= Repeated-measurement latent signal model
{title2=$X=T+\varepsilon_1,\quad Y=T+\varepsilon_2$}
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> $\varepsilon_1,\varepsilon_2$ 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> $s$. Then
$$
X=T+\varepsilon_1,\qquad Y=T+\varepsilon_2,\qquad
\mathbb E[X]=\mathbb E[Y]=\mu,\qquad
\operatorname{Var}(X)=\operatorname{Var}(Y)=s+v,\qquad
\operatorname{Cov}(X,Y)=s.
$$
Thus 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 expectation>
$$
\mathbb E[Y\mid X=x]=\mu+\rho(x-\mu).
$$
Since $0<\rho<1$, 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.
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