# CDF of a function of two random variables

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 $E\left\{Z\right\}$.

• 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. – Chamberlain Foncha Mar 6 '18 at 16:56
• Your derivation is wrong because you are assuming that $Z$ has the pdf $f(y_1)f(y_2)$ which is not true. – Chamberlain Foncha Mar 6 '18 at 16:59
• @ChamberlainFoncha I have edited the question. Does it make sense now? – din Mar 6 '18 at 17:23
• 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 $F(y)$. – whuber Mar 6 '18 at 17:44
• @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)? – din Mar 6 '18 at 18:04

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 $c=\log(a)/b$.

$\mathbf{E(Z)}$

Then

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

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.

$\mathbf{P(Z>z)}$

$z = p(y_1)p(y_1+y_2) = \begin{cases} a^2 \exp\left( - 2b y_1 - b y_2 \right), & \text{for$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: • 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 (?). – din Mar 14 '18 at 9:31
• 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 ? – din Mar 14 '18 at 9:34
• 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? – din Mar 15 '18 at 10:28
• Regarding the CCDF, would you please explain why we need: $\int \int f(y_1) f(y_2) dy_1 dy_2$ ? – din Mar 15 '18 at 13:46
• 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. – din Mar 15 '18 at 14:44