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.


1 Answer 1


Unfortunately, this is not true. For example, suppose $a,b,c\sim\mathcal N(0,1)$.


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.

  • $\begingroup$ Thanks. Can you then suggest how I can solve the expression $P[h> (A+B)]$? $\endgroup$
    – SJa
    Commented Oct 27, 2020 at 4:57
  • 1
    $\begingroup$ 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. $\endgroup$
    – PedroSebe
    Commented Oct 27, 2020 at 5:05

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