Power Priors With Unknown Outcome Variance

Why Unknown Variance Changes the Update

A normal power prior with a known outcome variance can discount historical observations by reducing their contribution to the mean update. When the variance is unknown, the historical likelihood also contributes information about the spread of outcomes. Raising that likelihood to a power discounts both contributions: the historical sample size and its residual variation. This matters in a clinical trial because a historical control dataset may inform not only the expected control outcome but also how variable outcomes are.

The model below treats the current and historical control outcomes as observations from a common normal distribution, while assigning the historical likelihood a fixed discount. The common-distribution assumption is substantive: after harmonizing the endpoint and its measurement scale, the model assumes the studies share a mean and variance. The discount reduces the historical data’s influence; it does not itself model a systematic difference between studies.

Definition: Let $y_{hj}$, $j=1,\ldots,n_h$, be historical control outcomes and $y_{cj}$, $j=1,\ldots,n_c$, be current control outcomes. Assume conditional independence and $$ y_{hj}\mid\mu,\sigma^2\sim N(\mu,\sigma^2),\qquad y_{cj}\mid\mu,\sigma^2\sim N(\mu,\sigma^2). $$ Here $\mu$ is the common outcome mean and $\sigma^2>0$ is the common, unknown outcome variance. For a fixed power-prior discount $a\in[0,1]$, the likelihood used for the update is $L_c(\mu,\sigma^2)L_h(\mu,\sigma^2)^a$, where $L_c$ and $L_h$ are the current and historical normal likelihoods.

Use the conjugate prior $$ \sigma^2\sim \operatorname{InverseGamma}(\alpha_0,\beta_0),\qquad \mu\mid\sigma^2\sim N\left(m_0,\frac{\sigma^2}{\kappa_0}\right), $$ with $\alpha_0,\beta_0,\kappa_0>0$. The inverse-gamma convention here is that its density at $v>0$ is proportional to $v^{-(\alpha_0+1)}\exp(-\beta_0/v)$. The prior parameters $m_0,\kappa_0,\alpha_0,\beta_0$ must be specified on the outcome’s scale. This proper normal-inverse-gamma prior is the conjugate assumption required for the closed-form update.

The Update From Study Summaries

Write $\bar{y}_h,\bar{y}_c$ for the historical and current sample means, and define their within-study sums of squares as $SS_h=\sum_j(y_{hj}-\bar{y}_h)^2$ and $SS_c=\sum_j(y_{cj}-\bar{y}_c)^2$. The discounted historical sample size is $a n_h$. The combined weight and weighted mean are

W=n_c+a n_h,\qquad \bar{y}_w=\frac{n_c\bar{y}_c+a n_h\bar{y}_h}{W}.

For $W>0$, the weighted residual sum of squares is

SS_w=SS_c+aSS_h+\frac{n_c\,a n_h}{W}(\bar{y}_c-\bar{y}_h)^2.

The last term accounts for the difference between the two study means. It is necessary: simply adding the within-study sums of squares would omit the variation associated with their separation. For example, if the means differ substantially, that separation contributes to uncertainty about a shared mean and variance.

The posterior remains normal-inverse-gamma, with parameters

\begin{aligned} \kappa_n&=\kappa_0+W,\\ m_n&=\frac{\kappa_0m_0+W\bar{y}_w}{\kappa_n},\\ \alpha_n&=\alpha_0+\frac{W}{2},\\ \beta_n&=\beta_0+\frac{1}{2}\left[ SS_w+\frac{\kappa_0W}{\kappa_n}(\bar{y}_w-m_0)^2 \right]. \end{aligned}

Thus $\sigma^2\mid\text{data}\sim\operatorname{InverseGamma}(\alpha_n,\beta_n)$, and $\mu\mid\sigma^2,\text{data}\sim N(m_n,\sigma^2/\kappa_n)$. These formulas use a fixed $a$; they do not update or estimate the discount.

Why the Formula Holds

The likelihood can be written in terms of the combined weighted mean and residual sum of squares:

L_cL_h^a\propto (\sigma^2)^{-W/2} \exp\left\{-\frac{SS_w+W(\mu-\bar{y}_w)^2}{2\sigma^2}\right\}.

This follows by expanding the two within-study sums of squares and completing the square in $\mu$. In particular, the weighted identity

n_c(\mu-\bar{y}_c)^2+a n_h(\mu-\bar{y}_h)^2 =W(\mu-\bar{y}_w)^2+ \frac{n_c\,a n_h}{W}(\bar{y}_c-\bar{y}_h)^2

produces the between-study term in $SS_w$. Multiplying the likelihood by the prior adds the prior quadratic term $\kappa_0(\mu-m_0)^2$ in the exponent. A second completion of the square gives

\kappa_0(\mu-m_0)^2+W(\mu-\bar{y}_w)^2 =\kappa_n(\mu-m_n)^2+ \frac{\kappa_0W}{\kappa_n}(\bar{y}_w-m_0)^2.

The coefficient of $(\mu-m_n)^2$ gives $\kappa_n$ and $m_n$. The remaining term adds to the inverse-gamma scale, giving $\beta_n$. The likelihood contributes a factor $(\sigma^2)^{-W/2}$, so the inverse-gamma shape increases by $W/2$, giving $\alpha_n$. This derivation also shows why the discount acts on information about the variance, not just on the mean.

Worked Example: Updating With a Half-Weighted Historical Study

Suppose the outcome is reduction in systolic blood pressure, measured in mmHg. A hypothetical historical control study has $n_h=4$, $\bar{y}_h=10$ mmHg, and $SS_h=20$ mmHg$^2$. A hypothetical current control group has $n_c=3$, $\bar{y}_c=12$ mmHg, and $SS_c=12$ mmHg$^2$. Use $a=0.5$ and prior parameters $m_0=5$ mmHg, $\kappa_0=2$, $\alpha_0=3$, and $\beta_0=8$ mmHg$^2$. Assume the two groups are independent conditional on the common $\mu,\sigma^2$, use the same endpoint scale, and are adequately represented by a common normal distribution.

The discounted historical size is $a n_h=0.5(4)=2$, so $W=3+2=5$. The weighted mean is

\bar{y}_w=\frac{3(12)+2(10)}{5}=\frac{56}{5}=11.2\text{ mmHg}.

The weighted residual sum of squares, including the between-study contribution, is

SS_w=12+0.5(20)+\frac{3(2)}{5}(12-10)^2 =12+10+4.8=26.8\text{ mmHg}^2.

The posterior parameters are

\begin{aligned} \kappa_n&=2+5=7,\\ m_n&=\frac{2(5)+5(11.2)}{7}=\frac{66}{7}=9.429\text{ mmHg},\\ \alpha_n&=3+\frac{5}{2}=5.5,\\ \beta_n&=8+\frac{1}{2}\left[26.8+\frac{2(5)}{7}(11.2-5)^2\right] =48.86\text{ mmHg}^2. \end{aligned}

For example, the prior-to-data contribution inside the scale update is $(10/7)(6.2)^2=54.91$ mmHg$^2$, so $\beta_n=8+0.5(26.8+54.91)=48.86$ mmHg$^2$, up to rounding. The posterior mean of $\mu$, conditional on $\sigma^2$, is $9.429$ mmHg. The posterior mean of $\sigma^2$ is $\beta_n/(\alpha_n-1)=48.86/4.5=10.86$ mmHg$^2$. This last expectation exists because $\alpha_n>1$.

What the Discount Changes

The endpoints $a=0$ and $a=1$ help clarify the model. At $a=0$, the historical likelihood contributes nothing and $W=n_c$; the update uses current outcomes and the prior. At $a=1$, historical and current observations contribute equally under the shared-parameter model, which is complete pooling. For $0<a<1$, historical observations receive fractional likelihood weight.

The discount changes more than the weighted mean. Both $a n_h$ and $aSS_h$ enter the update, and the separation between study means also depends on $a$. Consequently, a smaller discount reduces historical influence on the posterior for $\sigma^2$ as well as for $\mu$. It is not generally correct to compute a discounted mean but then use the variance estimate from the fully pooled data.

Worked Example: Comparing No Borrowing, Partial Borrowing, and Pooling

Use the data and prior from the preceding example. Compare $a=0$, $a=0.5$, and $a=1$. The groups still have $n_c=3$, $\bar{y}_c=12$, $SS_c=12$ and $n_h=4$, $\bar{y}_h=10$, $SS_h=20$; all quantities remain in the same units and the shared normal model is assumed.

With $a=0$, $W=3$, $\bar{y}_w=12$, and $SS_w=12$. Therefore

m_n=\frac{2(5)+3(12)}{2+3}=9.2\text{ mmHg},\qquad \alpha_n=3+\frac{3}{2}=4.5.

With $a=0.5$, the preceding calculation gives $m_n=9.429$ mmHg and $\alpha_n=5.5$.

With $a=1$, $W=3+4=7$, $\bar{y}_w=(36+40)/7=10.857$ mmHg, and

m_n=\frac{2(5)+7(10.857)}{2+7}=9.556\text{ mmHg},\qquad \alpha_n=3+\frac{7}{2}=6.5.

The historical mean is below the current mean but above the prior mean. As the discount increases, the posterior mean moves from $9.2$ toward the historical data’s influence, while $\alpha_n$ also increases. The shape parameter alone does not summarize all uncertainty about $\mu$: its posterior depends on $\beta_n$ and $\kappa_n$ as well. These computations show why the discount should be interpreted as a likelihood weight, not as a guarantee that the posterior mean moves in a particular direction in every dataset.

Diagnostics, Alternatives, and Limitations

Before applying the update, check whether its assumptions are plausible. Historical and current outcomes must use a harmonized definition, measurement scale, and analysis population; the common-$\mu$, common-$\sigma^2$ model must be defensible after that harmonization. Inspect study-level means and spreads, outliers, and distributional shape. Large mean differences or markedly different spreads are warnings that full pooling may be inappropriate; a fractional discount limits influence but does not explain why studies differ. Normality can also be consequential with small samples or strongly skewed outcomes.

The sample summaries are sufficient for this calculation under the stated normal model, but summaries must include within-study sums of squares as well as sample sizes and means. Means and sample sizes alone cannot determine the variance update. If $SS_h$ or $SS_c$ is missing, the displayed update cannot be completed from those summaries without additional assumptions or data.

The main alternatives answer different modeling questions. Setting $a=0$ avoids historical borrowing; setting $a=1$ fully pools the studies under a common distribution. A known-variance normal update is simpler but does not represent uncertainty in $\sigma^2$, so it is not a substitute when the variance is unknown. More flexible models can represent between-study differences, but they require additional modeling choices and are not the conjugate common-distribution update derived here. A fixed discount is a chosen input; this analysis does not account for uncertainty in that choice.

A common mistake is to say, “The historical sample size is four, so it adds four observations to the current data, and only the mean is discounted.” That reasoning incorrectly treats $a=0.5$ as full borrowing for the variance and omits the study-separation term. The correction is to use $a n_h$ in the combined weight and $aSS_h$ in the residual sum of squares, while retaining the between-study term $\frac{n_c a n_h}{W}(\bar{y}_c-\bar{y}_h)^2$. The power applies to the historical likelihood as a whole.

Key takeaway: Under a common normal mean and variance and a proper normal-inverse-gamma prior, a fixed power discount gives a closed-form update. It discounts the historical study’s contributions to both mean and variance information, including the contribution due to separation of the study means.

Check Your Understanding

Try each problem, then check the solutions below. Assume the conjugate model and notation defined above unless a question says otherwise.

If $a=0$, what happens to the historical study’s contribution to $W$ and $SS_w$?
Suppose $n_h=5$, $n_c=4$, and $a=0.6$. Calculate $W$ and the increment to $\alpha_n$.
For $n_h=5$, $n_c=4$, $\bar{y}_h=8$, $\bar{y}_c=9$, $SS_h=40$, $SS_c=18$, and $a=0.6$, calculate $\bar{y}_w$ and $SS_w$.
Why is adding $a n_h$ to the mean update but leaving historical data out of the variance update inconsistent with the stated power-prior model?

Solutions. (1) The historical contribution is zero: $a n_h=0$ and $aSS_h=0$, and the between-study term is also zero. (2) $W=4+0.6(5)=7$, so $\alpha_n$ increases by $W/2=3.5$. (3) The historical weight is $3$, so $W=7$, and $\bar{y}_w=[4(9)+3(8)]/7=60/7=8.571$. Then $SS_w=18+0.6(40)+[4(3)/7](9-8)^2=42+12/7=43.714$. (4) Raising the full historical likelihood to $a$ changes its power on the variance-dependent likelihood terms as well as on the mean terms. Leaving historical information out of the variance update would no longer be the update from $L_cL_h^a$.

Power Priors With Unknown Outcome Variance

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