Let there be two samples of size $n$, $x_i$ and $y_i$ from two different normal distributions.
What is $\operatorname{cov}(\bar X_n, \bar Y_n)$? And how can it be estimated?
The motivation for my question is to understand if there is a way to know if two paired samples are correlated in such away so that their expectancies "should" be compared used paired t-test.
Thanks.
\begin{eqnarray} \text{cov}(\bar X_n, \bar Y_n) &=& \text{cov}(1/n \sum X_i, 1/n \sum Y_j)\\ &=& 1/n^2 \cdot \text{cov}( \sum X_i, \sum Y_j)\\ &=& 1/n^2 \cdot \sum_i \sum_j \text{cov}( X_i, Y_j) \end{eqnarray}
To go further, we need to specify something about the covariances. If the samples are iid random samples where $\text{cov}(X_i,Y_j)$ is constant over all $i,j$:
\begin{eqnarray} \quad\quad &=& 1/n^2 \cdot n^2 \text{cov}( X, Y)\\ \quad\quad &=& \text{cov}( X, Y)\, . \end{eqnarray}
If instead (and as seems to be the case here) we're talking about paired data, where $X_i$ and $Y_j$ are only correlated when $i=j$ then:
\begin{eqnarray} \quad\quad &=& 1/n^2 \cdot \sum_i \sum_j \text{cov}( X_i, Y_j)\\ \quad\quad &=& 1/n^2 \cdot n \cdot \text{cov}( X_i, Y_i)\\ \quad\quad &=& 1/n \cdot \text{cov}( X_i, Y_i)\\ \quad\quad &=& 1/n \cdot \rho\, \sigma_x \sigma_y, \end{eqnarray}
where $\rho$ is the correlation between $X$ and $Y$ pairs.
Here is an answer derived using the theory of 'moments of moments', using power sum notation, and leaving the grunt work to mathStatica. In particular, in power sum notation, let:
$$s_{a,b}=\sum _{i=1}^n X_i^a Y_i^b$$
Then, $\operatorname{cov}(\bar X_n, \bar Y_n)$ = $\operatorname{cov}(\frac{s_{1,0}}{n}$, $\frac{s_{0,1}}{n}$) ... and since the covariance operator is just the {1,1} CentralMoment, the solution is:
where $\mu_{1,1}$ denotes the {1,1} central moment of the population ...
i.e. The solution is:
$$\operatorname{cov}(\bar X_n, \bar Y_n) = \frac{\operatorname{cov}(X, Y)}{n} $$
In the case of independence, $\operatorname{cov}(X,Y)$ is, of course, zero.
