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?
Best 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 $z x_1$.