# How to test equality of asymptotically gaussian estimators across 3 or more independent samples

Let $n \in \mathbb{N}$ and suppose we are given $K\geq 3$ independent samples $(\mathcal{X}_n^i)_{1\leq i \leq K}$ where $\mathcal{X}_n^i = (X_1^i,\dots,X_n^i)$ is a $n$-sample of i.i.d. real valued random variables with c.d.f. $F_i$. Let $S:\mathbb{R}^n \mapsto \mathbb{R}$ be a function, and for each $i$ let $\widehat{\mu_n^i} = S(\mathcal{X}_n^i)$.

$\widehat{\mu_n^i}$ is an estimator for some unknown quantity $\mu^i = T(F_i)$ where $T(F_i)$ is a functional of the underlying distribution $F_i$.

Now, suppose that there exists positive constants $(\sigma_i)_{1\leq i \leq K}$ (potentially different) such that: \begin{align*} \frac{\sqrt{n}}{\sigma_i}\left(\widehat{\mu_n^i}-\mu^i\right) \underset{n \rightarrow \infty}{\overset{d}{\longrightarrow}} \mathcal{N}(0,1). \end{align*}

Without further assumptions, how would one tests: \begin{align*} &H_0 : \mu^1 = \dots = \mu^K \\ &H_1 : \exists \text{ } i\neq j, \mu^i \neq \mu^j \end{align*}

• are your samples independent? Because that would simplify the question to a large extent. – MauOlivares Sep 13 '18 at 15:05
• Yes they are, forgot to write it down, I'll edit ! – Alfred F. Sep 13 '18 at 15:18

We are given that we have $k$ independent samples, all of them of size $n$. Since the estimators $\hat{\mu}_i$ are independent and asymptotically Gaussian, we know that the vector

$$\mathbf{m}=\begin{pmatrix}\hat{\mu}\\\vdots\\\hat{\mu}_k\end{pmatrix} \overset{d}{\rightarrow}\mathcal{N} \left(\mu,\Sigma\right)\:\:\text{ as }\:n\to\infty$$ the multivariate normal with mean vector $\mu'=(\mu_1\dots\mu_k)$ covariance matrix $\Sigma$. From here, you can construct a Wald test for the hypothesis of interest:

$$H_0: \mu_1=\dots=\mu_k=\mu^{0}$$ The test statistic is given by

$$W=n(\mathbf{m}-\mu^{0})'\Sigma^{-1}(\mathbf{m}-\mu^{0})\overset{d}{\rightarrow}\chi^2_{k} \:\:\text{ as }\:n\to\infty$$ and the test rejects $H_0$ if $W>\chi^2_{k,1-\alpha}$.

• Thanks for the answer. A couple remarks : In the statistic $W$, isnt there suppose to be to factor $n$? And the covariance matrix is to the power $-1$ instead of $-1/2$ right? Also, I am interested in testing equality, and I dont have access to $\mu^0$. So I can not implement this test numerically. Am I missing something? – Alfred F. Sep 13 '18 at 16:12
• You're right, I edited the answer. – MauOlivares Sep 13 '18 at 16:20
• Regarding the second question, there is a way to do this - a robust permutation test. I could write it but it would take some time, but if you want the quick answer, you can check this paper, check equation (3.3) – MauOlivares Sep 13 '18 at 16:23
• Thanks for this ref. The result seems to hold under the assumption of asymptotic linearity, which I unfortunately do not have... – Alfred F. Sep 13 '18 at 16:48
• Which type of estimator do you have in mind? – MauOlivares Sep 13 '18 at 20:20