# Bayesian Sample Size Determination, Part 2

(Part 1 is here.)

In part 1 I discussed choosing a sample size for the problem of inferring a proportion. Now, instead of a parameter estimation problem, let’s consider a hypothesis test.

This time we have two proportions $$\theta_0$$ and $$\theta_1$$, and two hypotheses that we want to compare: $$H_=$$ is the hypothesis that $$\theta_0 = \theta_1$$,  and $$H_{\ne}$$ is the hypothesis that $$\theta_0 \ne \theta_1$$. For example, we might be comparing the effects of a certain medical treatment and a placebo, with $$\theta_0$$ being the proportion of patients who would recover if given the placebo, and $$\theta_1$$ being the proportion of patients who would recover if given the medical treatment.

To fully specify the two hypotheses we must define their priors for $$\theta_0$$ and $$\theta_1$$. For appropriate values $$\alpha > 0$$ and $$\beta > 0$$ let us define

• $$H_=$$: $$\theta_0 \sim \mathrm{beta}(\alpha,\beta)$$ and $$\theta_1 = \theta_0$$.
• $$H_{\ne}$$: $$\theta_0 \sim \mathrm{beta}(\alpha, \beta)$$  and $$\theta_1 \sim \mathrm{beta}(\alpha, \beta)$$ independently.

($$\mathrm{beta}(\alpha,\beta)$$ is a beta distribution over the interval $$(0,1)$$.) Our data are

• $$n_0$$ and $$k_0$$: the number of trials in population 0 and the number of positive outcomes among these trials.
• $$n_1$$ and $$k_1$$: the number of trials in population 1 and the number of positive outcomes among these trials.

We also define $$n = n_0 + n_1$$ and $$k = k_0 + k_1$$.

To get the likelihood of the data conditional only on the hypothesis $$H_=$$ or $$H_{\ne}$$ itself, without specifying a particular value for the parameters $$\theta_0$$ or $$\theta_1$$, we must “sum” (or rather, integrate) over all possible values of the parameters. This is called the marginal likelihood. In particular, the marginal likelihood of $$H_=$$ is

\displaystyle \begin{aligned} & \Pr(k_0, k_1 \mid n_0, n_1, H_=) \\ = & \int_0^1 \mathrm{beta}(\theta \mid \alpha, \beta)\, \Pr\left(k_0 \mid n_0, \theta\right) \, \Pr\left(k_1 \mid n_0,\theta\right) \,\mathrm{d}\theta \\ = & c_0 c_1 \frac{\mathrm{B}(\alpha + k, \beta + n – k)}{\mathrm{B}(\alpha, \beta)} \end{aligned}

where $$\mathrm{B}$$ is the beta function (not the beta distribution) and

\displaystyle \begin{aligned}c_0 & = \binom{n_0}{k_0} \\ c_1 & = \binom{n_1}{k_1} \\ \Pr\left(k_i\mid n_i,\theta\right) & = c_i \theta^{k_i} (1-\theta)^{n_i-k_i}. \end{aligned}

The marginal likelihood of $$H_{\ne}$$ is

\displaystyle \begin{aligned} & \Pr\left(k_0, k_1 \mid n_0, n_1, H_{\ne}\right) \\ = & \int_0^1\mathrm{beta}\left(\theta_0 \mid \alpha, \beta \right)\, \Pr\left(k_0 \mid n_0, \theta_0 \right) \,\mathrm{d}\theta_0\cdot \\ & \int_0^1 \mathrm{beta}\left(\theta_1 \mid \alpha, \beta \right) \, \Pr\left(k_1 \mid n_1, \theta_1 \right)\,\mathrm{d}\theta_1 \\ = & c_0 \frac{\mathrm{B}\left(\alpha + k_0, \beta + n_0 – k_0\right)}{\mathrm{B}(\alpha, \beta)} \cdot c_1 \frac{\mathrm{B}\left(\alpha + k_1, \beta + n_1 – k_1\right)}{\mathrm{B}(\alpha, \beta)}. \end{aligned}

Writing $$\Pr\left(H_=\right)$$ and $$\Pr\left(H_{\ne}\right)$$ for the prior probabilities we assign to $$H_=$$ and $$H_{\ne}$$ respectively, Bayes’ rule tells us that the posterior odds for $$H_=$$ versus $$H_{\ne}$$ is

\displaystyle \begin{aligned}\frac{\Pr\left(H_= \mid n_0, k_0, n_1, k_1\right)}{ \Pr\left(H_{\ne} \mid n_0, k_0, n_1, k_1\right)} & = \frac{\Pr\left(H_=\right)}{\Pr\left(H_{\ne}\right)} \cdot \frac{\mathrm{B}(\alpha, \beta) \mathrm{B}(\alpha + k, \beta + n – k)}{ \mathrm{B}\left(\alpha + k_0, \beta + n_0 – k_0\right) \mathrm{B}\left(\alpha + k_1, \beta + n_1 – k_1\right)}. \end{aligned}

Defining $$\mathrm{prior.odds} = \Pr\left(H_{=}\right) / \Pr\left(H_{\ne}\right)$$, the following R code computes the posterior odds:

post.odds <- function(prior.odds, α, β, n0, k0, n1, k1) {
n <- n0 + n1
k <- k0 + k1
exp(
log(prior.odds) +
lbeta(α, β) + lbeta(α + k, β + n - k) -
lbeta(α + k0, β + n0 - k0) - lbeta(α + k1, β + n1 - k1))
}

We want the posterior odds to be either very high or very low, corresponding to high confidence that $$H_=$$ is true or high confidence that $$H_{\ne}$$ is true, so let’s choose as our quality criterion the expected value of the smaller of the posterior odds and its inverse:

$$E\left[\min\left(\mathrm{post.odds}\left(k_0,k_1\right), 1/\mathrm{post.odds}\left(k_0,k_1\right)\right) \mid n_0, n_1, \alpha, \beta \right]$$

where, for brevity, we have omitted the fixed arguments to $$\mathrm{post.odds}$$. As before, this expectation is over the prior predictive distribution for $$k_0$$ and $$k_1$$.

In this case the prior predictive distribution does not have a simple closed form, and so we use a Monte Carlo procedure to approximate the expectation:

• For $$i$$ ranging from $$1$$ to $$N$$,
• Draw $$h=0$$ with probability $$\Pr\left(H_=\right)$$ or $$h=1$$ with probability $$\Pr\left(H_{\ne}\right)$$.
• Draw $$\theta_0$$ from the $$\mathrm{beta}(\alpha, \beta)$$ distribution.
• If $$h=0$$ then set $$\theta_1 = \theta_0$$; otherwise draw $$\theta_1$$ from the $$\mathrm{beta}(\alpha, \beta)$$ distribution.
• Draw $$k_0$$ from $$\mathrm{binomial}\left(n_0, \theta_0\right)$$ and draw $$k_1$$ from $$\mathrm{binomial}\left(n_1, \theta_1\right)$$.
• Define $$z = \mathrm{post.odds}\left(k_0, k_1\right)$$.
• Assign $$x_i = \min(z, 1/z)$$.
• Return the average of the $$x_i$$ values.

The following R code carries out this computation:

loss <- function(prior.odds, α, β, n0, k0, n1, k1) {
z <- post.odds(prior.odds, α, β, n0, k0, n1, k1);
pmin(z, 1/z)
}
expected.loss <- function(prior.odds, α, β, n0, n1) {
# prior.odds = P.eq / P.ne = (1 - P.ne) / P.ne = 1/P.ne - 1
# P.ne = 1 / (prior.odds + 1)
N <- 1000000
h <- rbinom(N, 1, 1 / (prior.odds + 1))
θ0 <- rbeta(N, α, β)
θ1 <- rbeta(N, α, β)
θ1[h == 0] <- θ0[h == 0]
k0 <- rbinom(N, n0, θ0)
k1 <- rbinom(N, n1, θ1)
x <- loss(prior.odds, α, β, n1, k1, n0, k0)
# Return the Monte Carlo estimate +/- error estimate
list(estimate=mean(x), delta=sd(x)*1.96/sqrt(N))
}

Suppose that $$H_=$$ and $$H_{\ne}$$ have equal prior probabilities, and $$\alpha=\beta=1$$, that is, we have a uniform prior over $$\theta_0$$ (and $$\theta_1$$, if applicable). The following table shows the expected loss—the expected value of the minimum of the posterior odds against $$H_{\ne}$$ and the posterior odds against $$H_=$$—as $$n_0 = n_1$$ ranges from 200 to 1600:

$$n_0$$$$n_1$$Expected loss
2002000.130
4004000.095
6006000.078
8008000.069
100010000.062
120012000.057
140014000.053
160016000.050

We find that it takes 1600 trials from each population to achieve an expected posterior odds in favor of the (posterior) most probable results of 20 to 1.