0
$\begingroup$

suppose we have the situation like: $\text{Pr}\biggl(XY \leq \frac{\gamma_t(\mu_3Z+\mu_4)}{\zeta-\frac{\gamma_t \mu_1}{Y}-\frac{\gamma_t \mu_2}{X}}\biggr)$, where $X,Y,Z$ are independent random variables having CDF $F_X(x), F_Y(y), F_Z(z)$ and PDF $f_X(x), f_Y(y), f_Z(z)$ and all other things are constant, then can we represent it like this:

$\int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \text{Pr}\biggl(XY \leq \frac{\gamma_t(\mu_3Z+\mu_4)}{\zeta-\frac{\gamma_t \mu_1}{Y}-\frac{\gamma_t \mu_2}{X}}\biggr)f_X(x) f_Y(y) f_Z(z)\text{d}x \text{d}y \text{d}z ?$

Any help in this regard will be highly appreciated.

$\endgroup$
2
  • $\begingroup$ I guess you're assuming $X,Y,Z$ are non-negative. $\endgroup$
    – ploosu2
    Commented Jun 26, 2022 at 6:50
  • $\begingroup$ Yes sir. Thank for your answer. $\endgroup$
    – chaaru
    Commented Jun 26, 2022 at 7:33

1 Answer 1

1
$\begingroup$

Well, yes but that's quite trivial, since the first factor inside the integral is constant and what you get is just $1$ times it. What you could do, is put the indicator of the event that's inside the $Pr()$ in there:

$$ \mathbb{P}\biggl(XY \leq \frac{\gamma_t(\mu_3Z+\mu_4)}{\zeta-\frac{\gamma_t \mu_1}{Y}-\frac{\gamma_t \mu_2}{X}}\biggr) = \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \boldsymbol{1}_{xy \leq \frac{\gamma_t(\mu_3z+\mu_4)}{\zeta-\frac{\gamma_t \mu_1}{y}-\frac{\gamma_t \mu_2}{x}}} f_X(x) f_Y(y) f_Z(z)\text{d}x \text{d}y \text{d}z. $$

To solve the integrating regions solve the inequality. To simplify the constants, denote

  • $a = \frac{\mu_1}{\mu_3}$
  • $b = \frac{\zeta}{\gamma_t \mu_3}$
  • $c_1 = \frac{\mu_1}{\mu_3}$
  • $c_2 = \frac{\mu_2}{\mu_3}$

so the inequality becomes equivalent with

$$xy \leq \frac{z + a}{b-c_1/y - c_2/x}$$

Divide both sides by $xy$ to get

$$1 \leq \frac{z + a}{bxy - c_1x - c_2y}$$

I will assume $a\geq 0$. (If this isn't the case, you have to do it in two parts.) Then because $z+a\geq 0$, we must have that $z+a \geq bxy - c_1x - c_2y > 0$. From $bxy - c_1x - c_2y > 0$ we also get $(bx-c_2)y>c_1x$. I will also assume $c_1\geq 0$ so we must have $bx>c_2$ and $y>\frac{c_1x}{bx-c_2}$. (If $c_1<0$, the inequalities flip.)

We're assuming $X, Y, Z$ are exponential. Let them have rates $\lambda_X, \lambda_Y, \lambda_Z$, respectively. So we can write the integral as

$$ \lambda_X \lambda_Y \lambda_Z \int_0^\infty \int_0^\infty \int_0^\infty \boldsymbol{1}_{xy\leq \frac{z+a}{b-c_1x-c_2y}} e^{-\lambda_X x -\lambda_Y y -\lambda_Z z }dz dy dx \\ = \lambda_X \lambda_Y \lambda_Z \int_\frac{c_2}{b}^\infty \int_{\frac{c_1x}{bx-c_2}}^\infty \int_{\max(0, bxy-c_1x-c_2y-a)}^\infty e^{-\lambda_X x -\lambda_Y y -\lambda_Z z }dz dy dx \\ = \lambda_X \lambda_Y \int_\frac{c_2}{b}^\infty \int_{\frac{c_1x}{bx-c_2}}^\infty e^{-\lambda_X x -\lambda_Y y}e^{-\lambda_Z \max(0, bxy-c_1x-c_2y-a)} dydx \\ = \lambda_X \lambda_Y \left( \int_\frac{c_2}{b}^\infty \int_{\frac{c_1x}{bx-c_2}}^\frac{c_1x+a}{bx-c_2} e^{-\lambda_X x -\lambda_Y y} dydx + \int_\frac{c_2}{b}^\infty \int_{\frac{c_1x+a}{bx-c_2}}^\infty e^{-\lambda_X x -\lambda_Y y-\lambda_Z(bxy-c_1x-c_2y-a)} dydx \right) $$

But by simulation I get that this is wrong! The following SageMath simulation gives approximately $0.35$ (see values for the constants in the code)

def randExpo(l):
    return -float(log(1-random()))/l

def simu(a,b,c1,c2, ls):
    x,y,z = (randExpo(l) for l in ls)
    return 1 if x*y <= (z+a)/(b*x*y-c1*x-c2*y) else 0

a = 1.2
b = 3.1
c1 = 0.7
c2 = 0.35
ls = (1.3, 0.77, 2.34) #the rates for X, Y, Z
simuN = 10000
print (float(sum(simu(a,b,c1,c2,ls) for _ in range(simuN))/simuN))

But calculating the intgral here in Desmos gives the value $0.23$. Where is my error?

$\endgroup$
6
  • $\begingroup$ Thank you sir for your response. Say, $X,Y,Z$ are independent exponential random variables then how to proceed with the above integration? I am getting confused due to $\leq$ symbol. $\endgroup$
    – chaaru
    Commented Jun 26, 2022 at 7:48
  • $\begingroup$ You're welcome. You would have to solve the inequality to see over which region(s) you have to integrate. I would first divide by $xy$ (you can forget the case $xy=0$, it has zero prob). Then split into two cases depending whether the denominator is positive/negative. Multiply the denominator to other side and you should get something nice in terms of $z$. $\endgroup$
    – ploosu2
    Commented Jun 26, 2022 at 7:57
  • $\begingroup$ Ok sir. Could please write mathematically. $\endgroup$
    – chaaru
    Commented Jun 26, 2022 at 8:01
  • $\begingroup$ Thank you sir. But it has created more confusion now. $\endgroup$
    – chaaru
    Commented Jun 26, 2022 at 9:22
  • $\begingroup$ @chaaru No problem. I've tried to make it more clear. I renamed the constants (I don't use $d$ anymore in order to not mix with the integration $dx, dy$). But I'm getting an error! The value I get isn't confirmed by simulation. Can someone spot where I made a mistake? $\endgroup$
    – ploosu2
    Commented Jun 26, 2022 at 15:04

You must log in to answer this question.