The Canonical Joint Distribution · Tutorial 41 of 699

Conditional Distribution at a Later Look

Learn how the canonical correlation between nested statistics determines the later statistic’s conditional mean and variance, and how to use that distribution without confusing it with the marginal one.

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What You'll Learn

State the canonical joint-normal assumptions for two nested standardized statistics.
Derive the conditional mean and variance using a normal residual independent of the earlier statistic.
Calculate conditional distributions from information times and observed statistics.
Find later-look tail probabilities conditional on an earlier observed value.
Diagnose when the canonical conditional formula may not apply.
Distinguish conditional calculations from marginal probabilities and independence assumptions.

Why Condition on an Earlier Look?

At an interim analysis, the statistic at the first look is already observed, while the statistic at a later look is still uncertain. The marginal distribution of the later statistic describes its uncertainty before using the interim result. The conditional distribution describes that uncertainty after conditioning on the observed interim statistic. This distinction is useful when quantifying the chance of a later threshold being crossed, conditional on what has already happened.

The key feature is that cumulative statistics at nested looks share data. Under the canonical joint-normal model, their dependence is summarized by the ratio of their information times. Once that dependence and the marginal means are specified, the conditional distribution follows from a normal regression calculation.

Definition: Let $t_1$ and $t_2$ be information times with $0<t_1<t_2$, and let $Z_1$ and $Z_2$ be the standardized statistics at those looks. In the canonical joint-normal model, assume $$ \begin{pmatrix} Z_1 \\ Z_2 \end{pmatrix} \sim N\left( \begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix}, \begin{pmatrix} 1 & \rho \\ \rho & 1 \end{pmatrix} \right), \qquad \rho=\sqrt{\frac{t_1}{t_2}}. $$ Here $\mu_j=E(Z_j)$. Under $H_0$, both means are zero. The conditional distribution of $Z_2$ given $Z_1=z_1$ is normal with mean $\mu_2+\rho(z_1-\mu_1)$ and variance $1-\rho^2$.

The information times must be positive and strictly increasing, and the statistics must have unit marginal variances with the stated canonical correlation. The correlation formula reflects nested cumulative information: the earlier statistic is built from data also present in the later statistic. The earlier material on independent increments gives this relationship; it does not mean that the two cumulative $Z$ statistics are independent.

Deriving the Conditional Distribution

A direct derivation isolates the part of $Z_2$ not linearly explained by $Z_1$. Define the residual $$ R=Z_2-\mu_2-\rho(Z_1-\mu_1). $$ Because $Z_1$ and $Z_2$ are jointly normal, $R$ and $Z_1$ are jointly normal. Their covariance is $$ \operatorname{Cov}(R,Z_1) =\operatorname{Cov}(Z_2,Z_1)-\rho\operatorname{Var}(Z_1) =\rho-\rho=0. $$ For jointly normal variables, zero covariance implies independence. Thus $R$ is independent of $Z_1$. Its mean is zero, and its variance is $$ \begin{aligned} \operatorname{Var}(R) &=\operatorname{Var}\{Z_2-\rho Z_1\}\\ &=1+\rho^2-2\rho\operatorname{Cov}(Z_1,Z_2)\\ &=1+\rho^2-2\rho^2\\ &=1-\rho^2. \end{aligned} $$ Consequently $R\sim N(0,1-\rho^2)$, independently of $Z_1$. Rearranging the definition of $R$ gives $$ Z_2=\mu_2+\rho(Z_1-\mu_1)+R. $$ After conditioning on $Z_1=z_1$, the first two terms are fixed and the independent residual retains its distribution. Therefore $$ Z_2\mid Z_1=z_1 \sim N\left(\mu_2+\rho(z_1-\mu_1),\,1-\rho^2\right). $$ The second argument of $N$ is the variance. Substituting the canonical correlation yields the equivalent form $$ E(Z_2\mid Z_1=z_1)=\mu_2+\sqrt{\frac{t_1}{t_2}}(z_1-\mu_1), \qquad \operatorname{Var}(Z_2\mid Z_1=z_1)=1-\frac{t_1}{t_2}. $$

This calculation gives a complete conditional law, not just a conditional expectation. The conditional standard deviation is $\sqrt{1-t_1/t_2}$, which is less than the marginal standard deviation of 1. Conditioning provides information about the later statistic because the two looks are correlated.

Worked Example: A Later Upper-Tail Probability Under the Null

Suppose a hypothetical two-look design has information times $t_1=40$ and $t_2=100$. At the first look, the observed statistic is $z_1=1.50$. Under $H_0$, find the conditional distribution of $Z_2$ and the probability that it exceeds 2.00.

The canonical assumptions are that the pair is jointly normal, both marginal means are zero under $H_0$, both marginal variances are 1, and the covariance is the canonical correlation. The information times are positive and increasing, as required. The correlation is $$ \rho=\sqrt{\frac{40}{100}}=\sqrt{0.40}=0.6325. $$

Under the null, $\mu_1=\mu_2=0$. Hence the conditional mean and variance are $$ E(Z_2\mid Z_1=1.50)=0+0.6325(1.50)=0.9487, \qquad \operatorname{Var}(Z_2\mid Z_1=1.50)=1-0.40=0.6000. $$ Thus $$ Z_2\mid Z_1=1.50\sim N(0.9487,0.6000). $$

The conditional standard deviation is $\sqrt{0.6000}=0.7746$. Standardizing the threshold gives $$ P(Z_2>2.00\mid Z_1=1.50) =1-\Phi\left(\frac{2.00-0.9487}{0.7746}\right) =1-\Phi(1.357) \approx 0.0874. $$ So, under the stated model and conditional on this interim result, the chance that the later statistic exceeds 2.00 is about 8.74%. This is not the unconditional probability of exceeding 2.00.

Using Nonzero Marginal Means

The conditional formula does not require both marginal means to be zero. If the model specifies means $\mu_1$ and $\mu_2$, the earlier observation shifts the conditional mean by $\rho(z_1-\mu_1)$. The conditional variance remains $1-\rho^2$, because the covariance matrix is unchanged by shifting the means. This is a general bivariate-normal result; the means must be supplied by the model being used.

Worked Example: Conditioning With Specified Nonzero Means

Consider a hypothetical canonical joint-normal model with $t_1=40$, $t_2=80$, $\mu_1=0.20$, and $\mu_2=0.40$. Suppose the observed first-look value is $z_1=1.20$. Find the conditional mean, variance, and probability that $Z_2>1.50$. Assume the model has unit marginal variances and canonical correlation.

The information times are positive and increasing, and the model explicitly supplies the means and unit variances. The correlation is $$ \rho=\sqrt{\frac{40}{80}}=\sqrt{0.50}=0.7071. $$ The conditional mean and variance are $$ \begin{aligned} E(Z_2\mid Z_1=1.20) &=0.40+0.7071(1.20-0.20)\\ &=1.1071,\\ \operatorname{Var}(Z_2\mid Z_1=1.20) &=1-0.50=0.5000. \end{aligned} $$ Therefore $Z_2\mid Z_1=1.20\sim N(1.1071,0.5000)$, with conditional standard deviation $\sqrt{0.5000}=0.7071$. The requested probability is $$ P(Z_2>1.50\mid Z_1=1.20) =1-\Phi\left(\frac{1.50-1.1071}{0.7071}\right) =1-\Phi(0.5556) \approx 0.289. $$ Under this hypothetical model, the conditional probability is about 28.9%. The calculation uses the specified marginal means; replacing them with zero would describe a different model.

Worked Example: A Negative Interim Statistic

In a hypothetical design, let $t_1=64$, $t_2=100$, and suppose $H_0$ holds. If the first-look statistic is $z_1=-0.40$, find the conditional distribution of $Z_2$ and the probability that it exceeds 1.645.

The information times satisfy $0<64<100$, and under the null the canonical model has zero means, unit marginal variances, and correlation $$ \rho=\sqrt{\frac{64}{100}}=0.8000. $$ The conditional mean is $0+0.8000(-0.40)=-0.3200$, while the conditional variance is $1-0.8000^2=0.3600$. Therefore $$ Z_2\mid Z_1=-0.40\sim N(-0.3200,0.3600), $$ with standard deviation $0.6000$. The conditional tail probability is $$ P(Z_2>1.645\mid Z_1=-0.40) =1-\Phi\left(\frac{1.645-(-0.3200)}{0.6000}\right) =1-\Phi(3.275) \approx 0.00053. $$ The negative interim value pulls the conditional mean below zero. The later statistic still has random variation, but crossing 1.645 is very unlikely conditional on this observed value.

Interpretation, Checks, and Limitations

For a fixed observed $z_1$, the conditional mean is a prediction of the later statistic under the specified model, and the conditional variance quantifies its remaining variability. As $t_1/t_2$ increases, the correlation increases and the conditional variance decreases. If $t_1$ approaches $t_2$, the later statistic is increasingly determined by the earlier one; at equal information times, the formula gives zero conditional variance, consistent with there being no new information between identical looks.

Before applying the formula, check that the statistics are the canonical standardized cumulative statistics, the information times are correctly ordered, and the joint-normal approximation is suitable. In a design with a different covariance structure, the generic bivariate-normal formula still applies using the actual correlation, but substituting $\sqrt{t_1/t_2}$ would be unjustified. If the joint distribution is not adequately normal, this conditional normal law may only be an approximation.

A conditional normal calculation is often simpler than integrating a joint density over a region, because it reduces a two-dimensional probability calculation to a one-dimensional normal tail. Direct joint-density integration remains a valid alternative when the joint distribution is known. Simulation can also approximate conditional behavior, but it requires generating or selecting replicates consistent with the conditioning value; simply counting all simulated later threshold crossings estimates a marginal, not a conditional, probability.

Common mistake: Treating the later statistic as independent of the observed interim statistic because each statistic is standardized to variance 1. Standardization does not remove their shared-data dependence. Under the canonical model, their correlation is $\sqrt{t_1/t_2}$, so the conditional mean generally changes with $z_1$ and the conditional variance is $1-t_1/t_2$, not 1. A second mistake is to report a conditional tail probability as the marginal chance of crossing the threshold; the two probabilities answer different questions.

The conditional distribution is a model-based statement about repeated realizations under the specified joint law, conditional on an observed statistic. It does not assign a probability to $H_0$, and it does not by itself account for a monitoring rule or adjust an error rate. Its role here is narrower: it describes the distribution at a later look after conditioning on the value seen at an earlier look.

Check Your Understanding

Try each question first, then compare your reasoning with the solutions.

Under $H_0$, let $t_1=25$ and $t_2=100$. What are the canonical correlation and the conditional variance of $Z_2$ given $Z_1$?
Under the null, suppose $t_1/t_2=0.36$ and $z_1=2$. What is the conditional mean of $Z_2$?
For a bivariate normal pair with unit variances, means $\mu_1,\mu_2$, and correlation $\rho$, state the conditional mean and variance of $Z_2$ given $Z_1=z_1$.
Why is $P(Z_2>c\mid Z_1=z_1)$ generally not equal to $P(Z_2>c)$ in the canonical model?
If the observed correlation is not the canonical $\sqrt{t_1/t_2}$, what should be checked before using the conditional formula?

Solutions. (1) The correlation is $\sqrt{25/100}=0.50$, and the conditional variance is $1-0.50^2=0.75$. (2) The correlation is $\sqrt{0.36}=0.60$; under $H_0$, the conditional mean is $0+0.60(2-0)=1.20$. (3) The conditional mean is $\mu_2+\rho(z_1-\mu_1)$, and the conditional variance is $1-\rho^2$. (4) The later statistic and the earlier statistic are correlated, so conditioning changes the later distribution’s mean and reduces its variance; the marginal probability averages over all possible earlier values. (5) Check that the statistics are correctly defined and standardized, that the information times correspond to the nested data, and that the model’s covariance structure is appropriate. If the actual correlation differs, use a justified correlation in the bivariate-normal formula rather than assuming the canonical one.

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