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In a two-stage normal trial, conditioning on a first-stage sample mean exceeding $c$ raises that mean. With stage standard error $s$, mean parameter $\delta$, and combined mean $S$, the estimator bias after continuation is $E(S\mid X\ge c)-\delta =(s/2)\lambda ((c-\delta )/s)$, where $\lambda$ is the upper-tail Inverse Mills ratio. It is positive and tends to zero as continuation becomes virtually certain.

Let eventual outcome $I=j$ have probability ${\pi }_j$, with $\log T\mid I=j\sim N({\mu }_j,{\sigma }_j^2)$. An observed outcome at time $t$ contributes ${\pi }_j{f}_j(t)$ to the likelihood, while right censoring at $c$ contributes ${\sum }_j{\pi }_j{S}_j(c)$. The expectation-maximization algorithm imputes both the censored outcome and its latent log event time. Its E-step uses the displayed conditional weights and the first two moments of a truncated normal distribution above $\log c$. With $z=(\log c-{\mu }_j)/{\sigma }_j$ and Inverse Mills ratio $\psi (z)$, these moments are $m={\mu }_j+{\sigma }_j\psi (z)$ and $s=m^2+{\sigma }_j^2[1-\psi (z)(\psi (z)-z)]$. Observed outcomes use deterministic class weights and moments $\log t,(\log t{)}^2$. The M-step updates each class probability to its average weight, its mean to the weighted first moment, and its variance to the weighted second moment minus the squared new mean. Outcome-specific cumulative probabilities are cumulative incidence functions, not the conditional distributions $F(t\mid I=j)$ themselves.

This is an EM algorithm for a censored lognormal competing-risks mixture. Write $Y_i=\log T_i$ and suppose $Y_i\mid I_i=j\sim N({\mu }_j,{\sigma }_j^2)$, with mixing weights ${\pi }_1=\theta$, ${\pi }_2=1-\theta$. The missing data are $I_i$ and $Y_i$ for censored patients; neither is missing for recorded deaths or recoveries. The complete-data log-likelihood, apart from terms independent of the parameters, isThe lognormal Jacobian $-\log T_i$ is also parameter-independent and therefore does not affect the maximization.

At iteration $k$, set $z_{ij}=(\log c_i-{\mu }_j^{(k)})/{\sigma }_j^{(k)}$ for each censored patient and let $\overline{\Phi }(z)=1-\Phi (z)$. Bayes' rule gives the E-step class weightsGiven that class, $Y_i$ has a truncated normal distribution above $\log c_i$. With the Inverse Mills ratio $\psi (z)=\phi (z)/\overline{\Phi }(z)$, compute both moments needed for the E-step:Thus $\mathbb{E}[1_{\{I_i=j\}}\mid \text{data}]=w_{ij}$, $\mathbb{E}[1_{\{I_i=j\}}{Y}_i\mid \text{data}]=w_{ij}{m}_{ij}$, and the corresponding second moment is $w_{ij}{s}_{ij}$. For an observed outcome $j_i$ at time $t_i$, use $w_{ij}=1_{\{j=j_i\}}$, $m_{ij}=\log t_i$, and $s_{ij}=(\log t_i{)}^2$. These specify the full conditional expectation of ${\ell }_c$.

Put $W_j={\sum }_i{w}_{ij}$ and let $n$ be the total number of patients. Maximizing that expected complete-data log-likelihood gives the M-stepAll moments on the right are evaluated at the old parameters. Repeat the E-step and M-step until the observed log-likelihood stabilizes, using positive initial variances and mixing weights. The expectation-maximization algorithm increases that likelihood at each exact iteration, but need not find its global maximum; multiple starts help distinguish local optima. The resulting estimates give the case fatality probability $\hat{\theta }$ andThese are conditional event-time distributions, distinct from the cumulative incidence functions $\hat{\theta }\hat{F}(t\mid I=1)$ and $(1-\hat{\theta })\hat{F}(t\mid I=2)$.

For $U\sim N(\mu ,s^2)$ and $a=(T-\mu )/s$, substitution $U=\mu +sZ$ givessince ${\varphi }^{\prime }(z)=-z\varphi (z)$. Divide by $1-\Phi (a)$ to obtain the upper-truncated normal mean.

For $U={\hat{\delta }}_1$, set $\mu =\delta$, $s=1/\sqrt{I_1}$ and $T=f_1/\sqrt{I_1}$. Then $a=f_1-\delta \sqrt{I_1}$. Symmetry of the normal density and distribution yieldsThe ratio $r(x)=\varphi (x)/\Phi (x)$ is an Inverse Mills ratio. It is positive, and the pooled estimator's conditional bias is thereforeThis identity supplies both correction procedures below.

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

For a fixed observed colour ${\tilde{c}}_s$, Bayes' theorem givesCompleting the square shows that this is a truncated normal distribution with variance ${\sigma }_c^2$, lower endpoint zero, and untruncated locationHenceConditioning first on $E_s$ and using the law of total expectation yields the dusty colour-magnitude relationAs ${\tilde{c}}_s\to -\infty$, the Inverse Mills ratio implies ${\overline{E}}_s\to 0$ with vanishing derivative, so the asymptotic slope is $\beta$. As ${\tilde{c}}_s\to +\infty$, ${\overline{E}}_s\sim {\tilde{c}}_s-c_0-{\sigma }_c^2/\tau$, so the asymptotic slope is $R$.