Define $Y_1$ and $Y_2$ to be two positive and independent random variables, for which the pdf (probability density function) is the same and is given as: $f(y) = \beta \exp\left(- \beta y\right),\beta>0$.
Suppose that time is slotted, and $t$ ($=0,1,2,...$) is the time index. $Y_1$ is associated with time $t$ and $Y_2$ with time $t+1$.
For $a,b,c>0$, $a>1$ and $c=\frac{\log(a)}{b}$, we define $p(s)$ to be some error probability that is a function of $s$, which is given as follows \begin{equation} p(s) = \begin{cases} a \exp\left( - b s \right), & \text{for $s \ge c$} \\ 1, & \text{for $0 < s < c $} \end{cases} \end{equation} At time $t$, $s=y_1$. And at time $t+1$, $s=y_1+y_2$.
Let $E_1$ represent the event of having an error at time $t$; the corresponding probability is $p(y_1)$. Also, define $E_2$ to be the event that there is an error at time $t+1$; the corresponding probability is $p(y_1+y_2)$. For $E_2$ to happen (at $t+1$), a necessary condition is that $E_1$ happens (at $t$).

Let $E$ be the event that there is an error at time $t$ and $t+1$. So we have $\mathbb{P}\{ E \}= \mathbb{P}\{ E_1, E_2 \}= \mathbb{P}\{ E_1 \} \mathbb{P}\{ E_2 \mid E_1 \} = p(y_1) p(y_1+y_2)$. So if the realisations $y_1$ and $y_2$ of $Y_1$ and $Y_2$ are known, the (global) error probability is $p(y_1) p(y_1+y_2)$.

I am interested in the case where at $t$ and $t+1$ the realisations of, respectively, $Y_1$ and $Y_2$ are not known. Let $Z= p(Y_1) p(Y_1+Y_2)$. So in this case I want to derive the expected value of $Z$. I also want to derive the CCDF (or CDF) of $Z$.

Here is my first attempt of a solution \begin{eqnarray} \mathbb{E} \left\{Z\right\} &=&\int_0^\infty \int_0^\infty p(y_1) p(y_1+y_2) \, f(y_1) f(y_2) \,dy_1 dy_2 \\ &= &\int_0^c \int_0^c \beta \exp\left(- \beta y_1 \right) \beta \exp\left(- \beta y_2 \right) dy_2 dy_1\\ & +& \int_0^c \int_{c-y_1}^\infty a\exp\left(-b(y_1 +y_2)\right) \beta \exp\left(- \beta y_1 \right) \beta \exp\left(- \beta y_2 \right) dy_2 dy_1 \\ &+& \int_c^\infty \int_{0}^\infty a\exp\left(-b y_1\right) a\exp\left(-b(y_1 +y_2)\right) \beta \exp\left(- \beta y_1 \right) \beta \exp\left(- \beta y_2 \right) dy_2 dy_1. \end{eqnarray}
Is the above derivation correct?

CCDF $= Pr\left\{Z > z \right\} =$ ?

Please note that if explicit expressions are difficult to derive, I need to at least write these expressions as integrals function of $a$, $b$, $c$, and $\beta$; as done for $E\left\{Z\right\}$.

  • $\begingroup$ You need the distribution of $Z$ to make any expectation statements. Which starts by knowing the distribution of $Y=Y_1+Y_2$ which only exist if the support of $Y_1$ or $Y_2$ is bounded above by, say, $k$. In that case, $Y$ has pdf $k\beta f(y_1)$ as a convolution. $\endgroup$ Commented Mar 6, 2018 at 16:56
  • $\begingroup$ Your derivation is wrong because you are assuming that $Z$ has the pdf $f(y_1)f(y_2)$ which is not true. $\endgroup$ Commented Mar 6, 2018 at 16:59
  • $\begingroup$ @ChamberlainFoncha I have edited the question. Does it make sense now? $\endgroup$
    – din
    Commented Mar 6, 2018 at 17:23
  • $\begingroup$ The approach is correct, but the integral breaks into four parts, not three. You will have an easier time by using a more abstract notation at the outset, because many of the resulting univariate integrals will be directly computed by the CDF $F(y)$. $\endgroup$
    – whuber
    Commented Mar 6, 2018 at 17:44
  • $\begingroup$ @whuber I am only getting 3 parts: (1) $y_1<c$ and $y_1+y_2<c$, (2) $y_1<c$ and $y_1+y_2>c$, (3) $y_1>c$ and $y_1+y_2>c$. If I am not mistaken, the fourth part is $= 0$ since $y_1 >c$ and $y_1+y_2<c$ cannot happen. Do you have any idea on how to calculate CCDF(Z)? $\endgroup$
    – din
    Commented Mar 6, 2018 at 18:04

1 Answer 1


Starting point

$$z = p(y_1)p(y_1+y_2)$$ $$y_1 \sim f(y)$$ $$y_2 \sim f(y)$$ \begin{equation} p(s) = \begin{cases} a \exp\left( - b s \right), & \text{for $s \ge c$} \\ 1, & \text{for $0 < s < c $} \end{cases} \end{equation} $$f(y) = \beta \exp\left(- \beta y\right),\beta>0$$ with $c=\log(a)/b$.



$$E(Z) = \int_0^\infty \int_0^\infty p(y_1)p(y_1+y_2) f(y_1)f(y_2) dy_1 dy_2$$

but it should be split into different integrals (Whuber was right that you are missing a piece). You have the line $y_1>c$ and $y_1+y_2>c$ giving you

\begin{eqnarray} \mathbb{E} \left\{Z\right\} &= & \beta^2 \int_0^c \int_0^{c-y_1} \exp\left(- \beta y_1 - \beta y_2 \right) dy_2 dy_1\\ & +& a \beta^2 \int_0^c \int_{c-y_1}^\infty \exp\left(-b\left(y_1 +y_2\right)- \beta y_1 - \beta y_2 \right) dy_2 dy_1 \\ &+& a^2 \beta^2 \int_c^\infty \int_{0}^\infty \exp\left(-b y_1-b(y_1 +y_2)- \beta y_1 - \beta y_2 \right) dy_2 dy_1 \end{eqnarray}

The explicit expression is not difficult to derive from here, yet it is very cumbersome. Possibly certain simplifications are possible when you plug-in the value for $c$ in the final (integrated) expression.


$z = p(y_1)p(y_1+y_2) = \begin{cases} a^2 \exp\left( - 2b y_1 - b y_2 \right), & \text{for $y_1 \ge c$} \\ a \exp\left( - b y_1 - b y_2 \right), & \text{for $y_1 < c$ and $y_2 \ge c-y_1$} \\ 1, & \text{for $y_1 < c$ and $y_2 < c-y_1$} \end{cases} $

Then you need to integrate $\int\int f(y_1)f(y_2) d y_1 d y_2$ in the region where $Z>z$ by applying the appropriate boundaries:

$$\begin{array}\\ 1: \qquad &- b y_1 - b y_2 > \log(z/a)\\ 2: \qquad &- 2b y_1 - b y_2 > \log(z/a^2) \end{array}$$

This requires you, again, to split up the integral.

See also the below image for the level of $z$ as a function of $y_1$ and $y_2$. You will need to integrate using the two before mentioned borders, but it is a bit nasty to write it out in full so I will leave it here with the image:

z as function of y1 and y2

  • $\begingroup$ In the expression of $E\{Z\}$ you provide, I think that the integral limits in the first line should be as follows: $\int_0^c \int_0^c$. This corresponds to the case: $y_1<c$ and $y_1+y_2<c$, so $y_2<c-y_1<c$. As for $y_1 >c$ and $y_1+y_2>c$, it is represented in the third line. If my observations are correct, then the expression of $E\{Z\}$ that I provide is correct (?). $\endgroup$
    – din
    Commented Mar 14, 2018 at 9:31
  • $\begingroup$ Regarding $P\{ Z>z\}$, I am sorry but I am not sure I have understood. Would you please elaborate a bit more, or at least write $P\{ Z>z\}$ as a function of some integrals ? $\endgroup$
    – din
    Commented Mar 14, 2018 at 9:34
  • $\begingroup$ You mean: the case $y_1<c$ and $y_1+y_2<c$ yields $y_2<c-y_1$ ? But in this case, in the first line there should be $\int_0^c \int_0^{c-y_1}$ and not $\int_0^c \int_0^{y_1}$. Is it correct? $\endgroup$
    – din
    Commented Mar 15, 2018 at 10:28
  • $\begingroup$ Regarding the CCDF, would you please explain why we need: $\int \int f(y_1) f(y_2) dy_1 dy_2$ ? $\endgroup$
    – din
    Commented Mar 15, 2018 at 13:46
  • $\begingroup$ I am going to award you the bounty. But please, I need the integral bounds; I don't know how to split the integral for the CCDF expression you provide. $\endgroup$
    – din
    Commented Mar 15, 2018 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.