Fisher’s Z Transformation for a Sample Correlation

Why Transform a Correlation?

A sample correlation is always between $-1$ and $1$, but its sampling variability is not constant across that range. When the population correlation is near zero, its standard error is comparatively large; when the population correlation is near either endpoint, the standard error is smaller. This changing variability complicates normal approximations and interval construction on the original correlation scale.

For independent observations from a bivariate normal distribution, the first-order asymptotic variance has a particularly simple form. Fisher’s transformation, $ \operatorname{arctanh}(r) $, adjusts for that changing variance: its asymptotic variance is $1/n$, regardless of the population correlation. This makes it possible to construct an approximate interval on a scale where the standard error is simple, then return the endpoints to the correlation scale.

Sample Correlation and Its Asymptotic Variance

Let $(X_1,Y_1),\ldots,(X_n,Y_n)$ be independent, identically distributed pairs from a nondegenerate bivariate normal distribution. Write the population means as $\mu_X,\mu_Y$, the population standard deviations as $\sigma_X,\sigma_Y>0$, and the population correlation as $\rho$, where $-1<\rho<1$. Define the sample correlation by

r=\frac{S_{XY}}{\sqrt{S_{XX}S_{YY}}}, \qquad S_{XY}=\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar X)(Y_i-\bar Y), \qquad S_{XX}=\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar X)^2, \qquad S_{YY}=\frac{1}{n}\sum_{i=1}^{n}(Y_i-\bar Y)^2.

Using $n-1$ instead of $n$ in all three sample moments gives the same $r$, since the common factor cancels. The key asymptotic result is

Theorem: Under the independent, identically distributed bivariate normal model just specified, as $n\to\infty$, $$ \sqrt{n}(r-\rho)\xrightarrow{d}N\!\left(0,(1-\rho^2)^2\right). $$ Thus the first-order asymptotic variance of $r$ is $(1-\rho^2)^2/n$.

Here is a derivation. Standardize each observation by setting $U_i=(X_i-\mu_X)/\sigma_X$ and $V_i=(Y_i-\mu_Y)/\sigma_Y$. These variables have means zero, variances one, and correlation $\rho$. Let $A_n,B_n,C_n$ be the sample averages of $U_iV_i,U_i^2,V_i^2$, respectively, and let $\bar U,\bar V$ be the sample means. The correlation can be written

r=\frac{A_n-\bar U\bar V} {\sqrt{(B_n-\bar U^2)(C_n-\bar V^2)}}.

The law of large numbers gives $(A_n,B_n,C_n)\xrightarrow{p}(\rho,1,1)$, while $\bar U,\bar V=O_p(n^{-1/2})$. Consequently, the products of sample means in this expression are $O_p(n^{-1})$, so they do not affect the first-order, $n^{-1/2}$, fluctuation. Apply the multivariate central limit theorem to $(A_n,B_n,C_n)$, whose components have finite second moments. The function $h(a,b,c)=a/\sqrt{bc}$ is differentiable at $(\rho,1,1)$, with gradient $(1,-\rho/2,-\rho/2)$. The delta method therefore gives the first-order fluctuation as the average of

\psi(U,V)=UV-\frac{\rho}{2}(U^2+V^2).

For a centered bivariate normal pair with unit marginal variances, the required moments are $E[U^2V^2]=1+2\rho^2$, $E[U^3V]=E[UV^3]=3\rho$, and $E[U^4]=E[V^4]=3$. These follow, for example, by writing $V=\rho U+\sqrt{1-\rho^2}\,W$ with independent standard normal $U,W$ and expanding. In particular, $E[\psi(U,V)]=0$, and

\begin{aligned} \operatorname{Var}(\psi(U,V)) &=E[U^2V^2]-\rho E[UV(U^2+V^2)] +\frac{\rho^2}{4}E[(U^2+V^2)^2]\\ &=(1+2\rho^2)-6\rho^2 +\frac{\rho^2}{4}(8+4\rho^2)\\ &=1-2\rho^2+\rho^4=(1-\rho^2)^2. \end{aligned}

The central limit theorem applied to the average of $\psi(U_i,V_i)$ now gives the stated limit. The centering corrections above are negligible after multiplication by $\sqrt n$, which completes the derivation. $\square$

Worked Example: Comparing Variability at Different Correlations

Suppose the sample size is $n=100$. Compare the first-order standard error of the sample correlation when the population correlation is $\rho=0$ and when it is $\rho=0.8$.

At $\rho=0$, the asymptotic variance is $ (1-0^2)^2/100=0.01$, so the standard error is $\sqrt{0.01}=0.10$.

At $\rho=0.8$, $1-\rho^2=1-0.64=0.36$. The variance is $0.36^2/100=0.1296/100=0.001296$, and its square root is $0.036$.

The raw correlation’s asymptotic standard error is therefore substantially smaller at $0.8$ than at zero. A single standard error on the correlation scale cannot describe both cases.

Fisher’s Transformation Stabilizes the Variance

The transformation used for the sample correlation is the inverse hyperbolic tangent, also called Fisher’s Z transformation:

Definition: For $-1<t<1$, define $$ g(t)=\operatorname{arctanh}(t) =\frac{1}{2}\log\!\left(\frac{1+t}{1-t}\right). $$ Its range is the entire real line, and its derivative is $g'(t)=1/(1-t^2)$.

Because $-1<\rho<1$, the population value $g(\rho)$ is finite. Applying the scalar delta method to the theorem above gives the variance stabilization.

Theorem: Under the same bivariate normal sampling assumptions, $$ \sqrt n\bigl(\operatorname{arctanh}(r)-\operatorname{arctanh}(\rho)\bigr) \xrightarrow{d}N(0,1). $$ Equivalently, the first-order asymptotic variance of $\operatorname{arctanh}(r)$ is $1/n$, independent of $\rho$.

To see this directly, the delta method multiplies the limiting variance for $r$ by the square of the derivative at $\rho$:

\left(g'(\rho)\right)^2\frac{(1-\rho^2)^2}{n} = \frac{1}{(1-\rho^2)^2}\frac{(1-\rho^2)^2}{n} =\frac{1}{n}.

The transformation does not make the sampling distribution exactly normal in finite samples. It makes its first-order asymptotic variance constant, and the limit theorem supplies the corresponding normal approximation as $n$ grows. The strict condition $|\rho|<1$ matters: the derivative becomes unbounded as $\rho$ approaches either endpoint.

Worked Example: Checking the Stabilized Standard Error

For $n=100$, compare the approximate standard error of the transformed correlation at $\rho=0$ and $\rho=0.8$.

At $\rho=0$, the transformed variance from the delta method is $g'(0)^2(1-0^2)^2/100=1^2/100=0.01$.

At $\rho=0.8$, $g'(0.8)=1/(1-0.8^2)=1/0.36$. Thus the transformed variance is

\left(\frac{1}{0.36}\right)^2\frac{0.36^2}{100} =\frac{1}{100}=0.01.

In both cases the approximate standard error is $0.10=1/\sqrt{100}$. The original-scale standard errors differed in the preceding example; the transformation cancels that first-order dependence on $\rho$.

An Approximate Confidence Interval for the Correlation

Write $z_r=\operatorname{arctanh}(r)$. The limiting normal distribution says that, for large $n$, $z_r$ is approximately normal with center $\operatorname{arctanh}(\rho)$ and standard error $1/\sqrt n$. Let $z_{1-\alpha/2}$ denote the $1-\alpha/2$ quantile of the standard normal distribution. An approximate two-sided $100(1-\alpha)\%$ interval is first formed on the transformed scale:

\left[ z_r-\frac{z_{1-\alpha/2}}{\sqrt n}, \quad z_r+\frac{z_{1-\alpha/2}}{\sqrt n} \right].

The inverse transformation is $\tanh$, which is strictly increasing. Applying it to both endpoints therefore gives the corresponding approximate interval for $\rho$:

Approximate Fisher Z interval: For a sample correlation $r$ with $-1<r<1$, $$ \left[ \tanh\!\left(\operatorname{arctanh}(r)-\frac{z_{1-\alpha/2}}{\sqrt n}\right), \quad \tanh\!\left(\operatorname{arctanh}(r)+\frac{z_{1-\alpha/2}}{\sqrt n}\right) \right]. $$ Under the stated model and fixed $-1<\rho<1$, its coverage probability tends to $1-\alpha$ as $n\to\infty$.

The coverage statement follows because the transformed statistic, centered at $\operatorname{arctanh}(\rho)$ and scaled by $\sqrt n$, converges to a standard normal variable. The normal probability between $-z_{1-\alpha/2}$ and $z_{1-\alpha/2}$ is $1-\alpha$. Since $\tanh$ is the inverse of $\operatorname{arctanh}$, the transformed-scale event that contains $\operatorname{arctanh}(\rho)$ is exactly the event that the back-transformed interval contains $\rho$. This is an asymptotic, not an exact finite-sample, coverage guarantee.

Worked Example: A 95% Interval for a Positive Correlation

Suppose $n=100$ paired observations have sample correlation $r=0.60$. Construct the approximate 95% Fisher Z interval, using $z_{0.975}=1.96$.

First transform the estimate: $z_r=\operatorname{arctanh}(0.60)=\tfrac12\log(1.60/0.40)=\tfrac12\log(4)\approx0.6931$. The transformed standard error is $1/\sqrt{100}=0.10$, so the margin is $1.96(0.10)=0.196$. The interval on the transformed scale is $[0.6931-0.196,\;0.6931+0.196]=[0.4971,\;0.8891]$.

Now apply $\tanh$ to each endpoint. Using $\tanh(x)=(e^{2x}-1)/(e^{2x}+1)$, we obtain $\tanh(0.4971)\approx0.460$ and $\tanh(0.8891)\approx0.711$. The approximate 95% interval for the population correlation is therefore $[0.460,\;0.711]$.

The endpoints are not equally far from $0.60$. That is expected: the interval is symmetric on the transformed scale, and the inverse transformation is nonlinear.

Worked Example: An Interval for a Negative Correlation

Suppose $n=64$ observations give $r=-0.40$. Using $z_{0.975}=1.96$, find the approximate 95% Fisher Z interval.

The transformed estimate is $z_r=\tfrac12\log(0.60/1.40)\approx-0.4236$. The standard error is $1/\sqrt{64}=0.125$, and the margin is $1.96(0.125)=0.245$. Thus the transformed limits are $[-0.4236-0.245,\;-0.4236+0.245]=[-0.6686,\;-0.1786]$.

Back-transforming gives $\tanh(-0.6686)\approx-0.584$ and $\tanh(-0.1786)\approx-0.177$. The interval is approximately $[-0.584,\;-0.177]$. Both limits are negative, so this approximate interval excludes zero.

Assumptions, Uses, and a Common Pitfall

This method is tied to the sampling assumptions behind the variance result. The pairs must be independent and identically distributed, and the bivariate normal model with nonzero marginal variances must be a reasonable description of the data. Under that model, the relevant moments exist and the correlation is strictly inside $(-1,1)$. In practice, substantial outliers, dependence between pairs, or severe departures from bivariate normality can make this particular variance approximation unreliable. The interval is intended as a large-sample approximation, not a universal formula for every kind of paired data.

Common mistake: using the raw-scale standard error for a symmetric interval. A tempting calculation is to take $r$, add and subtract $1.96(1-r^2)/\sqrt n$, and call the result a 95% interval. This substitutes the estimate for $\rho$ in the raw-scale standard error, but it does not carry out the Fisher Z construction. In particular, it ignores the nonlinearity of the transformation and can produce endpoints outside $[-1,1]$. The correction is to compute $\operatorname{arctanh}(r)$, use standard error $1/\sqrt n$ on that scale, and apply $\tanh$ to the two limits.

The calculation also requires a finite transformed estimate. In the usual continuous bivariate normal setting with $n\geq3$, the sample correlation lies strictly between $-1$ and $1$ with probability one. If a data set gives exactly $r=1$ or $r=-1$, the transformation is not finite; this signals a degenerate or special sample configuration, not a value to insert into the formula.

Key takeaway: Under bivariate normal sampling, the sample correlation has first-order variance $(1-\rho^2)^2/n$. Fisher’s $ \operatorname{arctanh} $ transformation cancels this dependence, yielding asymptotic variance $1/n$ and a practical route to an approximate interval for $\rho$.

Check Your Understanding

Try these questions without looking back at the derivations.

State the sampling assumptions under which the asymptotic variance $(1-\rho^2)^2/n$ was derived.
Why does the asymptotic variance of the transformed sample correlation equal $1/n$? Identify the derivative that produces the cancellation.
For $n=25$, what is the approximate standard error of $\operatorname{arctanh}(r)$?
Why is a Fisher Z interval generally not symmetric around $r$ after its endpoints are transformed back?
What is wrong with treating $r\pm1.96(1-r^2)/\sqrt n$ as the Fisher Z interval?