There is still <conditional probability> $1-\Phi(1)\approx0.1587$ of $Y>2$, a movement farther from zero, where $\Phi$ is the <standard normal distribution function>. The line $0.6x$ describes a <conditional mean>, not an individual trajectory.
\Image[/regression-toward-mean-gaussian-scatter.png]
{title=An extreme first reading predicts a less extreme conditional mean, with substantial scatter}
{description=Original synthetic <scatter plot> of 1,800 generated pairs, explicitly not actual measurements. Each pair has unit marginal <variances> and <Pearson correlation coefficient> $0.6$; orange points satisfy $X>1$. The blue line is the <conditional expectation> $0.6x$, the band is one conditional <standard deviation> $0.8$ on either side, and the dashed line denotes $Y=X$. The band is not a <confidence interval> for an estimated line. The axes show a finite display window.}
A short derivation isolates $R=Y-0.6X$. Its <covariance> with $X$ is zero and its <variance> is $0.64$. Because $(X,R)$ has a <multivariate normal distribution>, zero <covariance> implies independence: $R\sim N(0,0.64)$ independently of $X$. This also explains why reversing the two measurements gives $\mathbb E[X\mid Y=y]=0.6y$. The <statistical association> does not identify a direction of <causal effect>.
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