One-Sample Multivariate Sign Tests

Testing a Specified Multivariate Center

A one-sample multivariate sign test asks whether a specified point is a center of symmetry for a distribution. It uses each observation’s direction from that point, rather than the observation’s distance from it. This makes the test resistant to changes in the magnitudes of extreme observations, while retaining information about whether directions tend to balance around the proposed center.

Let $X_1,\ldots,X_n$ be independent and identically distributed random vectors in $\mathbb{R}^p$, and let $\theta_0\in\mathbb{R}^p$ be the proposed center. The null hypothesis is central symmetry about $\theta_0$: the distribution of $X_i-\theta_0$ is the same as that of $-(X_i-\theta_0)$. The alternative is that this symmetry does not hold. This is a hypothesis about a center of symmetry; it is not automatically a test that the mean equals $\theta_0$.

Definition: The spatial sign of a residual $x-\theta_0$ is $$ U_{\theta_0}(x)= \begin{cases} \dfrac{x-\theta_0}{\|x-\theta_0\|}, & x\ne\theta_0,\\[6pt] 0, & x=\theta_0. \end{cases} $$ Write $U_i=U_{\theta_0}(X_i)$, $S_n=\sum_{i=1}^n U_i$, and $A_n=\sum_{i=1}^n U_iU_i^{T}$. The vector $S_n$ is the sign resultant: its direction indicates a directional imbalance, and its length measures the size of that imbalance.

Central symmetry implies $U_i$ and $-U_i$ have the same distribution. Consequently $E[U_i]=0$. The matrix $E[U_iU_i^T]$, however, need not be a multiple of the identity: the directions may be more common in some parts of the sphere than others. This distinction determines how the sign resultant should be calibrated.

Finite-Sample Calibration by Sign Flips

Under the null, the direction of a nonzero residual is as likely to point one way along its unoriented line as the opposite way. Conditioning on those lines removes the unknown distribution of directions. The resulting test is exact in finite samples, even when the angular distribution is not uniform.

Definition: For a nonzero sign $U_i$, its unoriented axis is the pair $\{U_i,-U_i\}$. Conditional on the observed axes, the null distribution of the signs can be generated by replacing each nonzero $U_i$, independently, by $\varepsilon_iU_i$, where $P(\varepsilon_i=1)=P(\varepsilon_i=-1)=1/2$. A zero sign remains zero.

When $A_n$ is positive definite, a useful statistic is

Q_n=S_n^T A_n^{-1}S_n.

The matrix $A_n$ is unchanged by any sign flips, so its inverse is the same throughout the conditional reference distribution. An exact conditional $p$-value is the fraction of all allowed sign assignments for which the recalculated statistic is at least as large as the observed $Q_n$. Including ties makes the test conservative when the discrete reference distribution cannot attain the desired significance level exactly. If $A_n$ is singular, this inverse-based statistic is not defined; one can instead use a statistic such as $\|S_n\|^2$, whose conditional distribution is still obtained by the same sign flips.

Theorem: Under independent sampling and central symmetry about $\theta_0$, the conditional sign-flip test has conditional rejection probability at most $\alpha$ at level $\alpha$, for any statistic computed from the signs and unchanged axes. Therefore its unconditional rejection probability is also at most $\alpha$.

Proof. For each observation, central symmetry gives the same probability to a nonzero residual direction and its opposite. Conditional on the unoriented axis, the orientation is therefore a fair sign. Independence of the observations means these orientations are conditionally independent across observations. Thus, given the axes, all $2^m$ sign assignments for the $m$ nonzero signs have equal probability. The conditional reference distribution enumerates precisely these assignments. Ranking the equally likely assignments by the statistic, with ties included in the upper tail, gives a conditional $p$-value that is super-uniform: its probability of being at most $\alpha$ is no greater than $\alpha$. Hence the conditional rejection probability is at most $\alpha$. Averaging this bound over the observed axes gives the same unconditional bound. $\square$

Worked Example: Exact Test With Six Bivariate Directions

Test central symmetry about $\theta_0=(0,0)^T$ using observations $(2,0),(1,0),(-3,0),(0,2),(0,1),(0,-4)$. Their spatial signs are three copies of $(1,0)^T$, one copy of $(-1,0)^T$, and two copies of $(0,1)^T$.

The sign resultant and sign matrix are

S_6=(2,2)^T,\qquad A_6= \begin{pmatrix} 4&0\\ 0&2 \end{pmatrix}.

Since $A_6$ is positive definite,

Q_6=(2,2) \begin{pmatrix} 1/4&0\\ 0&1/2 \end{pmatrix} \begin{pmatrix} 2\\2 \end{pmatrix} =1+2=3.

For the conditional reference distribution, the four horizontal signs have an absolute resultant of $4,2,$ or $0$, with probabilities $2/16,8/16,$ and $6/16$, respectively. The two vertical signs have an absolute resultant of $2$ or $0$, each with probability $2/4$. The horizontal and vertical flips are independent. A horizontal absolute resultant of $4$ gives a statistic of $4$, whatever the vertical signs, contributing probability $2/16=1/8$ to the upper tail. A horizontal resultant of $2$ contributes a statistic of at least $3$ exactly when the vertical resultant is $2$, adding $(8/16)(2/4)=1/4$. The other cases give statistics below $3$. Thus the exact conditional $p$-value is $1/8+1/4=3/8$. The observation does not provide evidence against the null at a $0.05$ level.

Asymptotic Chi-Square Calibration

Enumerating $2^m$ sign assignments becomes impractical when the sample is large. A large-sample calibration is available by estimating the covariance of the signs. Define $V=E[U_iU_i^T]$ under the null. If $V$ is positive definite, then the central limit theorem applies to the bounded, mean-zero signs, and the sample sign matrix estimates $V$.

Theorem: Suppose $X_1,\ldots,X_n$ are independent and identically distributed, centrally symmetric about $\theta_0$, and $V=E[U_iU_i^T]$ is positive definite. Then $$ Q_n=S_n^TA_n^{-1}S_n\xrightarrow{d}\chi^2_p. $$ Equivalently, the asymptotic level-$\alpha$ test rejects for $Q_n>\chi^2_{p,1-\alpha}$, where $\chi^2_{p,1-\alpha}$ is the $1-\alpha$ quantile of the chi-square distribution with $p$ degrees of freedom.

Proof. Since $U_i$ and $-U_i$ have the same distribution, $E[U_i]=0$. Each sign is bounded in norm by $1$, so it has a finite covariance matrix, equal to $V$. The multivariate central limit theorem gives

\frac{S_n}{\sqrt n}\xrightarrow{d}N_p(0,V).

Also, each entry of $U_iU_i^T$ is bounded. The law of large numbers therefore gives $A_n/n\xrightarrow{p}V$. Since $V$ is positive definite, matrix inversion is continuous at $V$, and $(A_n/n)^{-1}\xrightarrow{p}V^{-1}$. Rewrite the statistic as

Q_n= \left(\frac{S_n}{\sqrt n}\right)^T \left(\frac{A_n}{n}\right)^{-1} \left(\frac{S_n}{\sqrt n}\right).

By Slutsky’s theorem, its limiting distribution is $Z^TV^{-1}Z$, where $Z\sim N_p(0,V)$. The vector $V^{-1/2}Z$ has the standard $p$-variate normal distribution, so $Z^TV^{-1}Z=\|V^{-1/2}Z\|^2\sim\chi^2_p$. This proves the result. $\square$

Worked Example: Large-Sample Calibration in Two Dimensions

Suppose $n=100$ bivariate signs are observed: 36 equal $(1,0)^T$, 24 equal $(-1,0)^T$, 16 equal $(0,1)^T$, and 24 equal $(0,-1)^T$. Then

S_{100}=(36-24,16-24)^T=(12,-8)^T,\qquad A_{100}= \begin{pmatrix} 60&0\\ 0&40 \end{pmatrix}.

The statistic is

Q_{100}=\frac{12^2}{60}+\frac{(-8)^2}{40} =2.4+1.6=4.

For two degrees of freedom, the chi-square upper-tail probability at $4$ is $\exp(-4/2)=\exp(-2)$, approximately $0.1353$. This is the asymptotic $p$-value, not an exact finite-sample $p$-value. The exact conditional calibration would instead enumerate the sign flips of the 100 observed directions.

What the Sign Test Uses—and What It Does Not

Spatial signs discard radial distances but keep directions. Consequently, replacing an observation by another point on the same ray from $\theta_0$ leaves the test unchanged. This is useful when magnitudes may be extreme or poorly behaved, but it also means the test is not using all the information in the data. In particular, the sign test should not be interpreted as a general test against every possible departure from central symmetry: alternatives with little directional imbalance may be difficult for it to detect.

Worked Example: Changing an Outlier’s Magnitude

Consider the bivariate observations $(1,0),(1000,0),(0,2),(-4,0)$, tested about $(0,0)^T$. Their signs are $(1,0),(1,0),(0,1),(-1,0)$, so $S_4=(1,1)^T$ and $A_4=\begin{pmatrix}3&0\\0&1\end{pmatrix}$. Thus

Q_4=(1,1) \begin{pmatrix} 1/3&0\\ 0&1 \end{pmatrix} \begin{pmatrix} 1\\1 \end{pmatrix} =\frac{1}{3}+1=\frac{4}{3}.

If $(1000,0)$ is replaced by $(2,0)$, its sign remains $(1,0)^T$. The resultant, sign matrix, and statistic are all unchanged. This illustrates resistance to changes in radial magnitude; it does not imply insensitivity to changes in direction.

Common Mistake: Assuming the Sign Covariance Is Spherical

A tempting shortcut is to use $\|S_n\|^2$ and assume its null calibration is the same for every centrally symmetric distribution. That assumption is generally wrong. Central symmetry ensures a zero mean sign, but it does not ensure equal directional variability in every coordinate. The matrix $A_n$ adjusts the statistic for the observed sign covariance. If the directions are spherically distributed, then $V=I_p/p$; without that additional condition, this special form need not hold.

A second interpretive mistake is to call this automatically a test of the population mean. The null specifies symmetry about $\theta_0$. A mean, if it exists, may equal that center under central symmetry, but a mean-based hypothesis and a symmetry-based hypothesis are not interchangeable in general.

Key takeaway: A one-sample spatial sign test asks whether directions from a specified point balance under central symmetry. Conditional sign flips give finite-sample calibration; an estimated sign covariance gives an asymptotic chi-square calibration when that covariance is positive definite.

Check Your Understanding

Use these questions to check the construction, calibration, and interpretation of a one-sample multivariate sign test.

State the null hypothesis for a one-sample sign test about a specified point $\theta_0$. Why is it not automatically a test that the mean equals $\theta_0$?
For a nonzero residual, what information is retained in its spatial sign, and what information is discarded?
Why does conditioning on the unoriented axes yield independent fair sign flips under the null?
What condition on $V=E[U_iU_i^T]$ is needed for the stated chi-square limit, and where is that condition used in the derivation?
Why can the sign resultant’s unadjusted squared length have a different null calibration for different centrally symmetric distributions?