Let $n \in \mathbb{N}$, $n \geq 2$, assume that $X_1,\ldots, X_n$ are exchangeable, square integrable random variables with $\mathbf{E}\bigl[X^2_1\bigr] < \infty$. Prove that the following holds: $$\mathbf{Cov}[X_1, X_2] \geq -\frac{1}{n-1}\mathbf{Var}[X_1]\, .$$

Discuss when equality occurs. Are there non-trivial examples?

Solution: We know that (if the integral exists) $$\int_\Omega X_i \, X_j \, d\mathbf{P} = \int_{\mathbb{R}^2} x_1 x_2 \, d\mathcal{L}(X_i,X_j)\, \\ \int_\Omega X_i \, d\mathbf{P} = \int_{\mathbb{R}} x \, d\mathcal{L}(X_i)\,,$$ where $\mathcal{L}(X_i,X_j)$ denotes the joint probability distribution of $X_i$ and $X_j$.

So for $i\neq j, k \neq l$ we get $$\mathbf{E}[X_i] = \mathbf{E}[X_k], \quad\mathbf{Var}[X_i] = \mathbf{Var}[X_k],\quad \mathbf{Cov}[X_i, X_j] = \mathbf{Cov}[X_k, X_l] \, .$$

So:\begin{align*}0 &\leq \mathbf{Var}[X_1 + \cdots + X_n] = \sum_{i=1}^n \mathbf{Var}[X_i] + \sum_{\overset{i, j=1}{i\neq j}}^n\mathbf{Cov}[X_i, X_j] \\ &=\sum_{i=1}^n \mathbf{Var}[X_1] + \sum_{\overset{i, j=1}{i\neq j}}^n\mathbf{Cov}[X_1, X_2] \\ &= n\mathbf{Var}[X_1] + n(n-1)\mathbf{Cov}[X_1, X_2]\, . \end{align*} From which after dividing by $n(n-1)$ and rearranging the result follows. $\square$

Equality occurs for example for constant $X_1 = \cdots = X_n = c$ with $c \in \mathbb{R}$.

A more complicated example would be $X_1, X_2$ with values in $\{-1, 0, 1\}$ and \begin{align*} \mathbf{P}[X_1 = -1] = \mathbf{P}[X_1 = 0] = \mathbf{P}[X_1 = 1] = \frac{1}{3}\, , \quad X_2 = -X_1 \, . \end{align*} $X_1$ and $X_2$ are exchangeable (right??) and \begin{gather*} \mathbf{Cov}[X_1, X_2] = \frac{1}{3}\cdot 1 \cdot (-1) + \frac{1}{3}\cdot (-1) \cdot 1 + \frac{1}{3}\cdot 0 \cdot 0 = -\frac{2}{3} \\ \mathbf{Var}[X_1] = \frac{1}{3}\cdot 1^2 + \frac{1}{3}\cdot (-1)^2 + \frac{1}{3}\cdot 0^2 = \frac{2}{3}\, , \end{gather*} so since $n = 2$: $$ \mathbf{Cov}[X_1, X_2] = - \frac{1}{n-1}\mathbf{Var}[X_1]\, .$$

Is everything ok (proof and example)? Thanks!

PS: This question is different from:
"Covariance inequality for infinitely many exchangeable random variables"

Here we prove the result for $n$ exchangeable variables, that means for any finite number of exchangeable variables. In the other question we prove the result for infinitely many exchangeable variables.

  • 1
    $\begingroup$ I'm adding a proof verification tag, I'm not 100% clear on whether the finite case is different, but this post's work is significantly different from the linked thread, and the OP is asking for proof verification of this proof, not for a proof of the result. $\endgroup$ – Alan Apr 15 '15 at 15:49

Yes, the proof and example are good. You might want to include some further discussion about when equality occurs: under what condition does the inequality $0 \leq \text{Var}[X_1+\cdots+X_n]$ become an equality?


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