Sensitivity to the Mixture Weight

Why Vary the Mixture Weight?

A robust mixture prior represents uncertainty about whether historical information is sufficiently compatible with the current trial to borrow strongly. It combines a borrowing component, which encodes the historical information, with a protective component intended to limit borrowing when the data conflict with that information. The mixture weight is the prior probability assigned to the borrowing component. Even if the components are held fixed, changing this weight can affect both the posterior probability of borrowing and the posterior treatment estimate.

Sensitivity analysis varies the prespecified weight across a defensible range and recomputes the posterior quantities of interest. This is not the same as estimating a preferred weight from the current data. Rather, it asks how dependent the conclusions are on the prior odds of borrowing. The component marginal likelihoods determine how the observed data update those odds; the chosen prior weight determines where the updating starts.

Definition: Let $\theta$ be a treatment parameter, and let $B$ and $P$ label the borrowing and protective components. A robust mixture prior is $$ \pi(\theta)=w\pi_B(\theta)+(1-w)\pi_P(\theta), \qquad 0\leq w\leq 1. $$ Here $w=P(B)$ is the prespecified prior probability of the borrowing component. After observing data $y$, the posterior borrowing probability is $q(w\mid y)=P(B\mid y)$. These are distinct quantities: $w$ is specified before observing $y$, whereas $q(w\mid y)$ is updated using the data.

Posterior Updating and Its Dependence on the Weight

For component $j\in\{B,P\}$, define its marginal likelihood by $m_j(y)=\int f(y\mid\theta)\pi_j(\theta)\,d\theta$. It is the probability density or probability of the observed data after averaging over that component’s prior. Bayes’ rule gives the posterior borrowing probability

q(w\mid y)=\frac{w\,m_B(y)} {w\,m_B(y)+(1-w)m_P(y)}.

When both marginal likelihoods are positive, their ratio $BF_{B:P}=m_B(y)/m_P(y)$ is the Bayes factor comparing the borrowing component with the protective component. In odds form,

\frac{q(w\mid y)}{1-q(w\mid y)} = \frac{w}{1-w}BF_{B:P}.

Thus the data multiply the prior component odds by the Bayes factor. A small Bayes factor can make the posterior borrowing probability low even when $w$ is substantial; a large one can increase it. Sensitivity analysis examines how this update changes when the initial odds, and therefore $w$, change.

A Normal-Model Illustration

Suppose a treatment-effect estimate $y$ has the sampling model $Y\mid\theta\sim N(\theta,1)$, where the known sampling variance is one squared treatment-effect unit. Suppose the borrowing prior is $\theta\mid B\sim N(0,1)$, and the protective prior is $\theta\mid P\sim N(0,4)$. These are proper normal distributions, and the normal likelihood is used for the estimate with its stated known variance. Under either component, the marginal distribution of $Y$ is normal: its mean is zero and its variance is the sum of the sampling variance and prior variance.

Within each component, normal-normal conjugacy gives a posterior mean $m_j$ and variance $v_j$. The overall posterior is a mixture of these component posteriors, with posterior weights $q(w\mid y)$ and $1-q(w\mid y)$. Its mean is the weighted average of the component posterior means. Its variance also includes the variation between those means.

Worked Example: Conflicting Data and a Lower Borrowing Probability

Consider hypothetical trial data $y=3$ treatment-effect units under the normal model just specified. The estimate is farther from the common prior mean of zero than either prior component typically expects, so the component marginal likelihoods must be compared rather than assuming the borrowing prior remains dominant.

The marginal variances are $1+1=2$ under $B$ and $1+4=5$ under $P$. Substituting $y=3$ into the two normal densities gives

m_B(3)=\frac{1}{\sqrt{4\pi}}\exp\left(-\frac{9}{4}\right) \approx 0.02973, \qquad m_P(3)=\frac{1}{\sqrt{10\pi}}\exp\left(-\frac{9}{10}\right) \approx 0.07254.

The values use the stated marginal variances, including the factor $1/2$ in the normal density exponent. Their ratio is $BF_{B:P}=0.0297326/0.0725371\approx 0.4099$. The observed estimate therefore has about 0.4099 times the marginal density under the borrowing component as under the protective component.

The component posterior means and variances follow from normal-normal updating. For $B$, the posterior mean is $(1/(1+1))3=1.5$ units and the variance is $1/2=0.5$ squared units. For $P$, the posterior mean is $(4/(4+1))3=2.4$ units and the variance is $4/5=0.8$ squared units. Now vary $w$:

At $w=0.2$, $q=(0.2)(0.4099)/[0.8+(0.2)(0.4099)]\approx 0.0930$. The posterior mean is $(0.0930)(1.5)+(0.9070)(2.4)\approx 2.316$ units.
At $w=0.5$, $q=0.4099/[1+0.4099]\approx 0.2907$. The posterior mean is $(0.2907)(1.5)+(0.7093)(2.4)\approx 2.138$ units.
At $w=0.8$, $q=(0.8)(0.4099)/[0.2+(0.8)(0.4099)]\approx 0.6212$. The posterior mean is $(0.6212)(1.5)+(0.3788)(2.4)\approx 1.841$ units.

Increasing the prior borrowing weight increases posterior borrowing, but the Bayes factor below one continues to favor the protective component relative to the prior odds. The posterior mean moves toward the borrowing-component mean of 1.5 units as the borrowing weight increases.

How the Sensitivity Works

The dependence of $q$ on $w$ can be seen directly from the odds formula. Writing $BF=BF_{B:P}$, differentiation gives

q(w\mid y)=\frac{wBF}{1-w+wBF} =\frac{wBF}{1+w(BF-1)}, \qquad \frac{dq}{dw}= \frac{BF}{[1+w(BF-1)]^2}>0.

For positive marginal likelihoods, posterior borrowing probability increases with prior borrowing weight. The amount of change depends on the Bayes factor and the part of the weight range being examined. In log-odds form, $\operatorname{logit}(q)=\operatorname{logit}(w)+\log(BF)$: the data add the same log Bayes factor to the prior log odds at every interior value of $w$.

If the component posterior means are $m_B$ and $m_P$, the overall posterior mean is $\bar m(w)=q m_B+(1-q)m_P$. Holding the component priors and likelihood fixed, its sensitivity is

\frac{d\bar m}{dw}=(m_B-m_P)\frac{dq}{dw}.

The estimate therefore moves in the direction of the component mean that receives more posterior weight. This does not imply that every posterior summary changes monotonically: the mixture variance, for example, includes both within-component variances and a between-component term.

For a target posterior borrowing probability $q_*$ strictly between zero and one, the required prior weight can also be found from the odds equation:

w=\frac{q_*}{q_*+(1-q_*)BF_{B:P}}.

Worked Example: How Much Prior Weight Gives a Target Posterior Weight?

Use the conflicting-data Bayes factor $BF_{B:P}=0.4099$ from the previous calculation. What prior borrowing weight would produce a posterior borrowing probability of $0.5$ for those data?

Substitute $q_*=0.5$ in the formula:

w=\frac{0.5}{0.5+(0.5)(0.4099)} =\frac{0.5}{0.70495} \approx 0.7093.

A prior borrowing probability of about 0.7093 is needed for the posterior borrowing probability to reach 0.5. This does not mean the data favor borrowing: the Bayes factor remains below one. Instead, prior odds sufficiently favor borrowing to offset the data’s relative support for the protective component. The result is a calculation for this specified model and observed estimate, not a general recommendation to set $w$ to 0.7093.

Worked Example: Concordant Data and the Same Weight Range

Now consider a second hypothetical estimate, $y=0.5$ treatment-effect units, with the same likelihood and prior components. This estimate is closer to the components’ shared prior mean than $y=3$. The marginal densities are

m_B(0.5)=\frac{1}{\sqrt{4\pi}}\exp\left(-\frac{0.25}{4}\right) \approx 0.26499, \qquad m_P(0.5)=\frac{1}{\sqrt{10\pi}}\exp\left(-\frac{0.25}{10}\right) \approx 0.17401.

Thus $BF_{B:P}\approx 0.26499/0.17401=1.523$, which favors $B$ relative to $P$. The component posterior means are $m_B=(1/2)(0.5)=0.25$ units and $m_P=(4/5)(0.5)=0.4$ units. For $w=0.2$, $q=(0.2)(1.523)/[0.8+(0.2)(1.523)]\approx 0.2757$, giving a mixture mean of $(0.2757)(0.25)+(0.7243)(0.4)\approx 0.3586$ units. For $w=0.5$, $q=1.523/(1+1.523)\approx 0.6036$, giving a mean of approximately $0.3095$ units. For $w=0.8$, $q=(0.8)(1.523)/[0.2+(0.8)(1.523)]\approx 0.8590$, giving a mean of approximately $0.2712$ units.

Here the data and the prior odds both favor the borrowing component when $w$ exceeds one half. Yet the posterior estimate still changes across the range because the component posterior means differ. “Concordant” is not a claim that the component predictions are identical; it describes data that do not create strong conflict with the borrowing information.

Diagnostics, Comparisons, and Limitations

A useful sensitivity display reports $w$, $q(w\mid y)$, and the posterior treatment estimate together for each prespecified scenario. It should also show the component Bayes factor or otherwise make clear how the data update the prior odds. If the treatment estimate or a decision-relevant posterior probability changes materially across plausible weights, the conclusion is sensitive to the prespecified mixture weight and should be reported as such.

Check the calculations and model inputs if the posterior borrowing probability decreases as $w$ increases while the components and data are fixed: the derivative above shows this cannot occur for positive marginal likelihoods. Also check whether a reported posterior estimate is consistent with the component estimates and weights. A mixture mean must lie between the component means. Extreme sensitivity may reflect genuine component disagreement, a Bayes factor strongly favoring one component, or an unusually broad chosen range of $w$; it is a reason to inspect the assumptions, not by itself proof that the model is wrong.

Sensitivity to $w$ is distinct from sensitivity to component specifications. Changing the protective component’s scale changes its marginal likelihood and posterior distribution, as well as the Bayes factor; that is not a weight-only comparison. Keep the component models fixed when isolating sensitivity to $w$. Similarly, comparing the mixture with a single borrowing prior or with a single protective prior can help describe the role of the components, but those alternatives do not provide the mixture’s data-driven reweighting.

A robust mixture also differs from hard selection of one component: it retains both component posteriors and averages them using posterior probabilities. A separate approach could assign a probability distribution to the weight and integrate over it, but that changes the analysis from a fixed-weight sensitivity assessment. No sensitivity table can establish that a particular prior weight is objectively correct. The plausible range must be justified using scientific knowledge and trial design, and conclusions remain conditional on the likelihood and both component priors.

Common mistake: “The posterior borrowing probability is low, so the prior borrowing weight must have been low.” This is not necessarily true. A low $q(w\mid y)$ can result from a Bayes factor that strongly favors the protective component even when $w$ is substantial. Report the prior weight and the data-driven update separately.

Check Your Understanding

Use the normal-model conflicting-data example above, with $BF_{B:P}=0.4099$, $m_B=1.5$ units, and $m_P=2.4$ units. Try the questions, then check the solutions.

At $w=0.4$, calculate the posterior borrowing probability.
Using that value, calculate the mixture posterior mean in treatment-effect units.
For $BF_{B:P}=1$, what is the relationship between the posterior and prior borrowing probabilities?
Explain why raising $w$ cannot lower $q(w\mid y)$ when the components and data are fixed.
State one reason why a weight-only sensitivity analysis does not address all prior sensitivity.

Solutions. (1) $q=(0.4)(0.4099)/[0.6+(0.4)(0.4099)]=0.16396/0.76396\approx 0.2146$. (2) The mean is $(0.2146)(1.5)+(0.7854)(2.4)\approx 2.207$ treatment-effect units. (3) If $BF=1$, the data give equal marginal support to the two components, so the posterior odds equal the prior odds and $q=w$. (4) The derivative $dq/dw=BF/[1+w(BF-1)]^2$ is positive for a positive Bayes factor. (5) Changing a component’s prior distribution can change its marginal likelihood and posterior estimate; holding both components fixed while varying $w$ does not assess that source of sensitivity.