During developing a new statistical estimator, I faced the following problem.

Let $\mathbf{x}_i$ be a sequence of i.i.d. $d$-dimensional random vectors with \begin{align*} \mathbf{x}_i = \mathbf{O}_i \mathbf{\mu} + \mathbf{\varepsilon}_i, \end{align*} where $\mathbf{\mu}$ is a mean vector, $\mathbf{O}_i$ is an random orthogonal matrix (so $\mathbf{O}_i^\top \mathbf{O}_i = \mathbf{I}$), and $\mathbf{\varepsilon}_i$ is an i.i.d. mean-zero noise vector. Then the question is

Question Is there any way to test $H_0: \{\mathbf{\mu} = 0\} $?

Motivation

This problem is important in estimating the risk premium of the factor model with time-varying factor loading. From the PCA, we can only estimate the factor loading up to rotation, so conducting Fama-MacBeth regression with the aggregated factor loadings would be problematic since they have different rotations for each component.

Here is what I have done.

To find the answer, I have tried the simplest case: $d=1$ (a univariate case). For $d=1$, the orthogonal matrix $\mathbf{O}$ becomes just $+1$ or $-1$. Then we have \begin{align*} X_i = o_i \mu + \varepsilon_i, \qquad o_i = \begin{cases} +1 & \text{ with prob }p \\ -1 & \text{ with prob }1-p, \end{cases} \end{align*} where the variance of $\varepsilon_i$ is $\sigma^2$. Notice that $\bar{X} = n^{-1} \sum_{i=1}^n X_i$ is not a consistent estimator of $\mu$ (but $\bar{X} \xrightarrow{P} (2p - 1) \mu$), so we should consider another approach. Notice that $\mu \neq 0$ imples $\mu^2 > 0$, so it may be useful to use $n^{-1} \sum_{i=1}^n X_i^2$ for the test: \begin{align*} \frac{1}{n} \sum_{i=1}^n X_i^2 &= \frac{1}{n} \sum_{i=1}^n (o_i \mu + \varepsilon_i)^2 \\ &= \frac{1}{n} \sum_{i=1}^n \big(\mu^2 + 2 o_i \varepsilon_i + \varepsilon_i^2\big) \\ &\xrightarrow{P} \mu^2 + \sigma^2 \end{align*} by the law of large numbers. So we should debias $\sigma^2$. Notice also that the classical estimator of $\sigma^2$ is not consistent: $n^{-1} \sum_{i=1}^n (X_i - \bar{X})^2 \xrightarrow{P} \big( 1 - (2p-1)^2 \big) \mu^2 + \sigma^2$ from the fact $\bar{X} \xrightarrow{P} (2p - 1) \mu$, so it is not clear how to debias it. This problem is generically due to the degrees of freedom of sample first and second moment is $2$ while we have to infer $3$ parameters ($p, \mu, \sigma^2$).

(My previous answer was wrong.)

It would be really appreciated if you have any idea or comments.

Thank you for reading!

Seunghyeon