Repeated-measurement latent signal model ($X=T+{\epsilon }_1,Y=T+{\epsilon }_2$)

Imagine an instrument measuring the same fixed quantity twice on each unit. Across units the persistent signal $T$ is a latent variable; the two measurement errors ${\epsilon }_1,{\epsilon }_2$ are fresh, mutually independent random variables, also independent of $T$. Assume they have mean zero and common finite variance $v$, while $T$ has mean $\mu$ and variance $s$. ThenThus the Pearson correlation coefficient is $\rho =s/(s+v)$ when $s+v>0$. Both measurements retain the same marginal mean and variance. Repetition does not make the entire population collapse toward $\mu$; it changes what we expect after selecting on $X$.

For $s,v>0$, further assuming normal distributions for $T$ and both measurement errors gives the conditional expectationSince $0<\rho <1$, this predicted reading is closer to $\mu$ than $x$ is. Independence and zero mean errors alone do not justify this linear conditional expectation: without the normal distribution assumption the same expression is the linear least-squares projection, and the actual conditional expectation can be nonlinear. Correlated errors, a changing signal, or different error variances require their own model; a noisy reading by itself supplies no universal shrinkage factor.

Take the original synthetic model $T\sim N(0,0.6)$ and ${\epsilon }_1,{\epsilon }_2\sim N(0,0.4)$, 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 isFor 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 $1-\Phi (1)\approx 0.1587$ 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.

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.

Select units with first reading $X>1$, keep that membership fixed, and measure the same units again. The first reading now has a truncated normal distribution. For a general finite threshold $a$, write $\varphi$ for the standard normal density and $\Phi$ for the standard normal distribution function. Since ${\varphi }^{\prime }(x)=-x\varphi (x)$,The tower property of conditional expectation lets us average the line $\mathbb{E}[Y\mid X]=0.6X$ over the selected group.^[ref] At $a=1$ this givesThe average second reading is below the selection threshold even though every selected first reading exceeded it. This is a group expectation; any finite generated group's sample mean fluctuates around it. Selecting again on $Y>1$ would define a different group and answer a different question.

The latent variable decomposition makes the apparent drop especially clear. Gaussian conditioning gives $\mathbb{E}[T\mid X=x]=0.6x$ and $\mathbb{E}[{\epsilon }_1\mid X=x]=0.4x$. ThusThe first measurement error has zero unconditional mean, but positive selected mean. The second measurement error stays centered because it is independent of selection on $X$. The persistent signal $T$ never changed.