Let $x_1$, $x_2$ and $x_3$ be three independent and identically distributed (positive) random variables, for which the Probability Density Function (PDF) is as follows: \begin{equation} f(x) = \alpha \exp\left(- \alpha x\right). \end{equation} Define probability function $p(z)$ as \begin{equation} p(z) = \begin{cases} 1, & \text{when $0 < z < c $} \\ a \exp\left( - b z \right), & \text{when $z \ge c$} \end{cases} \end{equation} Note that $\alpha$, $a$, $b$, and $c$ are all positive constants. Also note that e.g. for $x_1+x_2 \ge c$ we have $p(x_1+x_2)= a \exp\left( - b (x_1 +x_2) \right)$.

Define $y=p(x_1) p(x_1+x_2) p(x_1+x_2+x_3)$.

I am trying to derive the Complementary Cumulative Distribution Function (CCDF) and the expected value of $y$; I am not necessarly looking for explicit expressions, but I need, at least, to write these expressions as integrals function of $\alpha$, $a$, $b$, and $c$.

  • Expected value of $y$:
    $E\left[y\right]= \int_0^\infty \int_0^\infty \int_0^\infty p(x_1) p(x_1+x_2) p(x_1+x_2+x_3)\, f(x_1) f(x_2) f(x_3) \,dx_1 dx_2 dx_3$
  • CCDF $= Pr(y > Y)$ ?
  • $\begingroup$ What is the relevance of $z$?Also, in the above your definition of $y$ is not a random variable, but a product of their PDFs. I think you can do some work to clarify your notation. $\endgroup$ – owen88 Feb 11 '18 at 16:33
  • $\begingroup$ @owen88 $z$ is just a variable. $y$ is the product of three functions that depend on $x_1$, $x_2$ and $x_3$, and since these are random variables, then $y$ is also a random variable. $\endgroup$ – din Feb 11 '18 at 16:46
  • $\begingroup$ Ok, if $p$ is just a function, then it is still not clear what $p( u \, | \, v)$ would mean (using $u,v$ to avoid confusion with $x_1,x_2,x_3,y,z$ the notation you already use). I assume that this is meant to be interpeted as a conditional probability, but this is still not clear from your notation. Can you do any more to explain why this variable y is of interest? $\endgroup$ – owen88 Feb 11 '18 at 16:50
  • $\begingroup$ @owen88 I have edited the question. I hope it's clearer now. $\endgroup$ – din Feb 11 '18 at 17:04
  • $\begingroup$ Unfortunately this does not make the question any clearer (in fact unfortunately, I think introducing the $h$ function makes it less clear). If $h$ or $p$ are functions it does not make sense to `condition' on $x_1$ being fixed. To do this type of conditioning you need a probability measure. Therefore as far as I can tell: $h(x_1 + x_2 | x_1) = h(x_1+x_2)$. $\endgroup$ – owen88 Feb 11 '18 at 17:10

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