Suppose we have two regimes (\(A\) and \(B\)) that people will recieve in a crossover fashion and suppose that \(A\) and \(B\) are equivalent in their efficacy so that the response to either regime, on average, is the same. Denote a subject’s response to regime \(A\) as \(X_A\) and the response to regime \(B\) as \(X_B\).

Now let’s assume that the response \(X_A\) or \(X_B\) exhibits variation over time that looks like a normal distribution with standard deviation \(\sigma\). Because \(A\) and \(B\) have identical efficacy, the probablity distributions of \(X_A\) and \(X_B\) are the same.

We say that “\(A\) is effective” if \(X_A > X_B + \delta\) for some positive margin \(\delta\) and we say that “\(B\) is effective” if \(X_B > X_A + \delta\). We say that “\(A\) and \(B\) are equivalent” if \(-\delta < X_A - X_B < \delta\).

QUESTION: If regimes \(A\) and \(B\) are in fact equivalent, how often should we expect to observe that \(A\) is effective? Or \(B\) is effective?

The statement “\(A\) is effective” is mathematically equivalent to \(X_A-X_B > \delta\). What we want to know is what is \(Pr(X_A-X_B > \delta)\)? Note that if \(\delta = 0\), then this probability is always \(1/2\) regardless of the value of \(\sigma\).

Assuming that \(X_A\) and \(X_B\) both have a Normal distribution with standard deviation \(\sigma\), we can say that \(X_A-X_B\) has a Normal distribution with mean 0 and standard deviation \(\sigma\sqrt{2}\). Then we can say that \[\begin{eqnarray*} Pr(X_A-X_B > \delta) & = & Pr\left(\frac{X_A-X_B}{\sigma\sqrt{2}} > \frac{\delta}{\sigma\sqrt{2}}\right)\\ & = & Pr\left(Z > \frac{\delta}{\sigma\sqrt{2}}\right)\\ & = & 1 - Pr\left(Z \leq \frac{\delta}{\sigma\sqrt{2}}\right)\\ & = & 1 - \Phi\left(\frac{\delta}{\sigma\sqrt{2}}\right) \end{eqnarray*}\] where \(\Phi\) is the cumulative distribution function for the standard Normal distribution.

We can plot this probability as a function of \(\delta\) and and for different values of \(\sigma\).

So in a large population with \(\sigma = 2\), then if \(\delta = 1\) we would expect about 36% of the subjects to show that \(A\) is effective. Because of symmetry, we would expect 36% of subjects to show that \(B\) is effective, and 28% to exhibit “no difference”.