Probability that two sums of random variables do not exceed their expectation at the same time

Question

I have say 20 identically distributed random variables $\xi_1,\ldots,\xi_{20}$. Each of them is $0$ with probability $0.5$ and $1$ with probability $0.5$. Then I can conclude that the probability that their sum do not exceed $10$ is non-zero.

But I want something stronger, that the probability that the sum of both the first $10$ and the last $10$ do not exceed $5$ at the same time is non-zero. What kind condition of do I need for that? For example, will it be true if I only know that the variables are positively correlated, i.e., that ${\rm P}[\xi_i=1|\xi_j=1]\geq 0.5$ for all $i,j$?

Answer 1

You definitely need something more than what you are currently assuming. Here is a counterexample: $$\xi_{i} = \begin{cases} \xi_1 &\text{ if } i \leq 10,\\ 1-\xi_1 &\text{ if } i > 10. \end{cases}$$

One really easy assumption is that for any $n > 1$ and $x = (x_1,\dots,x_{n-1}) \in \{0,1\}^{n-1}$, $$P(\xi_n = 1|\xi_1,\dots,\xi_{n-1} = x) > 0.$$ Let $\mathbf{0} = (0,0,\dots,0)\in \{0,1\}^{20}$. Then $$P\left(\sum_{i=1}^{10}\xi_i\text{ and } \sum_{i=11}^{20}\xi_i \leq 5\right) \geq P\left(\xi = \mathbf{0}\right) > 0.$$ If $\xi$ is additionally Markov and homogeneous, then we can remove the non-degeneracy assumption. If $P(\xi_n = 1|\xi_{n-1} = 1) = 1$, then $\xi_n = \xi_1$ for all $n$, so both sums are less than $5$ with probability $1/2 > 0$. Likewise, if $P(\xi_n = 1|\xi_{n-1} = 1) = 0$, then $\xi_n = 1 - \xi_{n-1}$, so both sums equal exactly 5 with probability 1.

Edit: As @freddy requested, I provide a counterexample for the case in which $\xi_1,\dots,\xi_{20}$ are pairwise positively correlated.

Let $X \sim \text{Unif}(\{1,2,3\})$. For any $k$ let $$\mathcal{X}(k) := \left\{\mathbf{x} \in \{0,1\}^{10}: \sum_{i=1}^{10}x_i = k\right\}.$$ I also use the shorthand $a:b = \{a,a+1,\dots,b-1,b\}$ so $\xi_{1:10} = (\xi_1,\dots,\xi_{10})$.

Let $\xi_{1:10}$ and $\xi_{11:20}$ be conditionally independent given $X$ and with the following distributions:

• if $X=1$, then $\xi_{1:10},\xi_{11:20}\sim\text{Unif}(\mathcal{X}(8))$.
• If $X=2$, then $\xi_{1:10}\sim \text{Unif}(\mathcal{X}(6))$ and $\xi_{11:20}\sim \text{Unif}(\mathcal{X}(1))$.
• If $X=3$, then $\xi_{1:10}\sim \text{Unif}(\mathcal{X}(1))$ and $\xi_{11:20}\sim \text{Unif}(\mathcal{X}(6))$.

Obviously, by definition, either $\sum_{i=1}^{10}\xi_i > 5$ or $\sum_{i=11}^{20} \xi_i > 5$ almost surely. This sequence of Bernoulli random variables also have the following properties. For any $i$, $$E[\xi_i] = \frac{1}{3}\left(\frac{8}{10} + \frac{6}{10} + \frac{1}{10}\right) = \frac{1}{2}.$$ If $i,j \leq 10$ or $i,j > 10$, then $$E[\xi_i\xi_j] = \frac{1}{3}\left(\frac{8}{10}\frac{7}{9} + \frac{6}{10}\frac{5}{9} + 0\right) = \frac{28.\overline{6}}{90} > \frac{22.5}{90} = \frac{1}{4} = E[\xi_i]E[\xi_j],$$ so $\xi_i$ and $\xi_j$ are positively correlated. Likewise, if $i \leq 10$ and $j > 10$, then $$E[\xi_i\xi_j] = \frac{1}{3}\left(\frac{8}{10}\frac{8}{10} + \frac{6}{10}\frac{1}{10} + \frac{1}{10}\frac{6}{10}\right) = \frac{25.\overline{3}}{100} > \frac{1}{4} = E[\xi_i]E[\xi_j].$$ It follows that $\xi_{1:20}$ are pairwise positively correlated.