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Let $Y_i \sim N(\mu, \Sigma)$ iid where $Y_i \in \mathcal{R}^J$ and $\Sigma$ is NOT diagonal.

Let $\bar{Y} = \begin{pmatrix} \frac{1}{N_1} \sum^{N_1}_{i=1} Y_{i1} \\ \frac{1}{N_2} \sum^{N_2}_{i=1} Y_{i2} \\ \vdots \\ \frac{1}{N_J} \sum^{N_J}_{i=1} Y_{iJ} \end{pmatrix}$

where $Y_{ij}$ is the $j^{th}$ element of vector $Y_i$ and $N_1 \neq N_2 \neq \dots \neq N_J$.

We know $\bar{Y} \sim N(\mu, \bar{\Sigma})$, but what's the value of $\bar{\Sigma}$?

My attempt:

Let $\sigma_{rc}$ be the element in the $r^{th}$ row and $c^{th}$ column of $\Sigma$.

It's easy to show that the $j^{th}$ element on the diagonal of $\bar{\Sigma}$ is $\frac{1}{N_j}\sigma_{jj}$.

For the off-diagonals,

\begin{equation} \sigma_{rc} = \text{Cov}(\frac{1}{N_r} \sum^{N_r}_{i=1} Y_{ir}, \frac{1}{N_c} \sum^{N_c}_{i=1} Y_{ic}) \\ = \frac{1}{N_rN_c} \text{Cov}(\sum^{N_r}_{i=1} Y_{ir}, \sum^{N_c}_{i=1} Y_{ic}) \\ = \frac{1}{N_rN_c} \sum^{N_r}_{i=1}\sum^{N_c}_{k=1} \text{Cov}(Y_{ir}, Y_{kc}) \end{equation} Now, $Y_{ir}$ and $Y_{kc}$ are independent (hence the covariwnce is 0) for all $k \neq i$ since these are observations from different independent draws of the vector $Y_i$. Summing over $N_r$ and $N_c$, the amount of times $i=k$ is the minimum of $N_r$ and $N_c$. So,

\begin{equation} \frac{1}{N_rN_c} \sum^{N_r}_{i=1}\sum^{N_c}_{k=1} \text{Cov}(Y_{ir}, Y_{kc}) \\ = \frac{\text{min}(N_r, N_c)}{N_rN_c} \text{Cov}(Y_{ir}, Y_{ic})\\ = \frac{1}{\text{max}(N_r, N_c)} \sigma_{rc} \end{equation}

Is my attempt correct?

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I think you can simplify the solution greatly. Define a matrix $A$ as follows

$A(i,j) = \frac{1}{N_{i}}$ if $j \leq N_{i}$ and 0 otherwise.

You will note that $\bar{Y} = AY$, and therefore the mean and variance of $\bar{Y}$ = $A\mu$ and $A \Sigma A^{T}$ respectively.

As for you solution, in the second last step, how did you go from the double summation to $MIN(N_r, N_c)$?

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  • $\begingroup$ The covariance is nonzero only when $k = i$ since this is the only time $Y_{ir}$ and $Y_{kc}$ are observations from the same independent draw and hence the only time they are correlated. In the double sum, $k=i$ $\text{min}(N_r, N_c)$ times. $\endgroup$ – TrynnaDoStat Feb 12 '18 at 14:50

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