# The CDF of the summation of independent and dependent random variables

I want to evaluate the following probability

$$\text{Pr}\left\{\frac{Y_1}{X_1}+\frac{Y_2}{X_2}\leq z\right\}$$

where the support of all random variables is $[0,\,\infty)$, but $Y_1\leq Y_2$, i.e., they are dependent random variables, but mutually independent from $X_1$ and $X_2$ (i.e., $X_1$ and $X_2$ are independent from $Y_1$ and $Y_2$). Also, $X_1$ and $X_2$ are independent and identically distributed (i.i.d.) random variables (RVs).

If $Y_1$ and $Y_2$ are i.i.d. RVs, then it is easy to find the above probability as

$$\int_{z_1=0}^{\infty}F_Z(z-z_1)f_Z(z_1)\,dz_1$$

where $Z_k=Y_k/X_k$ for $k=1,\,2$. and $F_Z(z)$ and $f_Z(z)$ are the CDF and PDF of the random variables $\{Z_k\}_{k=1}^2$.

But if $Y_1$ and $Y_2$ are dependent, how can I find the above probability?

• If $Y_1$ and $Y_2$ are dependent, the obvious question is what is known about their joint distribution (knowing $Y_1 \leqslant Y_2$ and their individual distributions is not enough). – metamorphy Jul 29 '18 at 2:24
• @metamorphy I just wanted the result in terms of their joint distribution as a general solution. But if it helps, all random variables (before ordering) are exponential random variables with parameter 1. I am not looking for a closed-form solution by the way. – BlackMath Jul 29 '18 at 2:37

## 1 Answer

In terms of the joint distribution of $Y_1$ and $Y_2$ we have $P\{\frac {Y_1} {X_1} +\frac {Y_2} {X_2} \leq z\}=\int \int \iint_{\{\frac u {x_1}+\frac v {x_2}\} \leq z} d_{Y_1,Y_2}(u,v)dF_1(x_1)dF_2(X_2)$. The region ${\{\frac u {x_1}+\frac v {x_2}\} \leq z}$ can be written as $\{u\leq v \leq x_2(z-\frac u {x_1}), 0\leq u \leq z x_1\}$. So yuo can integrate w.r.t $v$ from $u$ to $x_2(z-\frac u {x_1})$ and then w.r.t $u$ from $0$ to $z x_1$.

• How did you derive the limits? This is the part that I don't understand. – BlackMath Jul 29 '18 at 13:31
• Would you share your thoughts? As I mentioned, the limits are troublesome for me. I am not sure how to separate the four random variables. – BlackMath Jul 30 '18 at 2:10
• First fix $u$ see what the possible values of $v$ are. The inequality $u\leq v$ comes from $Y_1 \leq Y_2$. The inequality $v\leq x_2(z-\frac u {x_1})$ comes from $\frac u {x_1} +\frac v {x_2} \leq z$.. Finally, $u\leq zx_1$ comes from $\frac u {x_1} \leq \frac u {x_1} +\frac v {x_2} \leq z$. You have to make sure that you have not missed any constraint before you start integrating. – Kavi Rama Murthy Jul 30 '18 at 5:41
• Thank you. The $u\leq zx_1$ is what I was looking for. So, now we can write the limits as $$\int_{x_1=0}^{\infty}\int_{u=0}^{x_1z}\int_{x_2=0}^{\infty}\int_{v=0}^{x_2\left(z-\frac{u}{x_1}\right)}f_{Y_1, Y_2}(u,v)\,f_{X}(x_1)f_X(x_2)\,du\,dv\,dx_1\,dx_2$$ – BlackMath Jul 30 '18 at 14:49