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A bivariate probit model for a binary treatment and outcome uses latent equationswhere $(U,V)$ has a bivariate normal distribution. Its correlation parameter represents dependence between the two latent disturbances; when nonzero, treatment is endogenous in the outcome equation.

A bivariate standard normal distribution is a bivariate normal distribution whose two marginal means are zero and marginal variances are one; its remaining parameter is their correlation.

A subject-specific random intercept and random slope may have a joint bivariate normal distribution with arbitrary positive-definite covariance matrix $D$. With independent normal measurement errors of variance ${\sigma }^2$, the marginal covariance between a person's readings at $s,t$ is $D_{00}+(s+t)D_{01}+st{D}_{11}$, plus ${\sigma }^2$ on the diagonal. Different subjects remain independent. Allowing $D_{01}$ distinguishes this model from an independent random-intercept and random-slope model.

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 ${\varphi }_{\rho }$ denote the centered unit-variance bivariate normal distribution density,The complement of rejecting either null hypothesis is $W_1\le c,W_2\le c$. Therefore the probability of at least one rejection isWhen both null hypotheses are true, this is the familywise error rate. If one null hypothesis is false, a rejection of that hypothesis is not a Type I error, so the displayed probability is then an any-rejection probability rather than the familywise error rate.

Couple the Wald test statistics as $W_i=Z_i+{\delta }_i/s_i$, where the joint centered bivariate normal distribution of $(Z_1,Z_2)$ is fixed. Increasing either ${\delta }_i$ enlarges, pointwise in this coupling, the event $\{W_1>c\text{ or }{W}_2>c\}$. Its probability is consequently nondecreasing in each treatment effect.

If both null hypotheses hold, then ${\delta }_1,{\delta }_2\le 0$, so this probability is maximal at the joint boundary $(0,0)$. To check the familywise error rate across all configurations, suppose only hypothesis $i$ is true. A Type I error then occurs only if $W_i>c$, with probability at most $1-\Phi (c)$, where $\Phi$ is the standard normal distribution function. This is no larger than the any-rejection probability at $(0,0)$, since that joint event contains $\{Z_i>c\}$. If neither null hypothesis is true, the familywise error rate is zero. Hence the least-favourable null configuration for Type I error iswhere ${\Phi }_2$ is the centered bivariate normal distribution function with correlation coefficient $\rho$.

For every $u\in {\mathbb{R}}^2$,The right-hand side is a normal random variable by the definition of a bivariate normal distribution. Therefore every linear combination of the coordinates of $AX$ is normal, so $\boxed{AX\text{ is bivariate normal}}$.

The correlation $\rho$ measures dependence between two normalized latent disturbances; it is not a scale-free measure of the strength of one physical confounder. Different latent-variable constructions can induce the same $\rho$ while producing different treatment-outcome confounding, and the same omitted cause can produce different $\rho$ after changing thresholds or disturbance scales. Moreover, re-estimating $(\alpha ,\beta ,\gamma )$ at each value of $\rho$ changes the entire latent model, so the fitted models do not represent one fixed data-generating mechanism with only its confounder strength varied. At $\rho =\pm 1$ the bivariate normal distribution is singular as well. Varying $\rho$ is a model-based sensitivity analysis, but interpreting the interval $[-1,1]$ as an ordered range of strengths of a single unmeasured confounder is therefore logically unjustified.

The bivariate normal distribution has density

False. The Brownian motion produced by the Dambis-Dubins-Schwarz theorem need not be independent of its clock. Let $W$ be a standard Brownian motion and setIf the Brownian motion $B$ in $M_t=B_{[M{]}_t}$ were independent of the whole quadratic variation process, then conditioning on $[M]$ would give $\mathbb{E}[M_t[M{]}_t]=\mathbb{E}[B_{[M{]}_t}[M{]}_t]=0$. Instead, the fourth-moment formula for a bivariate normal distribution gives $\mathrm{Cov}(W_t^2,W_s^2)=2s^2$ for $s\le t$, and henceThus $B$ and $[M]$ are dependent.

With the notation from part (i), the total-variation process of $[M,N]$ isFor the centered bivariate normal distribution $(U,V)$ used there, $c=\mathbb{E}[UV]$. The integral triangle inequality givesIntegrating this pointwise inequality against $dC$ proves $V_t\le {\tilde{V}}_t$ for every $t\ge 0$.

Apply the realized absolute covariation theorem to each dyadic partition of an interval. More explicitly, use the continuous increasing clock $C=[M]+[N]$ and the Radon-Nikodym theorem to writeFor each time, let $(U,V)$ have the centered bivariate normal distribution with covariance matrix $\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)$, and defineThis process is continuous and increasing. To prove convergence, localize $M,N$, represent the pair as stochastic integrals against a two-dimensional Brownian motion, and approximate the integrands in $L^2(dC)$ by bounded step previsible processes. For step integrands, the result is the weak law of large numbers applied on each block to independent Gaussian random variables. The Burkholder-Davis-Gundy inequality and the Cauchy-Schwarz inequality make the error uniform on each compact interval in probability. Consequentlyin the sense of uniform convergence on compacts in probability.

For continuous local martingales $M,N$ and a sequence in which each term is a partition of an interval whose mesh tends to zero, the sumsconverge in the sense of uniform convergence on compacts in probability to a continuous increasing process. To identify the limit, put $C=[M]+[N]$ and choose Radon-Nikodym derivativesIf $(U,V)$ is a centered bivariate normal distribution with covariance matrix $\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)$, thenLocalizing, representing the pair as stochastic integrals against a two-dimensional Brownian motion, and approximating the integrands by bounded predictable step processes proves the convergence. The step-process case follows from the weak law of large numbers for independent Gaussian increments; the Burkholder-Davis-Gundy inequality controls the approximation error. Since $|\mathbb{E}[UV]|\le \mathbb{E}|UV|\le \sqrt{\mathbb{E}{U}^2\mathbb{E}{V}^2}$,