As the subject mentions, how can I find simplified form $$P = Pr [ min (a,b) > Th]$$

Note that both $a$ and $b$ are also random variables denoting the SINR of a wireless signal. $Th$ is the threshold value. Both $a$ and $b$ are independent.

Does it mean I have to find the Expected value of $P$? If yes, in this case, what can be the expected value of P

EDIT: Can I say the following:

$$ P = E[ P(min (a,b) > Th)] = E[P(a>Th)].E[P(b>Th)] \ \forall (0<{a,b} < 1)$$

  • $\begingroup$ What is $Th$? Are $a$ and $b$ independent? $\endgroup$ Commented Jan 23, 2018 at 4:56
  • $\begingroup$ but are they independent? $\endgroup$ Commented Jan 23, 2018 at 5:31
  • 1
    $\begingroup$ if they are independent then what you wrote is right because the probability $P(a>th)$ will be a constant and its expectation is also that constant $\endgroup$ Commented Jan 23, 2018 at 5:32
  • $\begingroup$ if $a$ and $b$ are not independent then you might have to use conditional probability relations $\endgroup$ Commented Jan 23, 2018 at 5:33
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    $\begingroup$ then they should be functions of some channel gains. Right? we need to know what is true about those channel gains $\endgroup$ Commented Jan 23, 2018 at 5:48

1 Answer 1


I may write the probability in a different way.

If $min(a,b)$ is greater than $Th$, then it means both $a$ and $b$ are greater than $Th$, so the probability may be rewritten as

$$ P[min(a,b)>Th] = P[ (a > Th) \cap ( b>Th)]$$

The events $min(a,b)>Th$ and $(a>Th) \cap (b>Th)$ are the same.

  • As an addition : this concept is quite known in Actuarial Mathematics, in terms of first-failure and last-survivor

Is this okay..?


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