# How to prove the probability inequality?

Given four mutually independent bounded random variables (RVs), denoted as $x_1,x_2,x_3,x_4$ and we have inequalities that $$\mathrm{Pr}(x_1<0)<\mathrm{Pr}(x_3<0) \\ \mathrm{Pr}(x_2<0)<\mathrm{Pr}(x_4<0)$$ How to judge the inequality $\mathrm{Pr}(x_1+x_2<0) \lessgtr \mathrm{Pr}(x_3+x_4<0)$? If possible, please show that.

• It's false in either form. Looks for counterexamples among variables that take two values.
– user147263
Commented Mar 23, 2015 at 3:12
• If we add more constraints, e.g. bounding interval of $x_1$ is $[a,b]$, bounding interval of $x_3$ is $[c,d]$, we have $a>c,b<d$. The same as to bounding interval of $x_2$ and $x_4$. How about the results? Commented Mar 24, 2015 at 17:28

Counterexample to $\mathbb{P}(x_1 + x_2 < 0) > \mathbb{P}(x_3 + x_4 < 0)$ under given conditions: Let $x_1, x_2 \sim \xi$ be independent and $$\xi = \begin{cases} 1, \text{ with prob. }1/2, \\ -1, \text{ with prob. } 1/2\end{cases}$$ and $x_3, x_4 \sim \zeta$ be independent and $$\zeta = \begin{cases} 1, \text{ with prob. }1/3, \\ -1, \text{ with prob. } 2/3\end{cases}$$ Then,
$$\mathbb{P}(x_1 + x_2 < 0) = \frac 1 4 < \frac 4 9 = \mathbb{P}(x_3 + x_4 < 0).$$
Counterexample to $\mathbb{P}(x_1 + x_2 < 0) < \mathbb{P}(x_3 + x_4 < 0)$ under given conditions:
Let $x_1, x_2 \sim \xi$ be independent and $$\xi = \begin{cases} 1, \text{ with prob. }1/2, \\ -100, \text{ with prob. } 1/2\end{cases}$$ and let $x_3, x_4$ be as defined above. Then, $$\mathbb{P}(x_1 + x_2 < 0) = \frac 3 4 > \frac 4 9 = \mathbb{P}(x_3 + x_4 < 0).$$