Spike-And-Slab Priors for a Point Null

Why Put Prior Probability at Exactly Zero?

A prior can express skepticism about the size of a treatment effect without asserting that the effect is exactly zero. A continuous skeptical prior does this by concentrating density near zero while assigning probability zero to any single exact value. A different scientific question is whether there is a genuine possibility of no effect at all. A spike-and-slab prior represents that possibility by assigning positive prior probability to an exact point null and spreading the remaining probability over a continuous alternative.

This distinction matters when the trial question includes the model $H_0:\theta=0$, not just whether an effect is small. Here $\theta$ denotes a treatment effect on a specified scale, and zero means no treatment effect on that scale. The units of $\theta$ must be stated in the application—for example, percentage points for a risk difference or millimetres of mercury for a blood-pressure difference.

Definition: A spike-and-slab prior for a treatment effect $\theta$ is a mixture of a point mass at zero and a continuous probability distribution: $$ \pi(\theta)=\pi_0\delta_0(\theta)+(1-\pi_0)g(\theta), $$ where $0<\pi_0<1$, $\delta_0$ is a point mass at $\theta=0$, and $g$ is a proper probability density for the alternative model. The prior probability of the point-null model is $\pi_0$; the prior probability of the continuous alternative model is $1-\pi_0$.

It is useful to name the two models $M_0$ and $M_1$. Under $M_0$, $\theta=0$ exactly. Under $M_1$, $\theta$ has density $g$. The slab density may itself express skepticism by putting most of its mass near zero, but it contains no point mass at zero. A point-null probability and concentration near zero are therefore separate prior choices.

Updating the Probability of the Point Null

Let $D$ denote the trial data. The marginal likelihood under each model is the probability density of the observed data after averaging over that model's parameter values. In particular, $m_0(D)=p(D\mid M_0)$, while $m_1(D)=\int p(D\mid\theta,M_1)g(\theta)\,d\theta$. Their ratio is the Bayes factor $BF_{10}=m_1(D)/m_0(D)$: values above one mean the data are better predicted by $M_1$, and values below one mean they are better predicted by $M_0$.

Theorem: For prior model probabilities $\pi_0=P(M_0)$ and $1-\pi_0=P(M_1)$, the posterior probability of the point null is $$ P(M_0\mid D) =\frac{\pi_0m_0(D)} {\pi_0m_0(D)+(1-\pi_0)m_1(D)} =\frac{\pi_0}{\pi_0+(1-\pi_0)BF_{10}}. $$ Equivalently, posterior odds for $M_1$ versus $M_0$ equal prior odds multiplied by $BF_{10}$.

The result follows directly from Bayes' rule. The joint probabilities of the observed data and each model are $P(M_0)m_0(D)$ and $P(M_1)m_1(D)$; dividing the first by their sum gives the posterior model probability. Dividing numerator and denominator by $m_0(D)$ gives the expression involving $BF_{10}$. The Bayes factor is not a posterior probability: the prior model probabilities are also needed to obtain one.

A Normal Summary-Likelihood Calculation

For transparent calculations, suppose a trial produces an estimate $y$ of $\theta$, with known standard error $SE$, and the sampling model is $y\mid\theta\sim N(\theta,v)$, where $v=SE^2$. Assume independent observations or an analysis that justifies this normal summary likelihood, and treat $SE$ as fixed. For illustration, take a zero-centred normal slab, $\theta\mid M_1\sim N(0,\tau^2)$, with $\tau>0$. Both the likelihood approximation and the slab are assumptions to assess for the particular endpoint and trial design.

m_0(y)=\frac{1}{\sqrt{2\pi v}}\exp\left(-\frac{y^2}{2v}\right), \qquad m_1(y)=\int \frac{1}{\sqrt{2\pi v}}\exp\left(-\frac{(y-\theta)^2}{2v}\right) \frac{1}{\sqrt{2\pi\tau^2}}\exp\left(-\frac{\theta^2}{2\tau^2}\right) \,d\theta

Under $M_1$, the estimate can be represented as $y=\theta+\varepsilon$, where $\theta\sim N(0,\tau^2)$ and $\varepsilon\sim N(0,v)$ are independent. Their sum is normal with mean zero and variance $\tau^2+v$, so $m_1(y)$ is the density of $N(0,v+\tau^2)$ evaluated at $y$. Taking the ratio of this density to $m_0(y)$ gives

BF_{10} =\sqrt{\frac{v}{v+\tau^2}} \exp\left\{\frac{y^2}{2} \left(\frac{1}{v}-\frac{1}{v+\tau^2}\right)\right\}.

The factor outside the exponential reflects how broadly the alternative predicts possible estimates; the exponential reflects how well the observed estimate fits each model. A large estimate can favour the alternative, but a very diffuse slab spreads its predictive probability over a wider range. Consequently, the Bayes factor depends on the specified slab, not only on the estimate and its standard error.

Conditional on $M_1$, the posterior distribution of the effect is also normal. Completing the square in the product of the likelihood and slab densities gives posterior variance $v\tau^2/(v+\tau^2)$ and posterior mean $\tau^2y/(v+\tau^2)$. Under the full spike-and-slab prior, this continuous posterior is only one part of the answer: the other part is the posterior point mass at zero.

Worked Calculations

Worked Example: A Modest Estimate and a Point-Null Probability

Hypothetical data. Suppose the estimated treatment effect is $y=0.20$ percentage points, with $SE=0.15$ percentage points. Use a slab $N(0,0.30^2)$ in percentage-point units and prior model probabilities $P(M_0)=P(M_1)=0.50$. Assume the normal summary likelihood above is appropriate.

Here $v=0.15^2=0.0225$ square percentage points and $\tau^2=0.09$ square percentage points. Substituting in the Bayes-factor formula,

BF_{10} =\sqrt{\frac{0.0225}{0.0225+0.09}} \exp\left\{\frac{0.20^2}{2} \left(\frac{1}{0.0225}-\frac{1}{0.1125}\right)\right\} =\sqrt{0.2}\exp(0.711111) \approx 0.91064.

The data are slightly more probable under the point null than under this specified slab, since $BF_{10}<1$. Updating the equal prior model probabilities gives

P(M_0\mid y) =\frac{0.50}{0.50+0.50(0.91064)} \approx 0.5234.

Thus the posterior probability of exactly no effect is about 52.3% under these two models and these prior probabilities. This is not a probability statement about the null being true in an unconditional sense; it is a posterior model probability conditional on the stated models, priors, and data. Under $M_1$, the conditional posterior mean is $0.09(0.20)/0.1125=0.16$ percentage points, and the conditional posterior variance is $0.09(0.0225)/0.1125=0.018$ square percentage points.

Worked Example: A Larger Estimate Can Shift Model Probability

Hypothetical data. Keep the same standard error, slab, and equal prior model probabilities, but suppose the trial estimate is $y=0.45$ percentage points. These inputs still use the same assumed normal summary likelihood.

The variance terms and square-root factor are unchanged. The exponent is $\frac{0.45^2}{2}(44.4444-8.88889)=3.6$, so

BF_{10} =\sqrt{0.2}\exp(3.6) \approx 16.3672.

The posterior probability of the point null is therefore

P(M_0\mid y) =\frac{0.50}{0.50+0.50(16.3672)} =\frac{1}{1+16.3672} \approx 0.05758.

Under these assumptions, the data substantially favour the continuous alternative over the point null, and the posterior point-null probability is about 5.76%. Conditional on $M_1$, the posterior mean is $0.09(0.45)/0.1125=0.36$ percentage points and the posterior variance remains $0.018$ square percentage points. The overall posterior is a mixture: about 5.76% of its probability is exactly at zero, with the remaining probability following this conditional normal distribution.

Worked Example: Exact Zero Versus Continuous Skepticism

Hypothetical data. Suppose $y=0$ percentage points and $SE=0.15$ percentage points, with the same $N(0,0.30^2)$ slab and equal prior model probabilities. The normal likelihood assumption is unchanged.

At $y=0$, the exponential term in the Bayes factor equals one:

BF_{10} =\sqrt{\frac{0.0225}{0.1125}}\exp(0) =\sqrt{0.2} \approx 0.44721.

Hence the posterior probability at the point null is

P(M_0\mid y) =\frac{0.50}{0.50+0.50(0.44721)} \approx 0.6910.

In this spike-and-slab analysis, the posterior assigns about 69.1% probability to exactly zero. By contrast, if the prior were only the continuous distribution $N(0,0.30^2)$, its posterior for these data would be $N(0,0.018)$, which assigns probability zero to the event $\theta=0$. The continuous prior can concentrate around zero, but neither it nor its posterior creates an atom at zero.

Assumptions, Diagnostics, and Alternatives

A point-null probability is meaningful only relative to the models being compared. Check that zero has a well-defined interpretation on the chosen effect scale, that $M_0$ and $M_1$ describe the relevant possibilities, and that the alternative prior is proper so its marginal likelihood is defined. In the normal calculation, also examine whether the estimate is adequately represented by a normal likelihood with the stated standard error. If the model has nuisance parameters, their treatment must be specified consistently in each model; the simple formula above does not remove that requirement.

Useful diagnostics include checking the implied prior predictive distribution of estimates under each model and recalculating the posterior over plausible choices of $\pi_0$ and slab spread. A prior predictive distribution that places implausibly little probability on clinically possible estimates signals a poor model specification. A posterior point-null probability that changes sharply under modest, defensible changes to the slab or prior model probability is sensitive to those choices and should be reported as such. The estimate and standard error alone cannot reveal this sensitivity.

The main alternatives answer different questions. A continuous skeptical prior addresses uncertainty about the size of the effect while excluding an exact point mass; it does not yield a posterior probability for the exact null. A spike-and-slab prior permits posterior model probabilities but requires an explicit probability for the point null and a slab. A test framed in terms of a negligible-effect region instead asks whether the effect lies within a specified range; that is not the same as asking whether it equals exactly zero.

A frequent mistake is to say, “The prior is concentrated near zero, so it gives a high probability that the effect is exactly zero.” That conclusion is incorrect for a continuous prior: any single point, including zero, has probability zero. To assign positive probability to exact zero, the prior must include a point mass. A second pitfall is to interpret $BF_{10}$ as the posterior probability of $M_1$. The Bayes factor is a ratio of marginal likelihoods; posterior model probability also depends on prior model odds.

Key takeaway: In a spike-and-slab analysis, the posterior has an exact point mass at zero and a continuous alternative component. The data update the prior probability of the point-null model through a Bayes factor, while the slab determines how the alternative predicts the data.

Check Your Understanding

Work through each question, then compare with the solutions.

Suppose $P(M_0)=0.75$ and the data give $BF_{10}=2$. Find the posterior probability of $M_0$.
In the normal summary model, let $y=0.15$ percentage points, $SE=0.15$ percentage points, $\tau=0.30$ percentage points, and $P(M_0)=0.50$. Calculate $BF_{10}$ and the posterior point-null probability.
A prior for $\theta$ is a continuous normal distribution centred at zero. What is $P(\theta=0)$ before and after observing data under a continuous likelihood?
In one or two sentences, explain why increasing the slab spread can change the Bayes factor even when the estimate and standard error do not change.

Solutions. (1) The prior alternative probability is $0.25$, so $P(M_0\mid D)=0.75/[0.75+0.25(2)]=0.60$. (2) Here $v=0.0225$, $v+\tau^2=0.1125$, and the exponent is $\frac{0.15^2}{2}(44.4444-8.88889)=0.4$. Thus $BF_{10}=\sqrt{0.2}\exp(0.4)\approx 0.6672$, and $P(M_0\mid y)=1/[1+0.6672]\approx 0.5998$. (3) Both probabilities are zero: a continuous distribution assigns zero probability to a single point, and updating with a continuous likelihood preserves a continuous posterior. (4) A broader slab spreads its predictive probability over a wider range of estimates. The marginal likelihood under $M_1$, and therefore the ratio of marginal likelihoods, changes as a result.