Consider the sequences of random variables $\{X_i\}_{i=1}^n$ and $\{Y_i\}_{i=1}^n$, as well as the corresponding sequence of products, $\{X_i Y_i\}_{i=1}^n$. All $X_i$ share the same mean value, $\mathrm{E}[X_i]= \mathrm{E}[X]$, and likewise, $\mathrm{E}[Y_i]= \mathrm{E}[Y]$. The random variables $X_i$ and $X_j$ are dependent with covariance $$ \mathrm{cov}(X_i, X_j) = \mathrm{E}[X_iX_j]-\mathrm{E}[X]^2, $$ and the same is true for $Y_i$ and $Y_j$, with corresponding covariance $$ \mathrm{cov}(Y_i, Y_j) = \mathrm{E}[Y_iY_j]-\mathrm{E}[Y]^2. $$ $X_i$ and $Y_j$, however, are independent, so that $\mathrm{cov}(X_i, Y_j) = 0$. Assume now that $$ \sum_{i\neq j} \mathrm{cov}(X_i, X_j) \leq 0 $$ and $$ \sum_{i\neq j} \mathrm{cov}(Y_i, Y_j) \leq 0 $$

The question then---given the above information---is the following always true: $$ \sum_{i\neq j} \mathrm{cov}(X_i Y_i, X_j Y_j) \leq 0? $$

ADDED LATER The answer below demonstrates that there are counterexamples, so a renewed formulation of the question would be

Under what conditions on the sequences $X_i$ and $Y_i$ is the following true: $$ \sum_{i\neq j} \mathrm{cov}(X_i Y_i, X_j Y_j) \leq 0? $$

This inequality can be reformulated. First, we write the covariance of $X_i Y_i$ and $X_j Y_j$ as $$ \mathrm{cov}(X_i Y_i, X_j Y_j) = \mathrm{E}[X_i X_j Y_i Y_j ]- E[X_i Y_i ] E[X_j Y_j ] $$ Because of independency, $$ \mathrm{E}[X_i X_j Y_i Y_j ] = \mathrm{E}[X_i X_j ] \mathrm{E}[Y_i Y_j ] $$ and therefore $$ \mathrm{cov}(X_i Y_i, X_j Y_j) =\mathrm{E}[X_i X_j ] \mathrm{E}[Y_i Y_j ] - E[X]^2 E[Y]^2 $$ Alternatively, we can use that $$ \mathrm{E}[X_iX_j] = \mathrm{cov}(X_i, X_j) + \mathrm{E}[X]^2 $$
and $$ \mathrm{E}[Y_iY_j] = \mathrm{cov}(Y_i, Y_j) + \mathrm{E}[Y]^2 $$
to write $$ \mathrm{cov}(X_i Y_i, X_j Y_j) = \mathrm{cov}(X_i,X_j) \mathrm{cov}(Y_i,Y_j) + \mathrm{cov}(X_i,X_j) \mathrm{E}[Y]^2 + \mathrm{cov}(Y_i,Y_j) \mathrm{E}[X]^2 $$ The above inequality can therefore be rewritten as

$$ \sum_{i\neq j} \mathrm{E}[X_i X_j ] \mathrm{E}[Y_i Y_j ] \leq n (n-1) E[X]^2 E[Y]^2 $$ or $$ \sum_{i\neq j} \mathrm{cov}(X_i,X_j) \mathrm{cov}(Y_i,Y_j) \leq \mathrm{E}[Y]^2 \sum_{i\neq j}|\mathrm{cov}(X_i,X_j)| + \mathrm{E}[X]^2 \sum_{i\neq j}|\mathrm{cov}(Y_i,Y_j)| $$

Do our starting assumptions provide enough information to guarantee these inequalities, or do we need to add further constraints on the properties of $X_i$ and $Y_i$?

Thanks for any help!

  • $\begingroup$ From the assumption, we can get $$\sum_{i\not=j}E(X_iX_j)\leq n(n-1)E(X)^2$$$$\sum_{i\not=j}E(Y_iY_j)\leq n(n-1)E(Y)^2$$And our aim is to $$\sum_{i\not=j}E(X_iX_j)E(Y_iY_j)\leq n(n-1)E(X)^2E(Y)^2$$I think we can consider from $n=2,3$. When $n=2$, it is obviously right. When $n=3$, I have no idea to prove it right or not. $\endgroup$ – gaoxinge Jan 30 '14 at 4:48

As a counterexample let $X$ and $Y$ be independent with $E(X)=E(Y)=0$ and let $n$ be even.

For each $i$ let $X_{2i}=X$ and $X_{2i+1}=-X$; similarly $Y_{2i}=Y$ and $Y_{2i+1}=-Y$.

Then $\sum_{i\ne j}\text{cov}(X_i,X_j)=0$ since the $\text{var}(X)$ and $-\text{var}(X)$ occurrences cancel eachother out, and similarly for the $Y_i$.

But $\sum_{i\ne j}\text{cov}(X_iY_i, X_jY_j)>0$ since $X_jY_j=X_iY_i$ for all $i\ne j$, using $(-1)^2=1$.

  • $\begingroup$ Thanks for instructive example. A small correction is that for even $n$ $$ \sum_{i\neq j} \mathrm{cov}(X_i, X_j) = -\frac{n}{2} X^2. $$ I think the example shows that a least condition for the suggested inequality to hold is that the independent sequences cannot share the same periodic structure as that forces positive covariance terms. Would you have any ideas for how to work that in towards a constructive proof? $\endgroup$ – user45947 Feb 3 '14 at 9:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.