Confidence interval for difference

Question

Let $X_1$ and $X_2$ be two i.i.d. random variables with zero mean and finite second moment. Let $l_1,l_2 \leq 0$ and $u_1,u_2 \geq 0$ be constants. Set $$ P(l_1 \leq X_1 \leq u_1) = p_1 \quad \text{and} \quad P(l_2 \leq X_2 \leq u_2)=p_2. $$ Does it follow that $$ P(l_1 - u_2 \leq X_1 - X_2 \leq u_1 - l_2) \geq (p_1+p_2)/2? $$

Answer 1

Since the intersection of events $$ \{l_1\leq X_1\leq u_1\}\cap \{l_2\leq X_2\leq u_2\} = \{l_1\leq X_1\leq u_1\}\cap \{-u_2\leq -X_2\leq -l_2\} $$ implies the event (and does not equal to, in general) $$ \{l_1-u_2 \leq X_1-X_2 \leq u_1-l_2\} $$ then $$ \mathbb P(l_1-u_2 \leq X_1-X_2 \leq u_1-l_2) \color{red}\geq \mathbb P(\{l_1\leq X_1\leq u_1\}\cap \{l_2\leq X_2\leq u_2\}) = \mathbb P(l_1\leq X_1\leq u_1)\cdot \mathbb P(l_2\leq X_2\leq u_2)=p_1p_2. $$ Since $$ 0\leq (p_1-p_2)^2 = p_1^2+p_2^2 - 2p_1p_2, $$ we have $$ p_1p_2 \color{red}{\leq} \frac{p_1^2+p_2^2}{2} \color{red}{\leq} \frac{p_1+p_2}{2} $$ As you can see, there are the opposite inequalities to what you need. So the inequality is most likely wrong in general.

Take, for example, two i.i.d random variables $X_1,X_2$ with discrete distribution $\mathbb P(X=0)=\mathbb P(X=2)=\frac16$, $\mathbb P(X=1)=\frac23$. Then $$ \mathbb P(0.9\leq X_1\leq 1.1)=\frac23, \quad \mathbb P(0.9\leq X_2\leq 1.1)=\frac23 $$ and $$ \mathbb P(0.9-1.1\leq X_1-X_2\leq 1.1-0.9)=\mathbb P(-0.2\leq X_1-X_2\leq 0.2) $$ $$= \mathbb P(X_1=X_2) = \frac{1}{6^2}+\frac{2^2}{3^2}+\frac{1}{6^2} = \frac12 < \frac{\frac23+\frac23}{2} $$