How to Derive the Power of the Wald Test

The power of a test is the probability that it rejects the null hypothesis at a particular parameter value. For a Wald test, power depends on the sample size, the alternative, and the significance level. There are two useful asymptotic results: under alternatives that approach the null at rate $1/\sqrt{n}$, power is described by a noncentral chi-square distribution; under a fixed alternative that violates the null, power tends to one.

Wald statistic for general restrictions

Suppose the parameter is $\theta\in\mathbb{R}^p$, and the null hypothesis imposes $q$ restrictions:

$$H_0:g(\theta)=0,$$

where $g$ maps into $\mathbb{R}^q$. Assume $g$ is differentiable and its derivative matrix has rank $q$ near the null. Let $\hat\theta$ estimate $\theta$, and suppose

$$\sqrt{n}(\hat\theta-\theta)\ \xrightarrow{d}\ N(0,V(\theta)).$$

Here $V(\theta)$ is the asymptotic covariance matrix of $\sqrt{n}\hat\theta$, not of $\hat\theta$ itself. Assume also that $V(\theta)$ is positive definite near the null. Let $G(\theta)$ be the $q\times p$ derivative matrix of $g$. A Wald statistic is

$$W_n=n\,g(\hat\theta)'\left[G(\hat\theta)\hat V G(\hat\theta)'\right]^{-1}g(\hat\theta),$$

where $\hat V$ consistently estimates $V(\theta)$. Equivalent formulas may use a consistent estimate of the covariance matrix of $g(\hat\theta)$ directly. Under the null, the statistic converges in distribution to a central chi-square random variable with $q$ degrees of freedom. Thus a level-$\alpha$ test rejects when $W_n>c_\alpha$, where $c_\alpha$ is the $1-\alpha$ quantile of $\chi^2_q$.

Deriving power under local alternatives

To obtain a nontrivial limiting power, consider alternatives that approach a null value $\theta_0$ as the sample size increases:

$$\theta_n=\theta_0+\frac{h}{\sqrt{n}},$$

where $g(\theta_0)=0$ and $h$ is a fixed vector. Write $G_0=G(\theta_0)$ and $V_0=V(\theta_0)$. A first-order Taylor expansion gives

$$\sqrt{n}\,g(\theta_n) =\sqrt{n}\left[g(\theta_0)+G_0\frac{h}{\sqrt{n}}+o(n^{-1/2})\right] \longrightarrow G_0h.$$

For this sequence of alternatives, assume the corresponding triangular-array regularity conditions: $\sqrt{n}(\hat\theta-\theta_n)\xrightarrow{d}N(0,V_0)$ and $\hat V$ consistently estimates $V_0$ along the sequence. Combining this with the expansion around $\theta_n$ yields

$$\sqrt{n}\,g(\hat\theta)\xrightarrow{d}N(G_0h,\,G_0V_0G_0').$$

The Wald statistic is the squared length of this vector after scaling by its covariance matrix. Therefore,

$$W_n\xrightarrow{d}\chi^2_q(\lambda), \qquad \lambda=(G_0h)'(G_0V_0G_0')^{-1}(G_0h).$$

The quantity $\lambda$ is the noncentrality parameter. When $h=0$, it is zero and the distribution is the central null distribution. When $G_0h\ne 0$, the local alternative moves away from the null restrictions to first order, and $\lambda$ is positive. The asymptotic power at that alternative is

$$\operatorname{Power}(h)=\Pr\!\left(\chi^2_q(\lambda)>c_\alpha\right).$$

This is the derivation: the estimator is asymptotically normal, the restrictions are linearized, and the resulting quadratic form has a noncentral chi-square distribution.

Worked example: a two-sided test of one mean

Suppose an estimator has standard error approximately $\sigma/\sqrt{n}$, and you test $H_0:\theta=\theta_0$ against a two-sided alternative at level $\alpha=0.05$. Consider the local alternative $\theta_n=\theta_0+\sigma/\sqrt{n}$, which is one standard error above the null on the estimator’s asymptotic scale. The standardized statistic

$$Z_n=\frac{\sqrt{n}(\hat\theta-\theta_0)}{\sigma}$$

converges to $N(1,1)$ under this alternative. The test rejects when $|Z_n|>1.96$. Its limiting power is therefore

$$\Pr(Z_n>1.96)+\Pr(Z_n<-1.96) =1-\Phi(0.96)+\Phi(-2.96) \approx 0.170.$$

So this alternative is rejected about 17% of the time in the large-sample approximation. For a one-restriction Wald statistic, $W_n=Z_n^2$ and the same calculation is the upper-tail probability of a noncentral chi-square distribution with one degree of freedom and noncentrality parameter $1$.

Fixed alternatives and consistency

If $\theta$ is fixed and $g(\theta)\ne 0$, then $g(\hat\theta)$ converges to a nonzero vector under standard consistency conditions. The factor $n$ in the Wald statistic makes $W_n$ diverge, so the rejection probability approaches one. This is the usual consistency result. The noncentral chi-square formula above is especially useful for alternatives that shrink toward the null; it should not be confused with a claim that a fixed alternative has a fixed noncentrality parameter as $n$ grows.

Common mistakes

• Using the wrong covariance scale. If $\hat V$ estimates the covariance of $\sqrt{n}\hat\theta$, the statistic includes the factor $n$. If you instead use an estimated covariance of $\hat\theta$, that factor is already accounted for in the inverse covariance.
• Using the central chi-square distribution for power. The central distribution sets the rejection threshold under the null. Under a local alternative, calculate power using the noncentral distribution.
• Confusing local and fixed alternatives. Local alternatives give a finite, informative limiting power; fixed alternatives violating the null are detected with probability tending to one.
• Ignoring regularity conditions. The argument needs asymptotic normality, consistent covariance estimation, differentiable restrictions, a full-rank restriction derivative, and a nonsingular restriction covariance. For local alternatives, the normality and covariance-consistency conditions must also hold along the local sequence. If these fail, the usual chi-square approximation may not apply.