The effect of thresholding on correlations

Michael Barkasi

July 8, 2026 (v1.0)

Theorem: The effect of thresholding on correlations between multivariate points

Given a finite set of m-dimensional real-valued vectors S\subset\mathbb{R}^m, suppose:

• S1. There are n vectors in S and these vectors are indexed such that S can be treated as an n\times m matrix with the vectors s\in S as rows.
• S2. The columns s_j=\langle s_{ij}\rangle_{1\leq i\leq n} of S follow Gaussian distributions, i.e., s_{ij}\sim\mathcal{N}(\mu_j,\sigma_j).
• S3. The columns of S have pairwise correlations given by R, an m\times m symmetric matrix with ones along the diagonal, i.e., R_{j_{1}j_{2}}=\mathrm{corr}(s_{j_1},s_{j_2}).

And define:

• D1. The real-valued vectors \mu_S,\sigma_S\in\mathbb{R}^m hold, respectively, the column means \mu_j and column standard deviations \sigma_j of S, i.e., \mu_S=\langle\mu_j\rangle_{1\leq j\leq m} and \sigma_S=\langle\sigma_j\rangle_{1\leq j\leq m}.
• D2. The diagonal matrix D is the m\times m matrix with all zeros, except the diagonal, which contains \sigma_S, i.e., D_{jj} = \sigma_j for all 1\leq j\leq m.
• D3. The matrix S^{*} is the result of thresholding S into binary values of 0 and 1 according to the rule s^{*}_{ij}=1 if s_{ij} > 0 and s^{*}_{ij}=0 otherwise.
• D4. For a square matrix M and integer vector v shorter than the sides of M, M_{v} is the square matrix formed from the rows and columns indexed by v.

It follows that:

• T1. The covariance matrix for the columns of S is \Sigma = DRD.
• T2. The column means \mu^{*}_j of S^{*} are given by the upper-tail Gaussian cumulative distribution function \Phi^{+}: \mu^{*}_j = \Phi^{+}(0,\mu_j,\sigma_j) = 1 - \Phi(0,\mu_j,\sigma_j)
• T3. The columns of S^{*} have pairwise correlations R^{*}_{j_{1}j_{2}} given by \Phi^{+} and the upper-tail bivariate Gaussian cumulative distribution function \Phi^{+}_2 as follows: \begin{equation*} R^{*}_{j_{1}j_{2}} = \frac{ \Phi^{+}_{2}(\langle 0,0\rangle,\langle\mu_{j_1},\mu_{j_2}\rangle,\Sigma_{\langle j_{1}, j_{2}\rangle}) - \mu^{*}_{j_1}\mu^{*}_{j_2} }{ \sqrt{\mu^{*}_{j_1}-{\mu^{*}_{j_1}}^2} \sqrt{\mu^{*}_{j_2}-{\mu^{*}_{j_2}}^2} } \end{equation*}

Proof (T1)

Let \Sigma^\prime=DR and note that by the definition of matrix multiplication: \begin{equation*} \Sigma^\prime_{ij} = \sum_{k=1}^{m}D_{ik}R_{kj} \end{equation*} By (D2), D_{ik} = 0 except when i=k, in which case D_{ik}=\sigma_i. Hence, \Sigma^{\prime}_{ij}=\sigma_{i}R_{ij}. Note that \Sigma=\Sigma^{\prime}D, and thus: \begin{equation*} \Sigma_{ij} = \sum_{k=1}^{m}\Sigma^{\prime}_{ik}D_{kj} \end{equation*} As before, D_{kj}=0 except when k=j, in which case D_{kj}=\sigma_{j}. Hence, \Sigma_{ij}=\Sigma^{\prime}_{ij}\sigma_{j}=\sigma_{i}R_{ij}\sigma_{j}. From the identity \mathrm{corr}(X,Y)\sigma_X\sigma_Y=\mathrm{cov}(X,Y) and (S3), follows that \Sigma_{ij}=\mathrm{cov}(s_i,s_j). Thus, \Sigma is the covariance matrix for the columns of S. \blacksquare

Proof (T2)

By the definition of \Phi^{+}, for 1\leq i\leq n: \begin{equation*} \Phi^{+}(0,\mu_j,\sigma_j)=\mathrm{P}(s_{ij}> 0\;|\;s_{ij}\sim\mathcal{N}(\mu_j,\sigma_j)) \end{equation*} Hence by (D3), \Phi^{+}(0,\mu_j,\sigma_j) is the number of values s^{*}_{ij} which equal one, over n. This value is, by definition (S2, D1), the mean \mu_j^{*}. \blacksquare

Proof (T3)

Consider again the identity: \begin{equation*} \mathrm{corr}(X,Y)=\frac{\mathrm{cov}(X,Y)}{\sigma_{X}\sigma_{Y}} \end{equation*} This identity implies that: \begin{equation*} R^{*}_{j_{1}j_{2}} = \frac{ \mathrm{cov}(s^{*}_{j_{1}},s^{*}_{j_{2}}) }{ \sigma^{*}_{j_{1}}\sigma^{*}_{j_{2}} } \end{equation*} Given that s^{*}_j\in\{0,1\}^{m} (D3), it follows that s^{*}_{ij}={s^{*}_{ij}}^2 for all 1\leq i\leq n, so \mu({s^{*}_j}^2)=\mu(s^{*}_j). Hence, given the identity: \sigma(X)=\sqrt{\mu(X^2)-{\mu(X)}^2} it follows that: \begin{equation*} \sigma^{*}_{j} = \sqrt{\mu^{*}_{j}-{\mu^{*}_{j}}^2} \end{equation*} Therefore: \begin{equation*} R^{*}_{j_{1}j_{2}} = \frac{ \mathrm{cov}(s^{*}_{j_{1}},s^{*}_{j_{2}}) }{ \sqrt{\mu^{*}_{j_1}-{\mu^{*}_{j_1}}^2} \sqrt{\mu^{*}_{j_2}-{\mu^{*}_{j_2}}^2} } \end{equation*} Given (T2) and the identity \mathrm{cov}(X,Y)=\mu(XY)-\mu(X)\mu(Y), it follows that \mathrm{cov}(s^{*}_{j_{1}},s^{*}_{j_{2}}) = \mu(s^{*}_{j_{1}}s^{*}_{j_{2}}) - \mu^{*}_{j_1}\mu^{*}_{j_2}. Similar to proof of (T2), by the definition of \Phi^{+}, for 1\leq i\leq n:
\begin{gather*} \Phi^{+}_{2}(\langle 0,0\rangle,\langle\mu_{j_1},\mu_{j_2}\rangle,\Sigma_{\langle j_{1}, j_{2}\rangle}) = \\ \mathrm{P}(s_{ij_{1}}> 0, s_{ij_{2}} > 0 \;|\; s_{ij_{1}}, s_{ij_{2}} \sim \mathrm{MVN}(\langle\mu_{j_1},\mu_{j_2}\rangle,\Sigma_{\langle j_{1}, j_{2}\rangle})) \end{gather*} Hence by (D3), \Phi^{+}_{2}(\langle 0,0\rangle,\langle\mu_{j_1},\mu_{j_2}\rangle,\Sigma_{\langle j_{1}, j_{2}\rangle}) gives the number of rows i such that s_{ij_{1}}=s_{ij_{2}}=1, over n. This value is equal to \mu(s^{*}_{j_{1}}s^{*}_{j_{2}}). \blacksquare