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A Gaussian practical-null mixture compares a narrow centered normal distribution prior with a wider centered normal distribution prior. Neither component is a point mass. With observed mean $y\mid \beta \sim N(\beta ,1/n)$ and prior precisions $q_0>q_1>0$, its model predictive variances are $V_i=1/n+1/q_i$, and the Bayes factor isThe overall posterior density is a Bayesian model averaging mixture with component means $ny/(n+q_i)$. A wide-prior penalty can favor the practical null near zero; its dominance depends quantitatively on both prior scales and the observation, not merely on the observation being of order $n^{-1/2}$.

Let $M\in \{1,2\}$ be one fixed model indicator with equal prior probabilities, and put $w_{m,t}=\mathbb{P}(M=m\mid y_1,\dots ,y_{t-1})$. Within each model the parameters are already updated using the same past data. The Bayesian model averaging forecast is $f_t(y)={\sum }_m{w}_{m,t}{f}_{mt}(y)$. Upon observing $y_t$, Bayes' theorem givesTaking the ratio cancels the denominator and gives the prescribed update because $f_{mt}(y_t)=e^{L_{mt}}$. Iteration yieldsThe weights are posterior model probabilities and their mixture is the full Bayesian predictive density. The indicator is fixed across days, rather than choosing a fresh model independently each morning.

With equal model prior probabilities, Bayes factor updating gives $w_0=B_{01}/(1+B_{01})$ and $w_1=1/(1+B_{01})$. The Bayesian model averaging posterior density is thereforeThis is a two-component mixture model of the normal distributions already derived. For the numerical observation above, $w_0\simeq 0.10829$ and $w_1\simeq 0.89171$. Consequently it is predominantly the wide-model posterior, not predominantly the narrow one.

Put ${\kappa }_i=n/(n+q_i)$. The posterior mean of the Gaussian practical-null mixture isNear zero, $B_{01}(0)>1$, so the narrow component has high posterior probability and ${\kappa }_0$ is small when $n_0\gg n$. This pulls the Bayesian model averaging posterior toward zero. For the given ratios, writing $z=\sqrt{n}y$ givesThus the order statement $y=O(n^{-1/2})$ alone does not guarantee strong pull toward zero: $z=3$ already has the opposite model preference.

As $|y|$ grows, $V_1>V_0$ makes $B_{01}(y)\to 0$, and the mixture model approaches $N({\kappa }_{1}y,1/(n+q_1))$. Its center is exactly ${\kappa }_{1}y$, and approximately $y$ only for a sufficiently diffuse wide prior, meaning $q_1\ll n$. Here ${\kappa }_1=100/101$, so the relative displacement is about one percent. Large observations alone do not remove this finite-prior shrinkage.

In a normal linear model, attach a continuous spike-and-slab prior to each regression coefficient. With suitably scaled predictors, introduce indicators ${\gamma }_j\sim \mathrm{Bernoulli}(\pi )$ and setThe narrow component describes practically negligible effects; the wide component permits substantial ones. Fit the joint Bayesian posterior of coefficients, indicators and any unknown residual variance. Bayesian model averaging gives shrinkage toward zero for poorly supported effects, while $\mathbb{P}({\gamma }_j=1\mid \mathcal{D})$ quantifies wide-component support. A shared beta distribution prior on $\pi$ can represent uncertainty about how many effects are substantial.

Select on scientifically meaningful effect size, for example a high $\mathbb{P}(|{\beta }_j|>{\delta }_j\mid \mathcal{D})$ for a prechosen threshold in meaningful predictor units. Membership in the wide component alone does not imply a large realized effect: its normal distribution still permits values near zero. Correlated predictors also require interpretation of the joint Bayesian posterior, rather than treating each coefficient as an isolated test.

With equal model prior probabilities, Bayesian model averaging gives the unnormalized densityNormalizing this expression over the allowed interval automatically incorporates each lens model's evidence and hence its posterior model probability.