# Distributive property of probabilistic inequalities involving random variables on both sides

Can I break down $$P(h \geq (A + B)$$, given all $$A,B,h$$ are all random variables. Will the following rule works?

$$P[h \geq (A + B)] = P(h\geq A) + P(h\geq B)$$

Actually, in one of my mathematical analysis, I end up with a complex expression which can be simplified to $$P[h \geq (A + B)]$$. I believe I can move forward if I can break it down somehow. Further explanations of variables are as below.

$$h \sim \exp(\lambda')$$ and $$g \sim \exp(\lambda'')$$

$$A = a(1 + e^{sh})$$, $$B = bg(1+e^{sh})$$

$$a,b,s, \lambda', \lambda''$$ are constants.

Unfortunately, this is not true. For example, suppose $$a,b,c\sim\mathcal N(0,1)$$.
$$P(a>b+c)=P(a-b-c>0)=1/2$$
Since $$a-b-c\sim\mathcal N(0,3)$$. But $$P(a>b)+P(a>c)=1/2+1/2=1$$, so we have a counterexample to your proposition.
• Thanks. Can you then suggest how I can solve the expression $P[h> (A+B)]$?
• Maybe it has to be solved computationally. I would turn to Monte Carlo methods. If you really need an exact solution, you would have to do something like $\int\int f_A(a(1 + e^{sh}))\cdot f_B(bg(1+e^{sh}))\cdot(1-e^{-h}) dh dg$, where $f_A,f_B$ are probability densities. Commented Oct 27, 2020 at 5:05