# Mean and covariance estimators for IDD normal

Suppose $X_1,\cdots, X_n$ are sampled iid from a multivariate Gaussian $\mathcal N(\mu, \Sigma)$. We denote the sample mean and covariance as follows \begin{align} \bar X &= \frac{1}{n} \sum_{i=1}^n X_i\\ S &= \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar X)(X_i - \bar X)^\mathrm{T }\end{align} One can prove the following theorem~(see Theorem 3.3.2 and Corollary 7.2.3 of ). Here $\mathcal W(V, n)$ is a Wishart distribution with $n$ degrees of freedom and scale matrix $V$.

Theorem 1. $\bar X$ and $S$ are independent and \begin{align} \bar X &\sim \mathcal N(\mu, \frac{\Sigma}{n}), \\ S &\sim \mathcal W(\frac{1}{n-1}\Sigma, n-1). \end{align}

Now we change a problem setting. Suppose $Y_i$ is sampled independently from a multivariate Gaussian $\mathcal N(\mu, \Sigma + T_i)$, $T_i$ PSD, for any $i = 1,\cdots, n$. We denote two estimators as follows \begin{align} \bar Y &= \frac{1}{n} \sum_{i=1}^n Y_i\\ Z &= \frac{1}{n-1} \sum_{i=1}^n (Y_i - \bar Y)(Y_i - \bar Y)^\mathrm{T } \end{align} Does the following theorem hold?

Theorem 2. $\bar Y$ and $Z$ are independent and \begin{align} \bar Y &\sim \mathcal N(\mu, \frac{1}{n}(\Sigma + \frac{1}{n}\sum_{i=1}^n T_i), \\ Z &\sim \mathcal W(\frac{1}{n-1}(\Sigma + \frac{1}{n} \sum_{i=1}^n T_i), n-1). \end{align}

If not, what is the distribution of $Z$? I know the covariance between $\bar Y$ and $Z$ is 0 . But are they independent?

From Anderson's textbook , it seems like one strategy of the proof is to construct $Z = \sum_{i=1}^{n-1} A_iA_i^{\mathrm{T}}.$ But can we apply the same construction as section 3.2, page 63-65 of  (1984 version)?

 Theodore Wilbur Anderson. An Introduction to Multivariate Statistical Analysis. Wiley New York, 1958.

 Mukhopadhyay, Nitis, and Mun S. Son. On the covariance between the sample mean and variance. Communications in Statistics—Theory and Methods 40.7 (2011): 1142-1148.

• It is IDD because independently differently distributed. I don't know why some people decided to change it to IID. There are partial solutions at math.stackexchange.com/questions/2329954/…. But it is still unclear if there is a distribution describing Z in an elegant way. In particular, will there be concentration bounds for Z? – Zi Wang May 17 '18 at 21:32
• No doubt that was done because "IDD" isn't a widely used/ standard abbreviation and you mention "iid" in the body text right near the start; my own reaction was to guess you had probably made a typo. I suggest you simply remove "IDD" from your title – Glen_b May 17 '18 at 23:48
• I think IDD is pretty obvious.. It was used in many papers in statistics. – Zi Wang Jul 26 '18 at 17:09
• It was certainly not obvious to me; my training is in statistics. If it's used in many papers in statistics, could you please point to a couple of papers in reasonably well-known statistics journals that use it? (My primary interest being -- when they use it, do they consider it obvious enough to use it without defining it?) – Glen_b Jul 26 '18 at 23:23
• – Zi Wang Jul 28 '18 at 20:40