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Mathematical Framework for latentcor

Main Framework

Latent Gaussian Copula Model for Mixed Data

latentcor utilizes the powerful semi-parametric latent Gaussian copula models to estimate latent correlations between mixed data types (continuous/binary/ternary/truncated or zero-inflated). Below we review the definitions for each type.

from latentcor import gen_data, latentcor

Definition of continuous model

A random $X\in\cal{R}^{p}$ satisfies the Gaussian copula (or nonparanormal) model if there exist monotonically increasing f = (fj)j = 1p with Zj = fj(Xj) satisfying Z ∼ Np(0, Σ), σjj = 1; we denote X ∼ NPN(0, Σ, f) :citefan2017high.

print(gen_data(n = 6, tps = "con")['X'])

Definition of binary model

A random $X\in\cal{R}^{p}$ satisfies the binary latent Gaussian copula model if there exists W ∼ NPN(0, Σ, f) such that Xj = I(Wj > cj), where I( ⋅ ) is the indicator function and cj are constants :citefan2017high.

print(gen_data(n = 6, tps = "bin")['X'])

Definition of ternary model

A random $X\in\cal{R}^{p}$ satisfies the ternary latent Gaussian copula model if there exists W ∼ NPN(0, Σ, f) such that Xj = I(Wj > cj) + I(Wj > cj), where I( ⋅ ) is the indicator function and cj < cj are constants :citequan2018rank.

print(gen_data(n = 6, tps = "ter")['X'])

Definition of truncated or zero-inflated model

A random $X\in\cal{R}^{p}$ satisfies the truncated latent Gaussian copula model if there exists W ∼ NPN(0, Σ, f) such that Xj = I(Wj > cj)Wj, where I( ⋅ ) is the indicator function and cj are constants :citeyoon2020sparse.

print(gen_data(n = 6, tps = "tru")['X'])

Mixed latent Gaussian copula model

The mixed latent Gaussian copula model jointly models W = (W1, W2, W3, W4) ∼ NPN(0, Σ, f) such that X1j = W1j, X2j = I(W2j > c2j), X3j = I(W3j > c3j) + I(W3j > c3j) and X4j = I(W4j > c4j)W4j.

X = gen_data(n = 100, tps = ["con", "bin", "ter", "tru"])['X'] print(X[ :6, : ])

Moment-based estimation of latent correlation matrix based on bridge functions

The estimation of latent correlation matrix Σ is achieved via the bridge function F which is defined such that E(τ̂jk) = F(σjk), where σjk is the latent correlation between variables j and k, and τ̂jk is the corresponding sample Kendall's τ.

Kendall's correlation

Given observed $\mathbf{x}_{j}, \mathbf{x}_{k}\in\cal{R}^{n}$,

$$\hat{\tau}_{jk}=\hat{\tau}(\mathbf{x}_{j}, \mathbf{x}_{k})=\frac{2}{n(n-1)}\sum_{1\le i<i'\le n}sign(x_{ij}-x_{i'j})sign(x_{ik}-x_{i'k}),$$

where n is the sample size.

latentcor calculates pairwise Kendall's τ̂ as part of the estimation process.

K = latentcor(X, tps = ["con", "bin", "ter", "tru"])['K'] print(K)

Using F and τ̂jk, a moment-based estimator is σ̂jk = F − 1(τ̂jk) with the corresponding Σ̂ being consistent for Σ :citefan2017high,quan2018rank,yoon2020sparse.

The explicit form of bridge function F has been derived for all combinations of continuous(C)/binary(B)/ternary(N)/truncated(T) variable types, and we summarize the corresponding references. Each of this combinations is implemented in latentcor.

Below we provide an explicit form of F for each combination.

Theorem (explicit form of bridge function)

Let $W_{1}\in\cal{R}^{p_{1}}$, $W_{2}\in\cal{R}^{p_{2}}$, $W_{3}\in\cal{R}^{p_{3}}$, $W_{4}\in\cal{R}^{p_{4}}$ be such that W = (W1, W2, W3, W4) ∼ NPN(0, Σ, f) with p = p1 + p2 + p3 + p4. Let $X=(X_{1}, X_{2}, X_{3}, X_{4})\in\cal{R}^{p}$ satisfy Xj = Wj for j=1,...,p_{1}, Xj = I(Wj > cj) for j = p1 + 1, ..., p1 + p2, Xj = I(Wj > cj) + I(Wj > cj) for j = p1 + p2 + 1, ..., p3 and Xj = I(Wj > cj)Wj for j = p1 + p2 + p3 + 1, ..., p with Δj = f(cj). The rank-based estimator of Σ based on the observed n realizations of X is the matrix with jj = 1, jk = kj = F − 1(τ̂jk) with block structure

$$\begin{aligned} \mathbf{\hat{R}}=\begin{pmatrix} F_{CC}^{-1}(\hat{\tau}) & F_{CB}^{-1}(\hat{\tau}) & F_{CN}^{-1}(\hat{\tau}) & F_{CT}^{-1}(\hat{\tau})\\\ F_{BC}^{-1}(\hat{\tau}) & F_{BB}^{-1}(\hat{\tau}) & F_{BN}^{-1}(\hat{\tau}) & F_{BT}^{-1}(\hat{\tau})\\\ F_{NC}^{-1}(\hat{\tau}) & F_{NB}^{-1}(\hat{\tau}) & F_{NN}^{-1}(\hat{\tau}) & F_{NT}^{-1}(\hat{\tau})\\\ F_{TC}^{-1}(\hat{\tau}) & F_{TB}^{-1}(\hat{\tau}) & F_{TN}^{-1}(\hat{\tau}) & F_{TT}^{-1}(\hat{\tau}) \end{pmatrix} \end{aligned}$$

$$\begin{aligned} F(\cdot)=\begin{cases} CC: & 2\sin^{-1}(r)/\pi \\\ \\\ BC: & 4\Phi_{2}(\Delta_{j},0;r/\sqrt{2})-2\Phi(\Delta_{j}) \\\ \\\ BB: & 2\{\Phi_{2}(\Delta_{j},\Delta_{k};r)-\Phi(\Delta_{j})\Phi(\Delta_{k})\} \\\ \\\ NC: & 4\Phi_{2}(\Delta_{j}^{2},0;r/\sqrt{2})-2\Phi(\Delta_{j}^{2})+4\Phi_{3}(\Delta_{j}^{1},\Delta_{j}^{2},0;\Sigma_{3a}(r))-2\Phi(\Delta_{j}^{1})\Phi(\Delta_{j}^{2})\\\ \\\ NB: & 2\Phi_{2}(\Delta_{j}^{2},\Delta_{k},r)\{1-\Phi(\Delta_{j}^{1})\}-2\Phi(\Delta_{j}^{2})\{\Phi(\Delta_{k})-\Phi_{2}(\Delta_{j}^{1},\Delta_{k},r)\} \\\ \\\ NN: & 2\Phi_{2}(\Delta_{j}^{2},\Delta_{k}^{2};r)\Phi_{2}(-\Delta_{j}^{1},-\Delta_{k}^{1};r)-2\{\Phi(\Delta_{j}^{2})-\Phi_{2}(\Delta_{j}^{2},\Delta_{k}^{1};r)\}\{\Phi(\Delta_{k}^{2})\\\ & -\Phi_{2}(\Delta_{j}^{1},\Delta_{k}^{2};r)\} \\\ \\\ TC: & -2\Phi_{2}(-\Delta_{j},0;1/\sqrt{2})+4\Phi_{3}(-\Delta_{j},0,0;\Sigma_{3b}(r)) \\\ \\\ TB: & 2\{1-\Phi(\Delta_{j})\}\Phi(\Delta_{k})-2\Phi_{3}(-\Delta_{j},\Delta_{k},0;\Sigma_{3c}(r))-2\Phi_{3}(-\Delta_{j},\Delta_{k},0;\Sigma_{3d}(r)) \\\ \\\ TN: & -2\Phi(-\Delta_{k}^{1})\Phi(\Delta_{k}^{2}) + 2\Phi_{3}(-\Delta_{k}^{1},\Delta_{k}^{2},\Delta_{j};\Sigma_{3e}(r)) \\\ & +2\Phi_{4}(-\Delta_{k}^{1},\Delta_{k}^{2},-\Delta_{j},0;\Sigma_{4a}(r))+2\Phi_{4}(-\Delta_{k}^{1},\Delta_{k}^{2},-\Delta_{j},0;\Sigma_{4b}(r)) \\\ \\\ TT: & -2\Phi_{4}(-\Delta_{j},-\Delta_{k},0,0;\Sigma_{4c}(r))+2\Phi_{4}(-\Delta_{j},-\Delta_{k},0,0;\Sigma_{4d}(r)) \\\ \end{cases} \end{aligned}$$

where Δj = Φ − 1(π0j), Δk = Φ − 1(π0k), Δj1 = Φ − 1(π0j), Δj2 = Φ − 1(π0j + π1j), Δk1 = Φ − 1(π0k), Δk2 = Φ − 1(π0k + π1k),

$$\begin{aligned} \Sigma_{3a}(r)= \begin{pmatrix} 1 & 0 & \frac{r}{\sqrt{2}} \\\ 0 & 1 & -\frac{r}{\sqrt{2}} \\\ \frac{r}{\sqrt{2}} & -\frac{r}{\sqrt{2}} & 1 \end{pmatrix}, \;\;\; \Sigma_{3b}(r)= \begin{pmatrix} 1 & \frac{1}{\sqrt{2}} & \frac{r}{\sqrt{2}}\\\ \frac{1}{\sqrt{2}} & 1 & r \\\ \frac{r}{\sqrt{2}} & r & 1 \end{pmatrix}, \end{aligned}$$

$$\begin{aligned} \Sigma_{3c}(r)= \begin{pmatrix} 1 & -r & \frac{1}{\sqrt{2}} \\\ -r & 1 & -\frac{r}{\sqrt{2}} \\\ \frac{1}{\sqrt{2}} & -\frac{r}{\sqrt{2}} & 1 \end{pmatrix}, \;\;\; \Sigma_{3d}(r)= \begin{pmatrix} 1 & 0 & -\frac{1}{\sqrt{2}} \\\ 0 & 1 & -\frac{r}{\sqrt{2}} \\\ -\frac{1}{\sqrt{2}} & -\frac{r}{\sqrt{2}} & 1 \end{pmatrix}, \end{aligned}$$

$$\begin{aligned} \Sigma_{3e}(r)= \begin{pmatrix} 1 & 0 & 0 \\\ 0 & 1 & r \\\ 0 & r & 1 \end{pmatrix}, \;\;\; \Sigma_{4a}(r)= \begin{pmatrix} 1 & 0 & 0 & \frac{r}{\sqrt{2}} \\\ 0 & 1 & -r & \frac{r}{\sqrt{2}} \\\ 0 & -r & 1 & -\frac{1}{\sqrt{2}} \\\ \frac{r}{\sqrt{2}} & \frac{r}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 1 \end{pmatrix}, \end{aligned}$$

$$\begin{aligned} \Sigma_{4b}(r)= \begin{pmatrix} 1 & 0 & r & \frac{r}{\sqrt{2}} \\\ 0 & 1 & 0 & \frac{r}{\sqrt{2}} \\\ r & 0 & 1 & \frac{1}{\sqrt{2}} \\\ \frac{r}{\sqrt{2}} & \frac{r}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1 \end{pmatrix}, \;\;\; \Sigma_{4c}(r)= \begin{pmatrix} 1 & 0 & \frac{1}{\sqrt{2}} & -\frac{r}{\sqrt{2}} \\\ 0 & 1 & -\frac{r}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\\ \frac{1}{\sqrt{2}} & -\frac{r}{\sqrt{2}} & 1 & -r \\\ -\frac{r}{\sqrt{2}} & \frac{1}{\sqrt{2}} & -r & 1 \end{pmatrix} \end{aligned}$$

and

$$\begin{aligned} \Sigma_{4d}(r)= \begin{pmatrix} 1 & r & \frac{1}{\sqrt{2}} & \frac{r}{\sqrt{2}} \\\ r & 1 & \frac{r}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\\ \frac{1}{\sqrt{2}} & \frac{r}{\sqrt{2}} & 1 & r \\\ \frac{r}{\sqrt{2}} & \frac{1}{\sqrt{2}} & r & 1 \end{pmatrix}. \end{aligned}$$

Estimation methods

Given the form of bridge function F, obtaining a moment-based estimation σ̂jk requires inversion of F. latentcor implements two methods for calculation of the inversion:

  • method = "original"
  • method = "approx"

Both methods calculate inverse bridge function applied to each element of sample Kendall's τ matrix. Because the calculation is performed point-wise (separately for each pair of variables), the resulting point-wise estimator of correlation matrix may not be positive semi-definite. latentcor performs projection of the pointwise-estimator to the space of positive semi-definite matrices, and allows for shrinkage towards identity matrix using the parameter nu.

Original method (`method = "original"`)

Original estimation approach relies on numerical inversion of F based on solving uni-root optimization problem. Given the calculated τ̂jk (sample Kendall's τ between variables j and k), the estimate of latent correlation σ̂jk is obtained by calling scipy.optimize.fminbound function to solve the following optimization problem:


jk = arg minr{F(r) − τ̂jk}2.

The parameter tol controls the desired accuracy of the minimizer and is passed to scipy.optimize.fminbound, with the default precision of 10 − 8.

estimate_original = latentcor(X, tps = ["con", "bin", "ter", "tru"], method = "original", tol = 1e-8)

Algorithm for Original method

Input: F(r) = F(r, Δ) - bridge function based on the type of variables j, k

  • Step 1. Calculate τ̂jk using (1).

print(estimate_original['K'])

  • Step 2. For binary/truncated variable j, set $\hat{\mathbf{\Delta}}_{j}=\hat{\Delta}_{j}=\Phi^{-1}(\pi_{0j})$ with $\pi_{0j}=\sum_{i=1}^{n}\frac{I(x_{ij}=0)}{n}$. For ternary variable j, set $\hat{\mathbf{\Delta}}_{j}=(\hat{\Delta}_{j}^{1}, \hat{\Delta}_{j}^{2})$ where Δ̂j1 = Φ − 1(π0j) and Δ̂j2 = Φ − 1(π0j + π1j) with $\pi_{0j}=\sum_{i=1}^{n}\frac{I(x_{ij}=0)}{n}$ and $\pi_{1j}=\sum_{i=1}^{n}\frac{I(x_{ij}=1)}{n}$.

print(estimate_original['zratios'])

  • Step 3 Compute F − 1(τ̂jk) as jk = argmin{F(r) − τ̂jk}2 solved via scipy.optimize.fminbound function with accuracy tol.

print(estimate_original['Rpointwise'])

Approximation method (`method = "approx"`)

A faster approximation method is based on multi-linear interpolation of pre-computed inverse bridge function on a fixed grid of points :citeyoon2021fast. This is possible as the inverse bridge function is an analytic function of at most 5 parameters:

  • Kendall's τ
  • Proportion of zeros in the 1st variable
  • (Possibly) proportion of zeros and ones in the 1st variable
  • (Possibly) proportion of zeros in the 2nd variable
  • (Possibly) proportion of zeros and ones in the 2nd variable

In short, d-dimensional multi-linear interpolation uses a weighted average of 2d neighbors to approximate the function values at the points within the d-dimensional cube of the neighbors, and to perform interpolation, latentcor takes advantage of the Python package scipy.interpolate.RegularGridInterpolator. This approximation method has been first described in :citeyoon2021fast for continuous/binary/truncated cases. In latentcor, we additionally implement ternary case, and optimize the choice of grid as well as interpolation boundary for faster computations with smaller memory footprint.

estimate_approx = latentcor(X, tps = ["con", "bin", "ter", "tru"], method = "approx") print(estimate_approx['Rpointwise'])

Algorithm for Approximation method

Input: Let  = h(g), pre-computed values F − 1(h − 1()) on a fixed grid $\check{g}\in\check{\cal{G}}$ based on the type of variables j and k. For binary/continuous case,  = (τ̌jk, Δ̌j); for binary/binary case,  = (τ̌jk, Δ̌j, Δ̌k); for truncated/continuous case,  = (τ̌jk, Δ̌j); for truncated/truncated case,  = (τ̌jk, Δ̌j, Δ̌k); for ternary/continuous case,  = (τ̌jk, Δ̌j1, Δ̌j2); for ternary/binary case,  = (τ̌jk, Δ̌j1, Δ̌j2, Δ̌k); for ternary/truncated case,  = (τ̌jk, Δ̌j1, Δ̌j2, Δ̌k); for ternay/ternary case,  = (τ̌jk, Δ̌j1, Δ̌j2, Δ̌k1, Δ̌k2).

  • Step 1 and Step 2 same as Original method.
  • Step 3. If |τ̂jk| ≤ ratio × τ̄jk( ⋅ ), apply interpolation; otherwise apply Original method.

To avoid interpolation in areas with high approximation errors close to the boundary, we use hybrid scheme in Step 3. The parameter ratio controls the size of the region where the interpolation is performed (ratio = 0 means no interpolation, ratio = 1 means interpolation is always performed). For the derivation of approximate bound for BC, BB, TC, TB, TT cases see @yoon2021fast. The derivation of approximate bound for NC, NB, NN, NT case is in the Appendix.

$$\begin{aligned} \bar{\tau}_{jk}(\cdot)= \begin{cases} 2\pi_{0j}(1-\pi_{0j}) & for \; BC \; case\\\ 2\min(\pi_{0j},\pi_{0k})\{1-\max(\pi_{0j}, \pi_{0k})\} & for \; BB \; case\\\ 2\{\pi_{0j}(1-\pi_{0j})+\pi_{1j}(1-\pi_{0j}-\pi_{1j})\} & for \; NC \; case\\\ 2\min(\pi_{0j}(1-\pi_{0j})+\pi_{1j}(1-\pi_{0j}-\pi_{1j}),\pi_{0k}(1-\pi_{0k})) & for \; NB \; case\\\ 2\min(\pi_{0j}(1-\pi_{0j})+\pi_{1j}(1-\pi_{0j}-\pi_{1j}), \\\ \;\;\;\;\;\;\;\;\;\;\pi_{0k}(1-\pi_{0k})+\pi_{1k}(1-\pi_{0k}-\pi_{1k})) & for \; NN \; case\\\ 1-(\pi_{0j})^{2} & for \; TC \; case\\\ 2\max(\pi_{0k},1-\pi_{0k})\{1-\max(\pi_{0k},1-\pi_{0k},\pi_{0j})\} & for \; TB \; case\\\ 1-\{\max(\pi_{0j},\pi_{0k},\pi_{1k},1-\pi_{0k}-\pi_{1k})\}^{2} & for \; TN \; case\\\ 1-\{\max(\pi_{0j},\pi_{0k})\}^{2} & for \; TT \; case\\\ \end{cases} \end{aligned}$$

By default, latentcor uses ratio = 0.9 as this value was recommended in @yoon2021fast having a good balance of accuracy and computational speed. This value, however, can be modified by the user

print(latentcor(X, tps = ["con", "bin", "ter", "tru"], method = "approx", ratio = 0.99)['R']) print(latentcor(X, tps = ["con", "bin", "ter", "tru"], method = "approx", ratio = 0.4)['R']) print(latentcor(X, tps = ["con", "bin", "ter", "tru"], method = "original")['R'])

The lower is the ratio, the closer is the approximation method to original method (with ratio = 0 being equivalent to method = "original"), but also the higher is the cost of computations.

Rescaled Grid for Interpolation

Since |τ̂| ≤ τ̄, the grid does not need to cover the whole domain τ ∈ [ − 1, 1]. To optimize memory associated with storing the grid, we rescale τ as follows:


τ̌jk = τjk/τ̄jk ∈ [ − 1, 1],

where τ̄jk is as defined above.

In addition, for ternary variable j, it always holds that

$$\Delta_{j}^{2}>\Delta_{j}^{1}` since :math:`\Delta_{j}^{1}=\Phi^{-1}(\pi_{0j})$$

and


Δj2 = Φ − 1(π0j + π1j).

Thus, the grid should not cover the the area corresponding to


Δj2 ≥ Δj1.

We thus rescale as follows:


Δ̌j1 = Δj1/Δj2 ∈ [0, 1];


Δ̌j2 = Δj2 ∈ [0, 1].

Adjustment of pointwise-estimator for positive-definiteness

Since the estimation is performed point-wise, the resulting matrix of estimated latent correlations is not guaranteed to be positive semi-definite. For example, this could be expected when the sample size is small (and so the estimation error for each pairwise correlation is larger).

X = gen_data(n = 6, tps = ["con", "bin", "ter", "tru"])['X'] print(latentcor(X, tps = ["con", "bin", "ter", "tru"])['Rpointwise'])

latentcor automatically corrects the pointwise estimator to be positive definite by making two adjustments. First, if Rpointwise has smallest eigenvalue less than zero, the latentcor projects this matrix to the nearest positive semi-definite matrix. The user is notified of this adjustment through the message (supressed in previous code chunk), e.g.

print(latentcor(X, tps = ["con", "bin", "ter", "tru"])['R'])

Second, latentcor shrinks the adjusted matrix of correlations towards identity matrix using the parameter \nu with default value of 0.001 (nu = 0.001), so that the resulting latentcor[0] is strictly positive definite with the minimal eigenvalue being greater or equal to \nu. That is


R = (1 − ν) + νI,

where \widetilde R is the nearest positive semi-definite matrix to Rpointwise.

print(latentcor(X, tps = ["con", "bin", "ter", "tru"], nu = 0.001)['R'])

As a result, R and Rpointwise could be quite different when sample size n is small. When n is large and p is moderate, the difference is typically driven by parameter nu.

X = gen_data(n = 100, tps = ["con", "bin", "ter", "tru"])['X'] out = latentcor(X, tps = ["con", "bin", "ter", "tru"], nu = 0.001) print(out['Rpointwise']) print(out['R'])

Appendix

Derivation of bridge function for ternary/truncated case

Without loss of generality, let j = 1 and k = 2. By the definition of Kendall's τ,

$$\tau_{12}=E(\hat{\tau}_{12})=E[\frac{2}{n(n-1)}\sum_{1\leq i\leq i' \leq n} sign\{(X_{i1}-X_{i'1})(X_{i2}-X_{i'2})\}].$$

Since X1 is ternary,

$$\begin{aligned} \begin{align} &sign(X_{1}-X_{1}') \nonumber\\ =&[I(U_{1}>C_{11},U_{1}'\leq C_{11})+I(U_{1}>C_{12},U_{1}'\leq C_{12})-I(U_{1}>C_{12},U_{1}'\leq C_{11})] \nonumber\\\ &-[I(U_{1}\leq C_{11}, U_{1}'>C_{11})+I(U_{1}\leq C_{12}, U_{1}'>C_{12})-I(U_{1}\leq C_{11}, U_{1}'>C_{12})] \nonumber\\\ =&[I(U_{1}>C_{11})-I(U_{1}>C_{11},U_{1}'>C_{11})+I(U_{1}>C_{12})-I(U_{1}>C_{12},U_{1}'>C_{12}) \nonumber\\\ &-I(U_{1}>C_{12})+I(U_{1}>C_{12},U_{1}'>C_{11})] \nonumber\\\ &-[I(U_{1}'>C_{11})-I(U_{1}>C_{11},U_{1}'>C_{11})+I(U_{1}'>C_{12})-I(U_{1}>C_{12},U_{1}'>C_{12}) \nonumber\\\ &-I(U_{1}'>C_{12})+I(U_{1}>C_{11},U_{1}'>C_{12})] \nonumber\\\ =&I(U_{1}>C_{11})+I(U_{1}>C_{12},U_{1}'>C_{11})-I(U_{1}'>C_{11})-I(U_{1}>C_{11},U_{1}'>C_{12}) \nonumber\\\ =&I(U_{1}>C_{11},U_{1}'\leq C_{12})-I(U_{1}'>C_{11},U_{1}\leq C_{12}). \end{align} \end{aligned}$$

Since X2 is truncated, C1 > 0 and

$$\begin{aligned} \begin{align} sign(X_{2}-X_{2}')=&-I(X_{2}=0,X_{2}'>0)+I(X_{2}>0,X_{2}'=0) \nonumber\\\ &+I(X_{2}>0,X_{2}'>0)sign(X_{2}-X_{2}') \nonumber\\\ =&-I(X_{2}=0)+I(X_{2}'=0)+I(X_{2}>0,X_{2}'>0)sign(X_{2}-X_{2}'). \end{align} \end{aligned}$$

Since f is monotonically increasing, sign(X2 − X2′) = sign(Z2 − Z2′),

$$\begin{aligned} \begin{align} \tau_{12}=&E[I(U_{1}>C_{11},U_{1}'\leq C_{12}) sign(X_{2}-X_{2}')] \nonumber\\ &-E[I(U_{1}'>C_{11},U_{1}\leq C_{12}) sign(X_{2}-X_{2}')] \nonumber\\\ =&-E[I(U_{1}>C_{11},U_{1}'\leq C_{12}) I(X_{2}=0)] \nonumber\\\ &+E[I(U_{1}>C_{11},U_{1}'\leq C_{12}) I(X_{2}'=0)] \nonumber\\\ &+E[I(U_{1}>C_{11},U_{1}'\leq C_{12})I(X_{2}>0,X_{2}'>0)sign(Z_{2}-Z_{2}')] \nonumber\\\ &+E[I(U_{1}'>C_{11},U_{1}\leq C_{12}) I(X_{2}=0)] \nonumber\\\ &-E[I(U_{1}'>C_{11},U_{1}\leq C_{12}) I(X_{2}'=0)] \nonumber\\\ &-E[I(U_{1}'>C_{11},U_{1}\leq C_{12})I(X_{2}>0,X_{2}'>0)sign(Z_{2}-Z_{2}')] \nonumber\\\ =&-2E[I(U_{1}>C_{11},U_{1}'\leq C_{12}) I(X_{2}=0)] \nonumber\\\ &+2E[I(U_{1}>C_{11},U_{1}'\leq C_{12}) I(X_{2}'=0)] \nonumber\\\ &+E[I(U_{1}>C_{11},U_{1}'\leq C_{12})I(X_{2}>0,X_{2}'>0)sign(Z_{2}-Z_{2}')] \nonumber\\\ &-E[I(U_{1}'>C_{11},U_{1}\leq C_{12})I(X_{2}>0,X_{2}'>0)sign(Z_{2}-Z_{2}')]. \end{align} \end{aligned}$$

From the definition of U, let Zj = fj(Uj) and Δj = fj(Cj) for j = 1, 2. Using sign(x) = 2I(x > 0) − 1, we obtain

$$\begin{aligned} \begin{align} \tau_{12}=&-2E[I(Z_{1}>\Delta_{11},Z_{1}'\leq \Delta_{12},Z_{2}\leq \Delta_{2})]+2E[I(Z_{1}>\Delta_{11},Z_{1}'\leq \Delta_{12},Z_{2}'\leq \Delta_{2})] \nonumber\\\ &+2E[I(Z_{1}>\Delta_{11},Z_{1}'\leq \Delta_{12})I(Z_{2}>\Delta_{2},Z_{2}'>\Delta_{2},Z_{2}-Z_{2}'>0)] \nonumber\\\ &-2E[I(Z_{1}'>\Delta_{11},Z_{1}\leq \Delta_{12})I(Z_{2}>\Delta_{2},Z_{2}'>\Delta_{2},Z_{2}-Z_{2}'>0)] \nonumber\\\ =&-2E[I(Z_{1}>\Delta_{11},Z_{1}'\leq \Delta_{12}, Z_{2}\leq \Delta_{2})]+2E[I(Z_{1}>\Delta_{11},Z_{1}'\leq \Delta_{12}, Z_{2}'\leq \Delta_{2})] \nonumber\\\ &+2E[I(Z_{1}>\Delta_{11},Z_{1}'\leq\Delta_{12},Z_{2}'>\Delta_{2},Z_{2}>Z_{2}')] \nonumber\\\ &-2E[I(Z_{1}'>\Delta_{11},Z_{1}\leq\Delta_{12},Z_{2}'>\Delta_{2},Z_{2}>Z_{2}')]. \end{align} \end{aligned}$$

Since $\{\frac{Z_{2}'-Z_{2}}{\sqrt{2}}, -Z{1}\}$, $\{\frac{Z_{2}'-Z_{2}}{\sqrt{2}}, Z{1}'\}$ and $\{\frac{Z_{2}'-Z_{2}}{\sqrt{2}}, -Z{2}'\}$ are standard bivariate normally distributed variables with correlation $-\frac{1}{\sqrt{2}}$, $r/\sqrt{2}$ and $-\frac{r}{\sqrt{2}}$, respectively, by the definition of Φ3(⋅,⋅,⋅; ⋅ ) and Φ4(⋅,⋅,⋅,⋅; ⋅ ) we have

$$\begin{aligned} \begin{align} F_{NT}(r;\Delta_{j}^{1},\Delta_{j}^{2},\Delta_{k})= & -2\Phi_{3}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},\Delta_{k};\begin{pmatrix} 1 & 0 & -r \\\ 0 & 1 & 0 \\\ -r & 0 & 1 \end{pmatrix} \right\} \nonumber\\\ &+2\Phi_{3}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},\Delta_{k};\begin{pmatrix} 1 & 0 & 0 \\\ 0 & 1 & r \\\ 0 & r & 1 \end{pmatrix}\right\}\nonumber \\\ & +2\Phi_{4}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},-\Delta_{k},0;\begin{pmatrix} 1 & 0 & 0 & \frac{r}{\sqrt{2}} \\\ 0 & 1 & -r & \frac{r}{\sqrt{2}} \\\ 0 & -r & 1 & -\frac{1}{\sqrt{2}} \\\ \frac{r}{\sqrt{2}} & \frac{r}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 1 \end{pmatrix}\right\} \nonumber\\\ &-2\Phi_{4}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},-\Delta_{k},0;\begin{pmatrix} 1 & 0 & r & -\frac{r}{\sqrt{2}} \\\ 0 & 1 & 0 & -\frac{r}{\sqrt{2}} \\\ r & 0 & 1 & -\frac{1}{\sqrt{2}} \\\ -\frac{r}{\sqrt{2}} & -\frac{r}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 1 \end{pmatrix}\right\}. \end{align} \end{aligned}$$

Using the facts that

$$\begin{aligned} \begin{align} &\Phi_{4}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},-\Delta_{k},0;\begin{pmatrix} 1 & 0 & r & -\frac{r}{\sqrt{2}} \\\ 0 & 1 & 0 & -\frac{r}{\sqrt{2}} \\\ r & 0 & 1 & -\frac{1}{\sqrt{2}} \\\ -\frac{r}{\sqrt{2}} & -\frac{r}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 1 \end{pmatrix}\right\} \nonumber\\ &+\Phi_{4}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},-\Delta_{k},0;\begin{pmatrix} 1 & 0 & r & \frac{r}{\sqrt{2}} \\\ 0 & 1 & 0 & \frac{r}{\sqrt{2}} \\\ r & 0 & 1 & \frac{1}{\sqrt{2}} \\\ \frac{r}{\sqrt{2}} & \frac{r}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1 \end{pmatrix}\right\} \nonumber\\\ =&\Phi_{3}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},-\Delta_{k};\begin{pmatrix} 1 & 0 & 0 \\\ 0 & 1 & r \\\ 0 & r & 1 \end{pmatrix}\right\} \end{align} \end{aligned}$$

and

$$\begin{aligned} \begin{align} &\Phi_{3}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},-\Delta_{k};\begin{pmatrix} 1 & 0 & 0 \\\ 0 & 1 & r \\\ 0 & r & 1 \end{pmatrix}\right\}+\Phi_{3}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},\Delta_{k};\begin{pmatrix} 1 & 0 & -r \\\ 0 & 1 & 0 \\\ -r & 0 & 1 \end{pmatrix} \right\} \nonumber\\\ =&\Phi_{2}(-\Delta_{j}^{1},\Delta_{j}^{2};0) =\Phi(-\Delta_{j}^{1})\Phi(\Delta_{j}^{2}). \end{align} \end{aligned}$$

So that,

$$\begin{aligned} \begin{align} F_{NT}(r;\Delta_{j}^{1},\Delta_{j}^{2},\Delta_{k})= & -2\Phi(-\Delta_{j}^{1})\Phi(\Delta_{j}^{2}) \nonumber\\\ &+2\Phi_{3}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},\Delta_{k};\begin{pmatrix} 1 & 0 & 0 \\\ 0 & 1 & r \\\ 0 & r & 1 \end{pmatrix}\right\}\nonumber \\\ & +2\Phi_{4}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},-\Delta_{k},0;\begin{pmatrix} 1 & 0 & 0 & \frac{r}{\sqrt{2}} \\\ 0 & 1 & -r & \frac{r}{\sqrt{2}} \\\ 0 & -r & 1 & -\frac{1}{\sqrt{2}} \\\ \frac{r}{\sqrt{2}} & \frac{r}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 1 \end{pmatrix}\right\} \nonumber\\\ &+2\Phi_{4}\left\{-\Delta_{j}^{1},\Delta_{j}^{2},-\Delta_{k},0;\begin{pmatrix} 1 & 0 & r & \frac{r}{\sqrt{2}} \\\ 0 & 1 & 0 & \frac{r}{\sqrt{2}} \\\ r & 0 & 1 & \frac{1}{\sqrt{2}} \\\ \frac{r}{\sqrt{2}} & \frac{r}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1 \end{pmatrix}\right\}. \end{align} \end{aligned}$$

It is easy to get the bridge function for truncated/ternary case by switching j and k.

Derivation of approximate bound for the ternary/continuous case

Let $n_{0x}=\sum_{i=1}^{n_x}I(x_{i}=0)$, $n_{2x}=\sum_{i=1}^{n_x}I(x_{i}=2)$, $\pi_{0x}=\frac{n_{0x}}{n_{x}}$ and $\pi_{2x}=\frac{n_{2x}}{n_{x}}$, then

$$\begin{aligned} \begin{align} |\tau(\mathbf{x})|\leq & \frac{n_{0x}(n-n_{0x})+n_{2x}(n-n_{0x}-n_{2x})}{\begin{pmatrix} n \\ 2 \end{pmatrix}} \nonumber\\\ = & 2\{\frac{n_{0x}}{n-1}-(\frac{n_{0x}}{n})(\frac{n_{0x}}{n-1})+\frac{n_{2x}}{n-1}-(\frac{n_{2x}}{n})(\frac{n_{0x}}{n-1})-(\frac{n_{2x}}{n})(\frac{n_{2x}}{n-1})\} \nonumber\\\ \approx & 2\{\frac{n_{0x}}{n}-(\frac{n_{0x}}{n})^2+\frac{n_{2x}}{n}-(\frac{n_{2x}}{n})(\frac{n_{0x}}{n})-(\frac{n_{2x}}{n})^2\} \nonumber\\\ = & 2\{\pi_{0x}(1-\pi_{0x})+\pi_{2x}(1-\pi_{0x}-\pi_{2x})\} \end{align} \end{aligned}$$

For ternary/binary and ternary/ternary cases, we combine the two individual bounds.

Derivation of approximate bound for the ternary/truncated case

Let x ∈ ℛn and y ∈ ℛn be the observed n realizations of ternary and truncated variables, respectively. Let $n_{0x}=\sum_{i=0}^{n}I(x_{i}=0)$, $\pi_{0x}=\frac{n_{0x}}{n}$, $n_{1x}=\sum_{i=0}^{n}I(x_{i}=1)$, $\pi_{1x}=\frac{n_{1x}}{n}$, $n_{2x}=\sum_{i=0}^{n}I(x_{i}=2)$, $\pi_{2x}=\frac{n_{2x}}{n}$, $n_{0y}=\sum_{i=0}^{n}I(y_{i}=0)$, $\pi_{0y}=\frac{n_{0y}}{n}$, $n_{0x0y}=\sum_{i=0}^{n}I(x_{i}=0 \;\&amp; \; y_{i}=0)$, $n_{1x0y}=\sum_{i=0}^{n}I(x_{i}=1 \;\&amp; \; y_{i}=0)$ and $n_{2x0y}=\sum_{i=0}^{n}I(x_{i}=2 \;\&amp; \; y_{i}=0)$ then

$$\begin{aligned} \begin{align} |\tau(\mathbf{x}, \mathbf{y})|\leq & \frac{\begin{pmatrix}n \\ 2\end{pmatrix}-\begin{pmatrix}n_{0x} \\ 2\end{pmatrix}-\begin{pmatrix}n_{1x} \\ 2\end{pmatrix}-\begin{pmatrix} n_{2x} \\ 2 \end{pmatrix}-\begin{pmatrix}n_{0y} \\ 2\end{pmatrix}+\begin{pmatrix}n_{0x0y} \\ 2 \end{pmatrix}+\begin{pmatrix}n_{1x0y} \\ 2\end{pmatrix}+\begin{pmatrix}n_{2x0y} \\ 2\end{pmatrix}}{\begin{pmatrix}n \\ 2\end{pmatrix}} \nonumber \end{align} \end{aligned}$$

Since n0x0y ≤ min (n0x, n0y), n1x0y ≤ min (n1x, n0y) and n2x0y ≤ min (n2x, n0y) we obtain

$$\begin{aligned} \begin{align} |\tau(\mathbf{x}, \mathbf{y})|\leq & \frac{\begin{pmatrix}n \\ 2\end{pmatrix}-\begin{pmatrix}n_{0x} \\ 2\end{pmatrix}-\begin{pmatrix}n_{1x} \\ 2\end{pmatrix}-\begin{pmatrix} n_{2x} \\ 2 \end{pmatrix}-\begin{pmatrix}n_{0y} \\ 2\end{pmatrix}}{\begin{pmatrix}n \\ 2\end{pmatrix}} \nonumber\\\ & + \frac{\begin{pmatrix}\min(n_{0x},n_{0y}) \\ 2 \end{pmatrix}+\begin{pmatrix}\min(n_{1x},n_{0y}) \\ 2\end{pmatrix}+\begin{pmatrix}\min(n_{2x},n_{0y}) \\ 2\end{pmatrix}}{\begin{pmatrix}n \\ 2\end{pmatrix}} \nonumber\\\ \leq & \frac{\begin{pmatrix}n \\ 2\end{pmatrix}-\begin{pmatrix}\max(n_{0x},n_{1x},n_{2x},n_{0y}) \\ 2\end{pmatrix}}{\begin{pmatrix}n \\ 2\end{pmatrix}} \nonumber\\\ \leq & 1-\frac{\max(n_{0x},n_{1x},n_{2x},n_{0y})(\max(n_{0x},n_{1x},n_{2x},n_{0y})-1)}{n(n-1)} \nonumber\\\ \approx & 1-(\frac{\max(n_{0x},n_{1x},n_{2x},n_{0y})}{n})^{2} \nonumber\\\ =& 1-\{\max(\pi_{0x},\pi_{1x},\pi_{2x},\pi_{0y})\}^{2} \nonumber\\\ =& 1-\{\max(\pi_{0x},(1-\pi_{0x}-\pi_{2x}),\pi_{2x},\pi_{0y})\}^{2} \end{align} \end{aligned}$$

It is easy to get the approximate bound for truncated/ternary case by switching x and y.