Asymptotic Distributions of Empirical Interaction Information

Abstract

Interaction Information is one of the most promising interaction strength measures with many desirable properties. However, its use for interaction detection was hindered by the fact that apart from the simple case of overall independence, asymptotic distribution of its estimate has not been known. In the paper we provide asymptotic distributions of its empirical versions which are needed for formal testing of interactions. We prove that for three-dimensional nominal vector normalized empirical interaction information converges to the normal law unless the distribution coincides with its Kirkwood approximation. In the opposite case the convergence is to the distribution of weighted centred chi square random variables. This case is of special importance as it roughly corresponds to interaction information being zero and the asymptotic distribution can be used for construction of formal tests for interaction detection. The result generalizes result in Han (Inf Control 46(1):26–45 1980) for the case when all coordinate random variables are independent. The derivation relies on studying structure of covariance matrix of asymptotic distribution and its eigenvalues. For the case of 3 × 3 × 2 contingency table corresponding to study of two interacting Single Nucleotide Polymorphisms (SNPs) for prediction of binary outcome, we provide complete description of the asymptotic law and construct approximate critical regions for testing of interactions when two SNPs are possibly dependent. We show in numerical experiments that the test based on the derived asymptotic distribution is easy to implement and yields actual significance levels consistently closer to the nominal ones than the test based on chi square reference distribution.

Appendix

Below we prove Lemma 7.

Proof

From the previous lemma we know that λ₁(D) = … = λ₄(D) = 1, λ₉(D) = 0.

By lengthy calculations we obtain a sequence of the following equalities:

$$\mathbf{D}^{2}=\mathbf{D}+\mathbf{Z},$$

$$\mathbf{D}^{3} = \mathbf{D} + \mathbf{Z} + \mathbf{Z}\mathbf{D},$$

$$\mathbf{D}^{4} = \mathbf{D} + \mathbf{Z} + 2\mathbf{Z}\mathbf{D} + \mathbf{Z}^{2},$$

where $\mathbf {Z}=(Z_{ij}^{i^{\prime }j^{\prime }})$ and:

$$Z_{ij}^{i^{\prime}j^{\prime}} = p_{i^{\prime}j^{\prime}}\Big(\frac{p_{i^{\prime}j}}{p_{i^{\prime}}p_{j}} + \frac{p_{ij^{\prime}}}{p_{i}p_{j^{\prime}}} -2 \Big),$$

$$\text{tr}(\mathbf{Z}) = 2\sum\limits_{i,j} \frac{p_{ij}^{2}}{p_{i}p_{j}} - 2 = 2H_{1},$$

$$\text{tr}(\mathbf{Z}\mathbf{D}) = - \text{tr}(\mathbf{Z}),$$

$$\text{tr}(\mathbf{Z}^{2}) = -4 + 2 \sum\limits_{i,j,i^{\prime},j^{\prime}} \frac{p_{ij}p_{i^{\prime}j}p_{i^{\prime}j^{\prime}}p_{ij^{\prime}}}{p_{i}p_{j}p_{i^{\prime}}p_{j^{\prime}}} + 2\sum\limits_{i,j} \frac{p_{ij}^{2}}{p_{i}p_{j}} = 2H_{1} + 2H_{2} ,$$

$$\text{tr}(\mathbf{D}) = (n_{X_{1}}-1)(n_{X_{2}}-1) = 4,$$

$$\text{tr}(\mathbf{D}^{2})=\text{tr}(\mathbf{D})+\text{tr}(\mathbf{Z}) = (n_{X_{1}}-1)(n_{X_{2}}-1) + 2\sum\limits_{i,j} \frac{p_{ij}^{2}}{p_{i}p_{j}} - 2 = 2 + 2\sum\limits_{i,j} \frac{p_{ij}^{2}}{p_{i}p_{j}} = 4 + 2H_{1},$$

$$\text{tr}(\mathbf{D}^{3}) = \text{tr}(\mathbf{D}) = (n_{X_{1}}-1)(n_{X_{2}}-1) = 4,$$

$$ \begin{array}{@{}rcl@{}} \text{tr}(\mathbf{D}^{4}) = (n_{X_{1}}-1)(n_{X_{2}}-1) - 2 + 2\sum\limits_{i,j,i^{\prime},j^{\prime}} \frac{p_{ij}p_{i^{\prime}j}p_{i^{\prime}j^{\prime}}p_{ij^{\prime}}}{p_{i}p_{j}p_{i^{\prime}}p_{j^{\prime}}} = 2 + 2\sum\limits_{i,j,i^{\prime},j^{\prime}} \frac{p_{ij}p_{i^{\prime}j}p_{i^{\prime}j^{\prime}}p_{ij^{\prime}}}{p_{i}p_{j}p_{i^{\prime}}p_{j^{\prime}}} \\ =4+ 2H_{2}. \end{array} $$

Note that e.g. $\text {tr}(\mathbf {D}) = (n_{X_{1}}-1)(n_{X_{2}}-1)$ follows from Lemmas 3-5 as it follows from Lemma 4 that tr(M) = −tr(C) ⋅ tr(D) and $tr(\mathbf {M})=(n_{X_{1}}-1)(n_{X_{2}}-1)(n_{Y}-1)$ and tr(D) = 1 − n_Y. As trace of square matrix is sum of its eigenvalues from the above equalities it follows that:

$$\left\{\begin{array}{ccccc} \lambda_{5}(\mathbf{D}) + \lambda_{6}(\mathbf{D}) + \lambda_{7}(\mathbf{D}) + \lambda_{8}(\mathbf{D}) = 0, \\ {\lambda_{5}^{2}}(\mathbf{D}) + {\lambda_{6}^{2}}(\mathbf{D}) + {\lambda_{7}^{2}}(\mathbf{D}) + {\lambda_{8}^{2}}(\mathbf{D}) = 2H_{1}, \\ {\lambda_{5}^{3}}(\mathbf{D}) + {\lambda_{6}^{3}}(\mathbf{D}) + {\lambda_{7}^{3}}(\mathbf{D}) + {\lambda_{8}^{3}}(\mathbf{D}) = 0, \\ {\lambda_{5}^{4}}(\mathbf{D}) + {\lambda_{6}^{4}}(\mathbf{D}) + {\lambda_{7}^{4}}(\mathbf{D}) + {\lambda_{8}^{4}}(\mathbf{D}) = 2H_{2}. \end{array}\right.$$

From the Newton-Girard identities we obtain:

$$\sum\limits_{5\le i_{1}<i_{2} \le 8} \lambda_{i_{1}}(\mathbf{D})\lambda_{i_{2}}(\mathbf{D}) = -H_{1},$$

$$\sum\limits_{5\le i_{1}<i_{2}<i_{3} \le 8} \lambda_{i_{1}}(\mathbf{D})\lambda_{i_{2}}(\mathbf{D})\lambda_{i_{3}}(\mathbf{D}) = 0,$$

$$\lambda_{5}(\mathbf{D})\lambda_{6}(\mathbf{D})\lambda_{7}(\mathbf{D})\lambda_{8}(\mathbf{D}) = \frac{1}{2}{H_{1}^{2}}-\frac{1}{2}H_{2}.$$

This means that λ₅(D),λ₆(D),λ₇(D),λ₈(D) are the roots of the polynomial:

$$ Q(x) = x^{4} -H_{1}x^{2}+ \frac{1}{2}{H_{1}^{2}}-\frac{1}{2}H_{2}. $$

(21)

Note that in view of its definition H₁ ≥ 0 and if X₁ and X₂ are independent then H₁ = H₂ = Δ = 0. From this the lemma follows. □