Heads up, I'm asking this question after having read Percentile of sum of random variables and other questions. Please don't close this as a duplicate. It's not the same question.
Suppose I have two positive random variables $X_1$ and $X_2$, and the following percentiles or upper bounds on them. The joint probability assumes independence.
\begin{aligned} P(X_1<35) &= 0.95 \\ P(X_2<40) &= 0.95 \\ P(X_1<35, X_2<40) &= 0.95^2 \end{aligned}
$X_1$ is concentrated beneath $35$, and $X_2$ beneath $40$. They're both under their respective limits with probability $0.0925$. That means that their sum can not reach $\ge70$ with $p=0.0925$, right? How can I prove this? I have a feeling I'm wrong. I know I'm skipping steps, yet don't know the steps.
$$P(X_1+X_2<35+40) = 0.95^2$$
No. You can't. Because it is not true.
Because it is possible to get less than 75 in other ways also. For example, 10+65 = 75.
$$P(X_1+X_2<35+40) = 0.95^2 $$ $$\implies P(X_1+X_2<35+40 \land X_1<35 \land X_2<40) $$$$+ P(X_1+X_2<35+40 \land X_1<35 \land X_2\ge40) $$$$+ P(X_1+X_2<35+40 \land X_1\ge35 \land X_2<40) $$$$+ P(X_1+X_2<35+40 \land X_1\ge35 \land X_2\ge40) $$$$= 0.95^2$$
In this, the fourth term is obviously 0.
In the fist term, $X_1<35 \land X_2<40 \implies X_1+X_2<35+40$. So the first term becomes $P(X_1<35 \land X_2<40)$ which is nothing but $ 0.95^2$.
Substituting the first and fourth term, we get $$ 0.95^2 + P(X_1+X_2<35+40 \land X_1<35 \land X_2\ge40) $$$$+ P(X_1+X_2<35+40 \land X_1\ge35 \land X_2<40) +0 $$$$ = 0.95^2 $$
Which means $$ P(X_1+X_2<35+40 \land X_1<35 \land X_2\ge40) + P(X_1+X_2<35+40 \land X_1\ge35 \land X_2<40) =0 $$
Which obviously not true. What could be true is $$P(X_1+X_2<35+40) > 0.95^2 $$