# Why does the often quoted formula for the Flemming O'Brien alpha spending function produce 'probabilities' close to 2?

Demets and Lan, STATISTICS IN MEDICINE, VOL. 13. 1341-1352 (1994), introduce the alpha-spending approach to interim analyses of clinical trials. Concretely, they introduce the idea of choosing a monotonically increasing function $\alpha(t)$ satisfying $\alpha(0) = 0$ and $\alpha(1) = \alpha$ and argue that when interim analysis are performed at time points $t_1, ..., t_K$ the rules for stopping for efficacy should be such that at each time point the overall probability under $H_0$ that the $H_0$ is rejected at that time point, or already has been rejected at an earlier interim analysis equals $\alpha(t_k)$. In particular the probability under $H_0$ that $H_0$ is rejected at the end of the trial is $\alpha$, as it always is.

This gives a uniform language to describe earlier approaches to interim analyse and gets rid of the need, present in many of these earlier approaches, to specify the number of interim analysis on forehand.

So far so good.

Next they give examples of how standard approaches, such as those of Flemming - O'Brian and Pockock can be reformulated in their language. The formula for the Flemming O'Brian approach that they give and which is reproduced in several more recent books and articles reads:

$\alpha(t^*) = 2 - 2 \Phi(z_{\alpha/2}/\sqrt{t^*})$

where $\Phi$ is the standard normal cummulative distribution function. ($z_{\alpha/2}$ is undefined in the original article but I tacitly assumed it to be the value in the domain of $\Phi$ for which $\Phi(z_{\alpha/2}) = \alpha/2$.)

[EDIT: after reading Whubers comment below I realize that the solution might be that $z_{\alpha/2}$ is actually the (positive) value for which $\Phi(z_{\alpha/2}) = 1 - \alpha/2$. Nevertheless I leave the rest of the original post here:]

Here $t^*$ is the fraction of the total information obtained at time of the interim analysis, but we really only need to know that $t^*$ is some number that equals 1 at the end of the trial in order to see that something is wrong. Plugging in $t^* = 1$ we find that $\alpha(t^*) = 2 - \alpha$, which is a far cry from the claimed value of $\alpha$. Also for smaller values of $t^*$ (corresponding to earlier interim analyses) we see that the value of $\alpha(t^*)$ is even closer to 2 which makes it even less credible that this number does indeed represent a probability.

Am I missing something? And what is the correct formula?

• One way to make sense of the formula is to understand the (undefined) function $\Phi$ to be the complementary cumulative distribution function of a standard Normal variable and $z_{\alpha/2}$ (also undefined here) to be the value for which $\Phi(z_{\alpha/2})=\alpha/2$. Then, since $\alpha \lt 1$, $\alpha/2 \lt 1/2$, whence $z_{\alpha/2}\gt 0$, entailing $z_{\alpha/2}/\sqrt{t^{*}}\gt 0$ and $\Phi(z_{\alpha/2}/\sqrt{t^{*}})\gt 1/2$ and finally $\alpha(t^{*})\lt 1$, as it should be. Could you therefore revisit your reference and tell us precisely what $\Phi$ and $z_{\alpha/2}$ mean? – whuber Nov 25 '15 at 20:01
• Thanks! See the edits. Phi is just what it always is, but symmetric to your comment we might probably solve the mystery by reversing the sign of $z_{\alpha/2}$. What do you think? – Vincent Nov 26 '15 at 8:29
• I think that would be a strange inconsistency, but as long as $z_{\alpha/2}$ and $\Phi$ are clearly and correctly defined, there should be nothing wrong with it. – whuber Nov 27 '15 at 15:05