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\} $?

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**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.

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**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: $n^{-1} \sum_{i=1}^n (X_i - \bar{X})^2 \xrightarrow{P} 4p (1-p) \mu^2 + \sigma^2$ from the fact $\bar{X} \xrightarrow{P} (2p - 1) \mu$. However, by combining all, we can have a system of equation such that
\begin{align*}
  \frac{1}{n} \sum_{i=1}^n \begin{bmatrix}
   X_i \\ X_i^2 \\ (X_i - \bar{X})^2
   \end{bmatrix} \ \xrightarrow{P} \ \begin{bmatrix}
  (2p-1) \mu  \\ \mu^2 + \sigma^2 \\ 4p (1-p) \mu^2 + \sigma^2
\end{bmatrix}
\end{align*}
would give a unique solution of $[p, \mu, \sigma^2]$. But it is not clear how to extend for the multivariate case.

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

Thank you for reading!

Seunghyeon