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: $n^{-1} \sum_{i=1}^n (X_i - \bar{X})^2 \xrightarrow{P} 4p (1-p) \mu^2 + \sigma^2$ 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

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