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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.

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.

Cognitive psychology 41, 49–100; latent variable analyses.

Persistent unmeasured differences between individuals can generate outcome dependence after adjusting for observed covariates. A random intercept is one representation. Conditional independence given that latent variable need not persist after it is integrated out; apparent history dependence can therefore occur without true state dependence.

A variational autoencoder is a generative model with a latent variable $z$ and a learned approximation to its posterior distribution given an observation $x$. Write the model as $p_{\theta }(x,z)=p(z)p_{\theta }(x\mid z)$. A recognition network, implemented as a neural network, produces the parameters of $q_{\varphi }(z\mid x)$; a generative decoder specifies $p_{\theta }(x\mid z)$. Using one recognition rule across observations is amortized inference, rather than solving a fresh independent optimization problem for every input.^[ref]

Training maximizes an evidence lower bound, often writtenThe first term scores reconstruction under the declared likelihood; the second compares the inferred latent law with the prior. The identityproves that it is a lower bound, since Kullback-Leibler divergence is nonnegative. It becomes tight when the recognition law equals the model posterior, under the usual density/support conditions.^[ref]

For a one-dimensional Gaussian example, let the prior be $N(0,1)$ and the recognition law be $N(\mu ,{\sigma }^2)$. Substituting their log densities and using $\mathbb{E}[z^2]={\mu }^2+{\sigma }^2$ givesAt $\mu =1$, $\sigma =2$, this is $2-\log 2\simeq 1.30685$. The value is a penalty for this particular latent law, not the whole training loss. Gaussian reparameterization writes $z=\mu +\sigma \epsilon$ with fixed-law standard-normal noise so that gradients can pass through the sampled latent value. After training, generation instead draws from the prior and then the generative decoder; it need not start from an observed input.

Software for Bayesian inference through Markov chain Monte Carlo in directed statistical models. Its model language uses dnorm(mean, precision), so the second argument is inverse variance; dcat indexes a supplied probability vector, and dbern introduces a binary latent variable. The model is declarative, so the order of deterministic node definitions does not impose procedural evaluation order.