Assume that $X_i$, $Y_k$, $i=0,\ldots,N$, $k=1,\ldots,K$ are non-negative independent non-identically distributed random variables. Let us define the random variable $Z$ as \begin{align} Z=\frac{aX_0} { \max\limits_{k=1,\ldots,K}Y_k\cdot \left( \sum \limits_{i=1}^{N} b_iX_i +1 \right)} \end{align} where $a$, $b_i$ are non-negative constants. The PDF of $X_i$ and $Y_k$ are defined as $f_{X_i}(x_i)$ and $f_{Y_k}(y_k)$, respectively. The CDF of $X_i$ and $Y_k$ are defined as $F_{X_i}(x_i)$ and $F_{Y_k}(y_k)$, respectively. How could I find the CDF of $Z$?
More specifically, I am going to find the CDF of $Z$ as follows: \begin{align} F_Z(z)=\Pr ( Z < z )=\int\limits_0^\infty \int \limits_0^\infty \Pr \left( \frac{aX_0} { y\left( x + 1 \right)} < z \right) f_Y(y) \; dy \; f_X(x) \; dx \; dy \end{align} where $Y=\max\limits_{k=1,\ldots,K}Y_k$ and $X=\sum \limits_{i=1}^N b_iX_i$, $f_Y(y)$ and $f_X(x)$ are PDFs of $X$ and $Y$, respectively. Note that all random variables are distributed following independent non-identically distributed exponential distributions.
I am not sure whether this formula is correct or not? Could you please verify it for me please?
In the end (see comments), it seems that the question is to write down the CDF $F_Z$ and/or the PDF $f_Z$ of $Z$ using the CDFs and/or the PDFs of $X_0$, $Y$ and $X$, where $$ Z=\frac{aX_0}{Y(X+1)}. $$ One assumes that $a\gt0$ and that $X_0$, $X$ and $Y$ are nonnegative and independent with respective CDFs $F_0$, $F_X$ and $F_Y$ and respective PDFs $f_0$, $f_X$ and $f_Y$.
For every $z\geqslant 0$, $$ [Z\leqslant z]=[aX_0\leqslant zY(X+1)], $$ hence $F_Z(z)=\mathrm P(aX_0\leqslant zY(X+1))$ is, by definition, $$ F_Z(z)=\int\!\!\!\iint_{au\leqslant zy(x+1)}f_0(u)f_Y(y)f_X(x)\mathrm dx\mathrm dy\mathrm du, $$ or, equivalently, $$ F_Z(z)=\iint F_0(a^{-1}zy(x+1))f_Y(y)f_X(x)\mathrm dx\mathrm dy. $$ Differentiating this with respect to $z$, one gets $$ f_Z(z)=a^{-1}\iint f_0(a^{-1}zy(x+1))\cdot yf_Y(y)\cdot (x+1)f_X(x)\cdot \mathrm dx\mathrm dy. $$