Covariance of two sample means

Question

I am trying to derive the covariance of two sample means and get confused at one point. Given is a sample of size $n$ with paired dependent observations $x_i$ and $y_i$ as realizations of RVs $X$ and $Y$ and sample means $\bar{x}$ and $\bar{y}$. I try to derive $cov(\bar{x},\bar{y})$.

I am relatively sure the result should be

$$cov(\bar{x},\bar{y})=\frac{1}{n}cov(X,Y)$$

However I arrive at

$$cov(\bar{x},\bar{y})=E(\bar{x}\bar{y})-\mu_x\mu_y = E\left(\frac{1}{n^2}\sum x_i \sum y_i\right) -\mu_x\mu_y =\frac{1}{n^2} n^2 E(x_i y_i) -\mu_x\mu_y=cov(X,Y)$$

I used

$$E\left(\frac{1}{n^2}\sum x_i \sum y_i\right)=\frac{1}{n^2} E\left(x_1y_1+x_2y_1+\cdots + x_ny_n\right)=\frac{1}{n^2} n^2 E(x_iy_i)$$

Somewhere should be a flaw in my thinking.

Answer 1

Covariance is a bilinear function meaning that $$ \operatorname{cov}\left(\sum_{i=1}^n a_iC_i, \sum_{j=1}^m b_jD_j\right) = \sum_{i=1}^n \sum_{j=1}^m a_i b_j\operatorname{cov}(C_i,D_j).$$ There is no need to mess with means etc.

Applying this to the question of the covariance of the sample means of $n$ independent paired samples $(X_i, Y_i)$ (note: the pairs are independent bivariate random variables; we are not claiming that $X_i$ and $Y_i$ are independent random variables), we have that \begin{align} \operatorname{cov}\left(\bar{X},\bar{Y}\right) &= \operatorname{cov}\left(\frac{1}{n}\sum_{i=1}^n X_i, \frac 1n\sum_{j=1}^n Y_j\right)\\ &= \frac{1}{n^2}\sum_{i=1}^n \sum_{j=1}^n \operatorname{cov} (X_i, Y_j)\\ &= \frac{1}{n^2}\sum_{i=1}^n \operatorname{cov} (X_i, Y_i) &\scriptstyle{\text{since $X_i$ and $Y_j$ are independent, and thus uncorrelated, for $i \neq j$}}\\ &= \frac 1n\operatorname{cov} (X, Y) \end{align}

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As noted below in a comment by flow2k, although $\operatorname{cov}(\bar{X},\bar{Y})$ is smaller than $\operatorname{cov}({X},{Y})$ by a factor of $n$, the (Pearson) correlation coefficients are the same: $\rho_{\bar{X},\bar{Y}} = \rho_{X,Y}$ !! Previously I had never given the correlation coefficients any thought at all.

Answer 2

I think the algebra issue is resolved with the following:

\begin{align}{1 \over n^2}E\left(\sum_{i=1}^n x_i \sum_{i=1}^n y_i\right)&={1 \over n^2}E\left(\sum_{i=1}^n x_i y_i +\sum_{i\ne j}x_i y_j\right)\\&={1 \over n^2}(n(Cov(x_i,y_i)+\mu_X \mu_Y)+n(n-1)\mu_X \mu_Y)\\&={1 \over n^2}(n Cov(x_i,y_i)+n^2 \mu_X \mu_Y))\\&=Cov(x_i,y_i)/n+ \mu_X \mu_Y\end{align}