Fisher Z Intervals for Partial Correlation

Why Transform a Partial Correlation?

A sample partial correlation estimates the association between two variables after linear adjustment for a set of controls. Like an ordinary correlation, it is bounded between $-1$ and $1$. Its sampling distribution can be notably asymmetric, especially when the population partial correlation is not near zero. Fisher’s transformation maps the correlation scale to the full real line, where a normal approximation is more useful for interval construction.

The number of controls matters because estimating their linear effects uses information from the sample. With more controls and the same sample size, the partial correlation is generally less precise. The interval below accounts for this through its standard error.

Definition: Let $Y_1,Y_2$ be two variables and let $Z=(Z_1,\ldots,Z_q)$ be a vector of $q$ control variables. Their population partial correlation, denoted $\rho_{12\cdot Z}$, is the correlation between the linear-regression residuals of $Y_1$ and $Y_2$ after each is regressed on the same controls $Z$, including an intercept. For a sample partial correlation $r_{12\cdot Z}$, Fisher’s transformation is $$ g(r_{12\cdot Z})=\frac{1}{2}\log\left(\frac{1+r_{12\cdot Z}}{1-r_{12\cdot Z}}\right)=\operatorname{atanh}(r_{12\cdot Z}), $$ with inverse $g^{-1}(w)=\tanh(w)$.

Here $q$ counts the controlled variables, not the two variables whose partial correlation is being estimated. The ordinary correlation is the special case $q=0$. In a sample of size $n$, residualizing on the controls and an intercept leaves $n-q-1$ residual degrees of freedom, provided the control design has full column rank.

The Fisher Z Interval

Assume that the observations are independent and identically distributed from a nonsingular multivariate normal distribution for $(Y_1,Y_2,Z)$. Suppose $q$ is fixed as $n$ increases, and the control design has full rank. For $n>q+3$, a commonly used approximate two-sided $100(1-\alpha)\%$ interval is

\left( \tanh\left[g(r_{12\cdot Z})-\frac{z_{1-\alpha/2}}{\sqrt{n-q-3}}\right], \quad \tanh\left[g(r_{12\cdot Z})+\frac{z_{1-\alpha/2}}{\sqrt{n-q-3}}\right] \right),

where $z_{1-\alpha/2}$ is the $1-\alpha/2$ quantile of the standard normal distribution. For a 95% interval, $z_{0.975}\approx1.96$. The standard error on the transformed scale is therefore $1/\sqrt{n-q-3}$.

The interval is formed in two stages: first make a symmetric interval around the transformed sample partial correlation, then apply the increasing inverse transformation to both endpoints. The resulting interval is generally asymmetric around the original sample partial correlation, and its endpoints remain between $-1$ and $1$.

Worked Example: A Positive Partial Correlation

In a sample of $n=120$, suppose the partial correlation between an outcome and a biomarker, controlling for $q=2$ covariates, is $r_{12\cdot Z}=0.46$. Construct an approximate 95% confidence interval.

Step 1: Calculate the transformed estimate. $$ g(0.46)=\frac{1}{2}\log\left(\frac{1.46}{0.54}\right) =\frac{1}{2}\log\left(\frac{73}{27}\right) \approx 0.4973113. $$

Step 2: Calculate the transformed-scale standard error and margin. $$ \operatorname{SE}=\frac{1}{\sqrt{120-2-3}}=\frac{1}{\sqrt{115}}\approx0.0932505, \qquad 1.96(0.0932505)\approx0.182771. $$

Step 3: Find the transformed endpoints. $$ 0.4973113-0.182771\approx0.314540, \qquad 0.4973113+0.182771\approx0.680082. $$

Step 4: Transform back. $$ \tanh(0.314540)\approx0.30456, \qquad \tanh(0.680082)\approx0.59157. $$ Thus the approximate 95% confidence interval is $(0.305,\,0.592)$, to three decimal places.

Why the Approximation Works

The central justification is an asymptotic normal approximation for the transformed partial correlation. The degrees of freedom remaining after removing the intercept and $q$ controls are $\nu=n-q-1$. Under multivariate normal sampling, the residual cross-product matrix for $Y_1,Y_2$, conditional on the controls, has a Wishart distribution with $\nu$ degrees of freedom. Consequently, the sample partial correlation has the same form as a correlation computed from a two-variable Wishart matrix with those degrees of freedom.

Theorem: Under the independent multivariate normal sampling assumptions above, with fixed $q$ and $|\rho_{12\cdot Z}|<1$, as $n\to\infty$, $$ \sqrt{\nu}\left[g(r_{12\cdot Z})-g(\rho_{12\cdot Z})\right]\xrightarrow{d}N(0,1), \qquad \nu=n-q-1. $$

To see the variance, standardize the conditional covariance matrix of the two residual variables to have diagonal entries 1 and off-diagonal entry $\rho=\rho_{12\cdot Z}$. Let $S$ be their residual cross-product matrix divided by $\nu$, with entries $S_{11},S_{22},S_{12}$. Wishart moment formulas give, to first order, the covariance structure of $S$ as

\begin{aligned} \operatorname{Var}(S_{11})&=\frac{2}{\nu},& \operatorname{Var}(S_{22})&=\frac{2}{\nu},& \operatorname{Var}(S_{12})&=\frac{1+\rho^2}{\nu},\\ \operatorname{Cov}(S_{11},S_{22})&=\frac{2\rho^2}{\nu},& \operatorname{Cov}(S_{11},S_{12})&=\frac{2\rho}{\nu},& \operatorname{Cov}(S_{22},S_{12})&=\frac{2\rho}{\nu}. \end{aligned}

The sample partial correlation is $h(S)=S_{12}/\sqrt{S_{11}S_{22}}$. At the population matrix, the gradient with respect to $(S_{11},S_{22},S_{12})$ is $(-\rho/2,-\rho/2,1)$. Applying the multivariate delta method with the displayed covariances yields

\nu\operatorname{Var}(r_{12\cdot Z})\longrightarrow (1-\rho^2)^2.

Since $g'(r)=1/(1-r^2)$, a second application of the delta method gives limiting variance 1 for $\sqrt{\nu}\{g(r_{12\cdot Z})-g(\rho)\}$. This proves the stated asymptotic normality.

Derivation of the interval: The theorem implies that $g(r_{12\cdot Z})$ is approximately normal with mean $g(\rho_{12\cdot Z})$ and variance $1/\nu$ for large samples. The customary Fisher finite-sample refinement uses $1/(\nu-2)=1/(n-q-3)$ in the standard error. Thus the transformed estimate plus or minus $z_{1-\alpha/2}/\sqrt{n-q-3}$ gives an approximate interval for $g(\rho_{12\cdot Z})$. Because $g$ is strictly increasing, applying $\tanh$ to each endpoint gives the stated interval for $\rho_{12\cdot Z}$. Its coverage is approximate, not exact. $\square$

The asymptotic argument establishes the leading large-sample variance. The $n-q-3$ denominator is a conventional finite-sample refinement, analogous to the familiar Fisher interval for an ordinary correlation. It is not a claim that the transformed statistic is exactly normal in finite samples.

Worked Example: A Negative Partial Correlation

Suppose $n=66$, there are $q=4$ controls, and the observed partial correlation is $-0.62$. Calculate the approximate 95% interval.

The transformed estimate is $$ g(-0.62)=\frac{1}{2}\log\left(\frac{0.38}{1.62}\right) =-\frac{1}{2}\log\left(\frac{81}{19}\right) \approx-0.7250051. $$ The standard error is $$ \frac{1}{\sqrt{66-4-3}}=\frac{1}{\sqrt{59}}\approx0.130189, $$ so the margin is $1.96(0.130189)\approx0.255170$. The transformed endpoints are therefore $$ -0.7250051-0.255170\approx-0.9802, \qquad -0.7250051+0.255170\approx-0.4698. $$

Back-transforming gives $\tanh(-0.9802)\approx-0.753$ and $\tanh(-0.4698)\approx-0.438$. The approximate 95% interval is $(-0.753,\,-0.438)$. It is entirely negative, which is consistent with a negative population partial correlation at this confidence level.

How the Number of Controls Changes Precision

For fixed $n$, increasing $q$ reduces $n-q-3$, increases the transformed-scale standard error, and typically widens the interval. This adjustment is one reason to report both the sample size and the number of controlled variables alongside a partial-correlation interval.

Worked Example: Comparing Two Numbers of Controls

Consider two analyses with the same sample size $n=100$ and the same observed partial correlation, $r=0.30$, but different numbers of controls. Compare their approximate 95% intervals for $q=2$ and $q=12$.

In both analyses, $$ g(0.30)=\frac{1}{2}\log\left(\frac{1.30}{0.70}\right) =\frac{1}{2}\log\left(\frac{13}{7}\right)\approx0.309520. $$

With $q=2$, the standard error is $1/\sqrt{95}\approx0.102598$, giving a margin of $1.96(0.102598)\approx0.201092$. The transformed endpoints are $0.108428$ and $0.510612$; after applying $\tanh$, the interval is approximately $(0.108,\,0.470)$.

With $q=12$, the standard error is $1/\sqrt{85}\approx0.108465$, giving a margin of $1.96(0.108465)\approx0.212592$. The transformed endpoints are $0.096928$ and $0.522112$; after applying $\tanh$, the interval is approximately $(0.097,\,0.479)$.

The second interval is wider because the same sample size is being used to estimate the partial association after controlling for more variables. This comparison illustrates the formula’s precision adjustment; in an actual analysis, the two observed partial correlations need not be equal.

Conditions, Interpretation, and Common Pitfalls

The interval is most defensible when the joint distribution is reasonably modeled by multivariate normal sampling, observations are independent, and the number of controls is small relative to the sample size. The controls must be linearly estimable: if the control design is rank-deficient, the usual residualization and degrees-of-freedom calculation do not apply without changing the model. The formula also requires $n-q-3>0$, and in practice a positive denominator alone does not ensure that a normal approximation is accurate.

Interpret the result as a range of population partial correlations compatible with the sample under the approximation and model assumptions. It describes linear association after adjustment for the specified controls; it is not automatically a causal effect. Changing the controls changes the estimand as well as the precision.

Using the wrong control count. The denominator is $n-q-3$, where $q$ is the number of controlled variables. Do not count $Y_1$ and $Y_2$ as controls.
Reporting transformed endpoints as correlations. The symmetric endpoints on the Fisher scale are not the final interval. Apply $\tanh$ to both before reporting.
Treating the formula as exact. The normal approximation and finite-sample variance refinement do not produce an exact confidence interval.
Assuming more controls only affect precision. Controls can also change the partial correlation being estimated; different control sets define different adjusted associations.

Key takeaway: Transform the sample partial correlation with $g(r)=\operatorname{atanh}(r)$, use the adjusted standard error $1/\sqrt{n-q-3}$, form a normal interval on the transformed scale, and back-transform both endpoints with $\tanh$. State the controls and treat the resulting coverage as approximate.

Check Your Understanding

Use the Fisher transformation and the stated assumptions to answer these questions.

In a sample of size $n=90$, a partial correlation controls for $q=5$ variables. What standard error does the Fisher interval use?
Why is the interval generally not symmetric around the observed partial correlation on the original scale?
If the number of controls increases while $n$ stays fixed, what happens to the transformed-scale standard error?
What is the role of the inverse transformation when constructing the reported interval?
Why should a Fisher Z interval for a partial correlation be described as approximate rather than exact?