Signed, Unsigned, And Signed-Hybrid Networks

Why the Sign of a Correlation Matters

A correlation-based co-expression network starts with pairwise gene-expression correlations and converts them into edge weights. The transformation determines what the network regards as a strong relationship. In particular, a negative correlation can represent a strong inverse association, a weak connection, or no connection, depending on the chosen network definition. Thus, “strong correlation” alone does not determine edge weight.

The earlier material on soft thresholding established how a tuning power controls the conversion from correlation magnitude to weight. Here the focus is the sign convention: unsigned networks treat correlations of equal magnitude alike; signed networks distinguish positive from negative correlations smoothly; and signed-hybrid networks retain positive correlations while setting negative correlations to zero.

Definition: Let $r_{ij}$ be the estimated correlation between the expression profiles of genes $i$ and $j$, with $-1\leq r_{ij}\leq 1$. For a tuning power $\beta>0$, an adjacency weight $a_{ij}$ is a nonnegative number assigned to the pair. For distinct genes, the common transformations are $$ a_{ij}^{\mathrm{unsigned}}=|r_{ij}|^\beta,\qquad a_{ij}^{\mathrm{signed}}=\left(\frac{1+r_{ij}}{2}\right)^\beta,\qquad a_{ij}^{\mathrm{hybrid}}= \begin{cases} r_{ij}^\beta,&r_{ij}>0,\\ 0,&r_{ij}\leq 0. \end{cases} $$ The diagonal is conventionally set to $a_{ii}=0$, since a gene is not treated as its own network edge.

All three transformations produce weights between zero and one. They differ in how they use correlation direction. The unsigned transformation uses only magnitude: $r_{ij}=0.8$ and $r_{ij}=-0.8$ receive the same weight. The signed transformation first maps correlation from $[-1,1]$ to $[0,1]$, so positive correlations receive larger weights than negative ones. The signed-hybrid transformation keeps the positive correlation magnitude but gives every nonpositive correlation weight zero.

How the Transformations Behave

Correlation pattern	Unsigned	Signed	Signed-hybrid
Strong positive	High weight	High weight	High weight
Strong negative	High weight, as for an equally strong positive correlation	Low weight	Zero weight
Zero	Zero weight	$(1/2)^\beta$	Zero weight
Small positive	Small weight	Above the signed weight for an equally large negative correlation	Small weight

The signed transformation’s nonzero weight at $r_{ij}=0$ is important. A zero correlation maps to $1/2$ before the power is applied, so it receives weight $2^{-\beta}$, not zero. This weight becomes small when $\beta$ is large, but it is positive for every finite $\beta$. By contrast, both unsigned and signed-hybrid transformations assign zero weight to a zero correlation.

Worked Example: Transforming Positive and Negative Correlations

Suppose a hypothetical expression analysis estimates correlations of $0.75$, $-0.75$, and $0$ for three gene pairs. Use $\beta=4$. The same power is applied to each pair in each network so that the transformations can be compared directly.

For $r_{ij}=0.75$, the unsigned weight is $|0.75|^4=0.3164$. The signed weight is $((1+0.75)/2)^4=0.875^4=0.5862$. The signed-hybrid weight is $0.75^4=0.3164$.

For $r_{ij}=-0.75$, the unsigned weight remains $|-0.75|^4=0.3164$. The signed weight is $((1-0.75)/2)^4=0.125^4=0.0002441$. The signed-hybrid weight is $0$, because the correlation is not positive.

For $r_{ij}=0$, the unsigned and signed-hybrid weights are both $0$. The signed weight is $((1+0)/2)^4=0.5^4=0.0625$. These values show that the same estimated correlation can imply very different edge strengths under different network definitions.

Properties and a Short Justification

For $\beta>0$, each transformation is nonnegative. The unsigned weight is symmetric in the sign of correlation: replacing $r_{ij}$ with $-r_{ij}$ leaves $|r_{ij}|^\beta$ unchanged. The signed weight increases as $r_{ij}$ increases, and the signed-hybrid weight increases over positive correlations but remains zero for nonpositive correlations. These properties follow directly from the formulas.

For the signed transformation, the map $g(r)=(1+r)/2$ is increasing on $[-1,1]$, taking values from zero to one. Raising a value in $[0,1]$ to a positive power preserves its order. Therefore, if $r_1<r_2$, then $g(r_1)^\beta\leq g(r_2)^\beta$. In particular, a more positive correlation cannot have a lower signed weight than a less positive one. For the unsigned transformation, the weight depends on $|r|$, so this ordering applies to magnitude, not to direction. For the signed-hybrid transformation, the weight is zero on $[-1,0]$, then increases with $r$ on $(0,1]$.

The power changes how concentrated weights are near zero and one. When $0<|r|<1$, increasing $\beta$ decreases $|r|^\beta$. For the signed transformation, the same effect applies to the mapped quantity $(1+r)/2$. The signed and signed-hybrid transformations therefore do not apply the power to the same quantity: the signed transformation powers a shifted and rescaled correlation, while the signed-hybrid transformation powers only positive correlations themselves.

Worked Example: Comparing Weights at a Fixed Threshold

Suppose a hypothetical analysis gives correlations $0.6$, $-0.6$, and $0$ for three gene pairs. Set $\beta=2$, and call a pair retained for display if its weight is at least $0.10$. This display cutoff is an extra choice, separate from the transformation.

For $r_{ij}=0.6$, the unsigned weight is $0.6^2=0.36$, the signed weight is $((1+0.6)/2)^2=0.8^2=0.64$, and the signed-hybrid weight is $0.6^2=0.36$. All three exceed $0.10$.

For $r_{ij}=-0.6$, the weights are $0.36$, $((1-0.6)/2)^2=0.2^2=0.04$, and $0$, respectively. The unsigned network retains the pair at this cutoff; the signed and signed-hybrid networks do not.

For $r_{ij}=0$, the weights are $0$, $(1/2)^2=0.25$, and $0$. Only the signed weight exceeds $0.10$. This does not mean the estimated correlation is positive: it reflects the signed transformation’s midpoint mapping and the chosen power. A weight cutoff should therefore be interpreted in light of the transformation, not as a universal test of whether a gene pair is associated.

Worked Example: Constructing Three Adjacency Matrices

Consider a hypothetical, valid correlation matrix for three genes. The correlations are $r_{AB}=0.6$, $r_{AC}=-0.2$, and $r_{BC}=0.1$, with a diagonal of ones. The determinant is $1+2(0.6)(-0.2)(0.1)-0.6^2-(-0.2)^2-0.1^2=0.566$, which is positive; the matrix is positive definite. Use $\beta=2$ and set the adjacency diagonal to zero.

The unsigned off-diagonal weights are $|0.6|^2=0.36$, $|-0.2|^2=0.04$, and $|0.1|^2=0.01$. The signed weights are $0.8^2=0.64$, $0.4^2=0.16$, and $0.55^2=0.3025$. The signed-hybrid weights are $0.6^2=0.36$, $0$, and $0.1^2=0.01$.

Using the gene order $A,B,C$, the resulting matrices are

A^{\mathrm{unsigned}}= \begin{pmatrix} 0&0.36&0.04\\ 0.36&0&0.01\\ 0.04&0.01&0 \end{pmatrix},\quad A^{\mathrm{signed}}= \begin{pmatrix} 0&0.64&0.16\\ 0.64&0&0.3025\\ 0.16&0.3025&0 \end{pmatrix},\quad A^{\mathrm{hybrid}}= \begin{pmatrix} 0&0.36&0\\ 0.36&0&0.01\\ 0&0.01&0 \end{pmatrix}.

Each matrix is symmetric because the input correlations are symmetric, and all off-diagonal weights are nonnegative. The matrices differ in what they encode: the unsigned matrix gives the inverse pair $A,C$ a positive weight; the signed matrix gives it a smaller but nonzero weight; and the signed-hybrid matrix removes it.

Choosing a Convention and Checking It

These transformations encode different scientific definitions of an edge, rather than three interchangeable calculations. An unsigned network treats coordinated increases and opposing expression changes as equally strong relationships when their correlation magnitudes match. A signed network orders pairs by direction as well as strength, while retaining a continuum of weights for negative correlations. A signed-hybrid network makes an edge represent positive co-expression only; a negative correlation is treated as no edge in that network.

Before interpreting a network, check the correlation input and the transformation together. Pairwise correlations should have been estimated from a clearly defined set of samples and expression values, using the chosen correlation measure consistently. The transformation does not repair outliers, dependence among samples, missing-data problems, or unwanted technical and biological confounding. A large correlation caused by a batch effect remains a large correlation after transformation.

Check that the correlation matrix is symmetric, that its entries lie in $[-1,1]$, and that the adjacency diagonal is set to zero.
Record the correlation method, the chosen $\beta$, and the transformation. A reported edge weight cannot be interpreted without them.
Compare the distribution of input correlations with the distribution of transformed weights, including the positive and negative portions separately.
For signed networks, check how many weights arise from correlations near zero. These can be nonzero even when the correlations themselves are near zero.
Repeat key summaries under reasonable alternatives for the transformation or power. Large changes indicate that conclusions depend on this modeling choice.

A useful comparison is with a hard threshold, which assigns an edge according to whether a correlation or its magnitude passes a cutoff. Hard thresholds create a binary decision at that cutoff; the three transformations here create graded weights, although a later analysis might impose an additional cutoff. The choice among signed, unsigned, and signed-hybrid weights is different from the choice of hard versus soft thresholding: one specifies how sign is handled, and the other concerns whether edge strength is binary or graded.

Common Mistakes and Limitations

Assuming “signed” means negative edge weights. In these definitions, all adjacency weights are nonnegative. “Signed” refers to how the correlation’s sign affects the weight.
Equating an unsigned edge with positive co-expression. An unsigned weight can be large because of a strong negative correlation. Its direction is not represented in the weight.
Assuming a zero correlation always gives zero weight. That holds for unsigned and signed-hybrid transformations, but not for the signed transformation at finite $\beta$.
Comparing weights across transformations as if they had the same scale and meaning. A weight of $0.3$ under one transformation need not represent the same correlation pattern as $0.3$ under another. State the transformation and $\beta$.
Treating an edge as evidence of causation. Correlation-based weights describe pairwise association in the analyzed data. They do not establish that one gene regulates another.

The transformations also do not establish that an observed association is statistically precise or biologically meaningful. They operate on estimated correlations; uncertainty in those estimates, sample composition, and confounding can all affect the resulting weights. Choosing a convention is therefore part of defining the network estimand, not a substitute for assessing the data and research question.

Key takeaway: Unsigned weights depend on correlation magnitude, signed weights depend on a shifted correlation, and signed-hybrid weights retain positive correlations only. State the convention and $\beta$ whenever reporting or interpreting adjacency weights.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For $r_{ij}=-0.5$ and $\beta=2$, calculate the unsigned, signed, and signed-hybrid weights.
Two pairs have correlations $0.7$ and $-0.7$. Which transformations give them equal weights, and why?
Why can a signed network assign positive weight to a pair with correlation zero? Does that weight imply a positive estimated correlation?
A researcher wants negative correlations to count as no edge, while positive correlations retain graded weights. Which transformation matches this goal? State one limitation of interpreting the resulting edges.

Solutions. (1) Unsigned: $|-0.5|^2=0.25$. Signed: $((1-0.5)/2)^2=0.25^2=0.0625$. Signed-hybrid: $0$, since the correlation is nonpositive. (2) The unsigned weights are equal because $|0.7|^ \beta=|-0.7|^\beta$ for any $\beta>0$. The signed and signed-hybrid weights generally differ because they treat the signs differently. (3) The signed map sends zero to $1/2$, so its weight is $(1/2)^\beta>0$ for finite $\beta>0$. No: the estimated correlation is still zero; the positive weight is a consequence of the transformation. (4) The signed-hybrid transformation matches that goal. Its edges still represent association, not causation, and may be affected by confounding or uncertainty in the estimated correlations.