Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 32 / 1

For a one-sided Wald test use the signed normal Wald statistic, rather than its square. The maximum-likelihood estimate is the sample mean ${\bar{Y}}_n$, with exact variance ${\sigma }^2/n$, soA sum of independent normal random variables is normal, giving ${\bar{Y}}_n\sim N(\delta ,{\sigma }^2/n)$ and hence $W_n\sim N(\sqrt{n}\delta /\sigma ,1)$. Under the null hypothesis its mean is zero:Thus its null distribution is exactly the standard normal distribution, without an asymptotic approximation. If the squared Wald statistic convention is used, $W_n^2$ has a chi-squared distribution with one degree of freedom; the signed form is needed to distinguish the two directions.

The same affine transformation of the normal distribution givesIts variance stays one while its mean moves positively. This is the exact alternative distribution used to calculate statistical power. The square, if used instead, has a noncentral chi-squared distribution with one degree of freedom and noncentrality $n({\delta }^{\ast }{)}^2/{\sigma }^2$.

Let $z_p={\Phi }^{-1}(p)$ denote a quantile of the standard normal distribution. Reject when $W_n>z_{1-\alpha }$. The statistical power at the specified positive effect isEquating this to $1-\beta$ and using $z_{\beta }=-z_{1-\beta }$ gives $\sqrt{n}{\delta }^{\ast }/\sigma =z_{1-\alpha }+z_{1-\beta }$. Thus, for the usual target $1-\beta >\alpha$,Rounding up ensures at least the target statistical power. This normal-mean sample size calculation assumes a positive integer sample size; if a requested power is at most $\alpha$, every positive sample size already exceeds that target for ${\delta }^{\ast }>0$, and one should not square a negative quantile sum to impose an unnecessary lower bound.

Write $s=\sigma /\sqrt{n}$, $\mu =\delta /s$, and let $Z_1,Z_2$ be independent standard normal random variables obtained by centering and scaling the separate stage means. ThenThe second statistic uses all patients, so the statistics are correlated even though the stages' new observations are independent. Their covariance is $1/\sqrt{2}$ and each variance is one. This gives the exact bivariate normal distributionUnder the null hypothesis both means are zero. At $\delta ={\delta }^{\ast }$ replace $\mu$ by $\sqrt{n}{\delta }^{\ast }/\sigma$; the mean vector is $(\sqrt{n}{\delta }^{\ast }/\sigma ,\sqrt{2n}{\delta }^{\ast }/\sigma {)}^T$. In this group sequential design the full-sample statistic may be viewed as a potential statistic from the underlying sequence of outcomes, even on paths where recruitment stops.

Set $z=z_{1-\alpha }$. Under the null hypothesis rejection requires both $W_1\ge f$ and $W_2>z$. Since $W_2$ has the standard normal distribution,The bivariate normal distribution in the preceding calculation has a nonsingular covariance matrix and strictly positive statistical probability density everywhere. For every finite futility boundary $f$ and $0<\alpha <1$, the open rectangle $\{W_1<f,W_2>z\}$ has positive probability. ThereforeFor an explicit expression, conditional on $W_1=w$ the full-sample statistic is $N(w/\sqrt{2},1/2)$ under the null hypothesis, yieldingThe Type I error is reduced because some otherwise rejecting paths stop for futility. Equality is approached as $f\to -\infty$, but does not hold at a finite futility boundary.

Let $X$ and $Y$ denote the first- and second-stage sample means, let $s=\sigma /\sqrt{n}$, and write $S=(X+Y)/2$. Continuation is the selection event $\mathcal{C}=\{X\ge sf\}$. The second-stage sample mean stays independent of $\mathcal{C}$, so $E(Y\mid \mathcal{C})=\delta$. The first-stage sample mean has a truncated normal distribution. With $a=f-\delta /s$ and the upper-tail Inverse Mills ratio $\lambda (a)=\varphi (a)/(1-\Phi (a))$,The positive conditional selection bias after futility continuation comes from selecting unusually large first-stage outcomes. The unconditional sample mean of a fixed $2n$ observations would be unbiased; that is a different sampling distribution from the one restricted to continued trials.

As $\delta \to \infty$, $a\to -\infty$, $\varphi (a)\to 0$ and $1-\Phi (a)\to 1$, so the estimator bias tends to zero. It decreases with $\delta$: differentiating gives ${\lambda }^{\prime }(a)=\lambda (a)(\lambda (a)-a)>0$, because $\lambda (a)=E(Z\mid Z>a)>a$ for a standard normal random variable. Thus

Use a Rao-Blackwell estimator after interim selection, which is exactly conditionally unbiased. The second-stage sample mean $Y$ alone is unbiased conditional on continuation, but discards the earlier observations. Apply the Rao-Blackwell theorem by averaging $Y$ conditional on the combined sample mean $S$ and the fact of continuation.

Set $c=sf$ and $v=s/\sqrt{2}=\sigma /\sqrt{2n}$. Before truncation, $X\mid S$ is $N(S,v^2)$, a distribution whose mean no longer involves the unknown $\delta$. After imposing $X\ge c$, its mean is $S+v\lambda ((c-S)/v)$. Since $Y=2S-X$, the resulting estimator isIt uses the outcomes from both stages through their combined sample mean. By iterated expectation, $E(\tilde{\delta }\mid \mathcal{C})=E(Y\mid \mathcal{C})=\delta$, so its conditional estimator bias is zero, compared with the strictly positive estimator bias above. Its conditional variance is no larger than that of the second-stage-only estimate. This does not assert a smaller mean squared error than every biased estimator.