Power with Known Covariance

From a Rejection Rule to Power

A significance level describes how often a test rejects when its null hypothesis is true. Power asks a different question: if the mean vector has a particular value away from the null, how often will the test reject? For a multivariate mean test with known covariance, the answer can be calculated exactly under a Gaussian model. The key is that the test statistic has a noncentral chi-square distribution under the alternative.

Assume $X_1,\ldots,X_n$ are independent and identically distributed $p$-dimensional Gaussian vectors with mean vector $\mu$ and known positive-definite covariance matrix $\Sigma$. The null hypothesis is $H_0:\mu=\mu_0$, and the alternative allows any $\mu\ne\mu_0$. Write $\bar X=n^{-1}\sum_{i=1}^n X_i$. The known-covariance test rejects for a sufficiently large covariance-standardized squared distance between $\bar X$ and $\mu_0$.

Definition: For significance level $\alpha$, let $c_{p,\alpha}$ be the upper-tail critical value satisfying $P(\chi^2_p>c_{p,\alpha})=\alpha$, where $\chi^2_p$ denotes a central chi-square random variable with $p$ degrees of freedom. The known-covariance mean test rejects $H_0$ when $$ Q=n(\bar X-\mu_0)^\mathsf{T}\Sigma^{-1}(\bar X-\mu_0)>c_{p,\alpha}. $$

Power is the rejection probability at a specified true mean $\mu$: $P_\mu(Q>c_{p,\alpha})$. It depends on the alternative mean, the sample size, the covariance matrix, the dimension, and the test's significance level. It is not a single number for a composite alternative unless a particular $\mu$ is specified.

Why the Alternative Gives a Noncentral Chi-Square

The sample mean has distribution $\bar X\sim N_p(\mu,\Sigma/n)$. Standardizing it relative to the null gives $$ Z=\sqrt{n}\,\Sigma^{-1/2}(\bar X-\mu_0). $$ Here $\Sigma^{-1/2}$ is the symmetric inverse square root of $\Sigma$. Under the alternative, $Z$ has identity covariance and mean $\sqrt{n}\,\Sigma^{-1/2}(\mu-\mu_0)$; moreover $Q=Z^\mathsf{T}Z$. The squared length of a standard-covariance Gaussian vector with nonzero mean has a noncentral chi-square distribution.

Theorem: Under the Gaussian model above, for any fixed $\mu$, $$ Q\sim\chi^2_p(\lambda),\qquad \lambda=n(\mu-\mu_0)^\mathsf{T}\Sigma^{-1}(\mu-\mu_0). $$ The quantity $\lambda$ is the noncentrality parameter. It is zero under $H_0$ and positive for every $\mu\ne\mu_0$.

Proof. Set $a=\sqrt{n}\,\Sigma^{-1/2}(\mu-\mu_0)$, so that $Z\sim N_p(a,I_p)$ and $a^\mathsf{T}a=\lambda$. For $t<1/2$, the moment-generating function of $Q=\sum_{j=1}^p Z_j^2$ is $$ E_\mu[e^{tQ}]=(1-2t)^{-p/2}\exp\left(\frac{\lambda t}{1-2t}\right). $$ This is the moment-generating function of a noncentral chi-square random variable with $p$ degrees of freedom and noncentrality $\lambda$. Since the moment-generating functions agree on an open interval containing zero, they determine the same distribution. $\square$

The noncentrality has a useful interpretation: it is $n$ times the squared Mahalanobis distance between the true mean and the null mean. Thus a larger sample size increases $\lambda$ linearly when the alternative and covariance are fixed. The same Euclidean mean difference can produce different $\lambda$ values in different covariance directions; the covariance standardization matters.

Exact Power and Its Mixture Calculation

Let $F_{p,\lambda}$ denote the cumulative distribution function of $\chi^2_p(\lambda)$. The exact power at $\mu$ follows immediately from the theorem:

Theorem: Under the stated Gaussian assumptions, the power of the level-$\alpha$ known-covariance test at mean $\mu$ is $$ \pi(\mu)=P_\mu(Q>c_{p,\alpha}) =1-F_{p,\lambda}(c_{p,\alpha}),\qquad \lambda=n(\mu-\mu_0)^\mathsf{T}\Sigma^{-1}(\mu-\mu_0). $$

The formula is an exact finite-sample calculation, not a large-sample approximation. One way to evaluate it is the Poisson mixture representation. If $K$ has a Poisson distribution with mean $\lambda/2$, then a noncentral chi-square with $p$ degrees of freedom is a mixture of central chi-square distributions with degrees of freedom $p+2K$. Consequently, $$ \pi(\mu)=\sum_{k=0}^{\infty}P(K=k)\,P(\chi^2_{p+2k}>c_{p,\alpha}) =\sum_{k=0}^{\infty}e^{-\lambda/2}\frac{(\lambda/2)^k}{k!} P(\chi^2_{p+2k}>c_{p,\alpha}). $$

To verify the mixture, the moment-generating function of a central $\chi^2_{p+2K}$, averaged over $K$, is $$ (1-2t)^{-p/2}E[(1-2t)^{-K}] =(1-2t)^{-p/2}\exp\left\{\frac{\lambda}{2}\left((1-2t)^{-1}-1\right)\right\} =(1-2t)^{-p/2}\exp\left(\frac{\lambda t}{1-2t}\right). $$ This is the noncentral chi-square moment-generating function derived above, for $t<1/2$; hence the mixture has the required distribution. The series is also practical numerically: its nonnegative terms can be summed until the remaining Poisson probability is negligible or bounded by a chosen tolerance.

Worked Example: Power for a Two-Dimensional Alternative

Suppose $p=2$, $\Sigma=I_2$, $\mu_0=(0,0)^\mathsf{T}$, $n=4$, and the true mean is $\mu=(0.5,0.5)^\mathsf{T}$. Use a level-$0.05$ test.

The squared standardized distance per observation is $(\mu-\mu_0)^\mathsf{T}\Sigma^{-1}(\mu-\mu_0)=0.5^2+0.5^2=0.5$. Therefore $\lambda=4(0.5)=2$. For two degrees of freedom, $c_{2,0.05}=-2\log(0.05)\approx5.9915$. The mixture index $K$ is Poisson with mean $\lambda/2=1$.

For $p=2$, the central tail in mixture term $k$ is $P(\chi^2_{2+2k}>c)=e^{-c/2}\sum_{j=0}^k(c/2)^j/j!$. Since $e^{-c/2}=0.05$, the first central tails, for $k=0,\ldots,5$, are approximately $0.0500, 0.1998, 0.4241, 0.6482, 0.8160,$ and $0.9165$. Multiplying by the corresponding Poisson probabilities gives $$ \begin{aligned} \pi(\mu) &=\sum_{k=0}^{\infty}P(K=k)P(\chi^2_{2+2k}>5.9915)\\ &\approx (0.3679)(0.0500)+(0.3679)(0.1998) +(0.1839)(0.4241)+(0.0613)(0.6482)\\ &\quad +(0.0153)(0.8160)+(0.0031)(0.9165)+\cdots\\ &\approx 0.2255. \end{aligned} $$ The exact power is about $22.6\%$. The omitted terms are small; for example, truncating after $k=8$ leaves a total omitted Poisson weight of less than $1.2\times10^{-6}$, which also bounds the omitted power.

Sample Size Changes the Alternative Distribution

With $\mu,\mu_0,\Sigma$, and $p$ fixed, increasing $n$ does not change the null critical value $c_{p,\alpha}$, but it increases the alternative noncentrality. Thus the alternative distribution moves away from the null distribution. The following calculation makes that change explicit for the same two-dimensional mean difference as above.

Worked Example: Comparing Two Sample Sizes

Keep $p=2$, $\Sigma=I_2$, $\mu_0=(0,0)^\mathsf{T}$, and $\mu=(0.5,0.5)^\mathsf{T}$, with $\alpha=0.05$. The standardized squared difference remains $0.5$, so $\lambda=0.5n$.

For $n=2$, $\lambda=1$, and the mixture index has a Poisson distribution with mean $0.5$. Using the same critical value $5.9915$, the mixture sum is approximately $$ (0.6065)(0.0500)+(0.3033)(0.1998)+(0.0758)(0.4241) +(0.0126)(0.6482)+(0.0016)(0.8160)+\cdots =0.1327. $$ The power is about $13.3\%$.

For $n=8$, $\lambda=4$, and $K$ is Poisson with mean $2$. The corresponding sum begins $$ (0.1353)(0.0500)+(0.2707)(0.1998)+(0.2707)(0.4241) +(0.1804)(0.6482)+(0.0902)(0.8160)+\cdots \approx0.4154. $$ The power is about $41.5\%$. Both calculations use the same null critical value; the increase comes from the larger noncentrality at the larger sample size.

Worked Example: Accounting for a Nonidentity Covariance

Let $p=2$, $\mu_0=(0,0)^\mathsf{T}$, $n=6$, and $\Sigma=\begin{pmatrix}4&1\\1&1\end{pmatrix}$. Consider the alternative $\mu=(1,0)^\mathsf{T}$ and a level-$0.05$ test.

The determinant of $\Sigma$ is $4(1)-1(1)=3>0$, and its leading diagonal entry is positive, so it is positive definite. Its inverse is $\Sigma^{-1}=\frac{1}{3}\begin{pmatrix}1&-1\\-1&4\end{pmatrix}$. With $\delta=\mu-\mu_0=(1,0)^\mathsf{T}$, $$ \delta^\mathsf{T}\Sigma^{-1}\delta =(1,0)\frac{1}{3}\begin{pmatrix}1&-1\\-1&4\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix} =\frac{1}{3}. $$ Therefore $\lambda=6/3=2$. The test has $p=2$, so its critical value is again $5.9915$, and the Poisson-mixture calculation from the first example gives power approximately $0.2255$. Using the unstandardized squared difference $1$ instead would incorrectly give $\lambda=6$; the covariance matrix changes the distance relevant to the test.

Assumptions, Interpretation, and a Common Pitfall

The exact power calculation depends on the model assumptions that produced the distribution of $Q$: independent Gaussian observations, a common known positive-definite covariance matrix, and a specified alternative mean. If those assumptions do not hold, the noncentral chi-square power formula is not automatically exact. In particular, this calculation treats $\Sigma$ as known; replacing it with a sample covariance changes the test statistic's distribution and requires a different calibration.

A common mistake is to use the central chi-square tail $P(\chi^2_p>c_{p,\alpha})=\alpha$ to report power under an alternative. That probability is the rejection probability under the null, not under the specified $\mu$. The correction is to calculate $\lambda$ from the alternative and use $P(\chi^2_p(\lambda)>c_{p,\alpha})$. A related error is to omit the factor $n$ from $\lambda$; doing so would make the calculated power unchanged when sample size changes, contrary to the distribution of the sample mean.

Power is useful for comparing how readily a test detects different mean departures under a stated model. The noncentrality parameter condenses the alternative's size, direction relative to covariance, and sample size into one quantity. Reporting the alternative and these assumptions alongside a power value makes clear what that value does—and does not—describe.

Check Your Understanding

Use the definitions and calculations above to answer the following questions.

For fixed $\mu,\mu_0,\Sigma$, how does the noncentrality parameter change when $n$ is doubled?
Why is the central chi-square tail at the test's critical value not the power at an alternative mean?
If the alternative equals $\mu_0$, what value does $\lambda$ take, and what is the rejection probability?
For a fixed mean difference, why can changing $\Sigma$ change power even when $n$ stays the same?
What model assumption is needed for the noncentral chi-square power calculation to be exact in finite samples?

What You'll Learn