Correlated endpoints: simulation, modeling, and extreme correlations

Abstract

Modeling and simulation of correlated random variables are important for evaluating operating characteristics of experimental designs in various applications, of which clinical trials with multiple endpoints provide an important example. There exist efficient algorithms to address the problem of generating multivariate distributions with given marginals and correlation structure. For model fitting as well as for simulation, it is important to know the feasible range of pairwise correlations, which can be much narrower than the interval $[-\,1,+\,1]$. We provide closed-form expressions for extreme correlations for several classes of bivariate distributions that involve both discrete and continuous endpoints, as well as an algorithm for the construction of such distributions in the discrete case.

Appendix

1.1 A.1. Proof of formulas (14), (15), Example 7

Let $f_{-1}=0$ and note that the set $\{f_i\}$ defined in (13) satisfies

$$\begin{aligned} 0 = f_{-1} \le f_0 \le f_1 \le f_2 \le \dots , ~~\lim _i f_i = 1. \end{aligned}$$

Therefore,

$$\begin{aligned} \mathrm{E}(Y_1 Y_2)= & {} \mathrm{E}[G^{-1}(U) F^{-1}(U)] = \int _0^1 G^{-1}(u) F^{-1}(u) du\nonumber \\= & {} \sum _{i=0}^{\infty } \int _{f_{i-1}}^{f_i} G^{-1}(u) F^{-1}(u) du. \end{aligned}$$

(33)

If $u\in (f_{i-1},f_i)$, then $F^{-1}(u) \equiv i$ by definition of the inverse function. Thus, (33) implies that

$$\begin{aligned} \mathrm{E}(Y_1 Y_2) = \sum _{i=0}^{\infty } i~\int _{f_{i-1}}^{f_i} G^{-1}(u) du = \sum _{i=1}^{\infty } i~\int _{f_{i-1}}^{f_i} G^{-1}(u) du . \end{aligned}$$

(34)

Using the transformation of variables $x=G^{-1}(u)$, and noting that $du = g(x) dx$, one gets

$$\begin{aligned} \mathrm{E}(Y_1 Y_2) = \sum _{i=1}^{\infty } i \int _{G^{-1}(f_{i-1})}^{G^{-1}(f_i)} x g(x) dx, \end{aligned}$$

which implies (14).

To prove (15), remark that

$$\begin{aligned} \mathrm{E}(\bar{Y}_1 Y_2) = \sum _{i=1}^{\infty } i \int _{f_{i-1}}^{f_i} G^{-1}(1-u) du. \end{aligned}$$

If $x=G^{-1}(1-u)$, then $du = -g(x) dx$ and

$$\begin{aligned} \mathrm{E}(\bar{Y}_1 Y_2) = - \sum _{i=1}^{\infty } i \int _{G^{-1}(1-f_{i-1})}^{G^{-1}(1-f_i)} x g(x) dx = \sum _{i=1}^{\infty } i \int _{G^{-1}(1-f_i)}^{G^{-1}(1-f_{i-1})} x g(x) dx, \end{aligned}$$

which proves (15). Note that $f_i$ is a non-decreasing function of i and, thus, $G^{-1}(1-f_i) \le G^{-1}(1-f_{i-1})$.

1.2 A.2. Proof of formulas (16), (17), Example 8

To prove (16), we will need the following property of the standard normal distribution:

$$\begin{aligned} I_{a,b} = \int _a^b \varPhi ^{-1}(u) du = \phi [\varPhi ^{-1}(a)] - \phi [\varPhi ^{-1}(b)]. \end{aligned}$$

(35)

Indeed, make the transformation of variables, $x = \varPhi ^{-1}(u)$, so that

$$\begin{aligned} u = \varPhi (x),~du = \phi (x) dx. \end{aligned}$$

(36)

Now it follows from (9) and (36) that

$$\begin{aligned} I_{a,b} = \int _{\varPhi ^{-1}(a)}^{\varPhi ^{-1}(b)} x \phi (x) dx = - \int _{\varPhi ^{-1}(a)}^{\varPhi ^{-1}(b)} d[\phi (x)] = \phi [\varPhi ^{-1}(a)] - \phi [\varPhi ^{-1}(b)], \end{aligned}$$

which proves (35).

To prove the first formula in (16), since $\mathrm{E}(X_2)= 0$ and $\mathrm{Var}(X_2) = 1$, it is sufficient to show that

$$\begin{aligned} I = \mathrm{E}[\varPhi ^{-1}(U) F^{-1}(U)] = \sum _{i=0}^{\infty } \phi [\varPhi ^{-1}(f_i)]. \end{aligned}$$

(37)

It follows from (34) that

$$\begin{aligned} I = \sum _{i=1}^{\infty } i~\int _{f_{i-1}}^{f_i} \varPhi ^{-1}(u) du , \end{aligned}$$

and, therefore, (35) implies:

$$\begin{aligned} I= & {} \sum _{i=1}^{\infty } i~\{\phi [\varPhi ^{-1}(f_{i-1})] - \phi [\varPhi ^{-1}(f_i)]\} \nonumber \\= & {} \phi [\varPhi ^{-1}(f_{0})] - \phi [\varPhi ^{-1}(f_1)] + 2 \{\phi [\varPhi ^{-1}(f_{1})] - \phi [\varPhi ^{-1}(f_2)]\} \nonumber \\&+\,3 \{\phi [\varPhi ^{-1}(f_{2})] - \phi [\varPhi ^{-1}(f_3)]\} + ~\dots ~= \sum _{i=0}^{\infty } \phi [\varPhi ^{-1}(f_i)], \end{aligned}$$

(38)

which proves (37) and thus proves (16).

Formula (17) also follows from the equalities in (38) since $f_k = 1$ when $X_1$ has a finite support with $P_i=0$ for all $i>k$.

1.3 A.3. Proof of formula (22)

We present the expectation $\mathrm{E}(X)$ as

$$\begin{aligned} \mathrm{E}(X) = \int _0^1 F^{-1}_X(u) du = \int _0^{1-p} F^{-1}_X(u) du + \int _{1-p}^1 F^{-1}_X(u) du . \end{aligned}$$

(39)

If $Y_1 = F^{-1}_X(u), ~Y_2 = Bern^{-1}_p(U)$ and $Y_2^- = Bern^{-1}_{1-p}(1-U)$, then it follows from (20) that

$$\begin{aligned} \mathrm{E}(Y_1 Y_2) = \int _{1-p}^1 F^{-1}_X(u) du, ~\text{ and }~~\mathrm{corr}(Y_1,Y_2) = \frac{\mathrm{E}(Y_1 Y_2) -p\mathrm{E}(X)}{\sqrt{p(1-p)\mathrm{Var}(X)}}. \end{aligned}$$

On the other hand, $Y_2^- = I(1-U > p) = I(U < 1- p)$ and it follows from (39) that $\mathrm{E}(Y_1^- Y_2) = \mathrm{E}(X) - \mathrm{E}(Y_1 Y_2)$ and, consequently,

$$\begin{aligned} \mathrm{corr}(Y_1^-, Y_2)= & {} \frac{\mathrm{E}(X)-\mathrm{E}(Y_1 Y_2) - (1-p)\mathrm{E}(X)}{\sqrt{p(1-p)\mathrm{Var}(X)}} = \frac{-\mathrm{E}(Y_1 Y_2) + p\mathrm{E}(X)}{\sqrt{p(1-p)\mathrm{Var}(X)}} \\= & {} -\mathrm{corr}(Y_1,Y_2), \end{aligned}$$

which proves (22).

1.4 A.4. Example 15. Code for computing matrix {$\pi _{ij}$} in (27) and implementing formula (28)

Matlab code for (27), (28) (the text after % is a comment)

R code for (27), (28)