# Is it possible to derive the CDF of $Z$?

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?

• What do you know? What did you try? What similar simpler problems can you solve? For example, how would you compute the PDF/CDF of $M=\max\limits_kY_k$? – Did Mar 25 '12 at 14:34
• Hi Didier Piau, Thanks Didier Piau for your concern. I have made my question more clear. Could you please give me a hint ? – Tran Tam Mar 25 '12 at 16:06
• Quote: For example, how would you compute the PDF/CDF of $M=\max\limits_kY_k$? Unquote. – Did Mar 25 '12 at 16:33
• Hi, here I calculate $f_Y(y)=K f_{Y_k}(y_k)\biggl(F_{Y_k}(y_k)\biggr)^{K-1}$. I already have $f_{X}(x)$ also. I want to verify that the above formula to derive the CDF of $Z$ is correct or not ? Thanks for your concern! – Tran Tam Mar 25 '12 at 17:56

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 $$f_Z(z)=a^{-1}\iint f_0(a^{-1}zy(x+1))\cdot yf_Y(y)\cdot (x+1)f_X(x)\cdot \mathrm dx\mathrm dy.$$