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



Estimation on rotationally-disturbed random vectors
$\newcommand\ep\varepsilon$This is impossible to do even for $d=1$, as your [model is not identifiable](https://en.wikipedia.org/wiki/Identifiability#Definition), that is, the parameters of the model are not identifiable even if the distribution of the $X\_i$'s is fully known.


Indeed, let $R\_1,\dots,R\_n$'s be independent Rademacher random variables, with $P(R\_i=\pm1)=1/2$. Then the following two scenarios result in the same distribution of the $X\_i$'s:


(i) $\mu=0$ and $\ep\_i=R\_i$ for all $i$;


(ii) $\mu=1$, and $O\_i=R\_i$ and $\ep\_i=0$ for all $i$.


We see that even the function $1(\mu=0)$ of the parameter $\mu$ is not identifiable, that is, the value of
$1(\mu=0)$ cannot be determined even if the distribution of the $X\_i$'s is fully known.


