I am working in wireless communication and some times we use the PDFs and CDFs of random variables.
So I have read a paper and I found the derivation of CDF and PDF of random variable, but I did not understand the steps of the proof.
Let $X_1$ and $X_2$ two random variable with PDFs
$$
f_{X_i}(x)=\frac{m_i^{m_i}}{\Omega_i^{m_i}\Gamma(m_i)}x^{m_i-1}e^{-\frac{m_ix}{\Omega_i}}.
$$
i want to find the CDF of random variable $Z$ define by
$$
Z=\frac{X_iX_j}{2X_i+X_j}
$$
Now the athours of paper use the following steps:
\begin{align}\label{}
F_Z(z)=&\mathbb{P}\left\{Z\leq z\right\} \\
=&\mathbb{P}\left\{\frac{X_iX_j}{2X_i+X_j}\leq z\right\} \\
\stackrel{(a)}{=}&\int_{0}^{\infty}\mathbb{P}\left\{\frac{X_iX_j}{2X_i+X_j}\leq z|X_i\right\}f_{X_i}(X_i)dX_i\\
\stackrel{(b)}{=}& \int_{0}^{z}
\mathbb{P}\left\{X_j\geq\frac{2zX_i}{X_i-z}|X_i\right\}
f_{X_i}(X_i)dX_i\\
&+
\int_{z}^{\infty}
\mathbb{P}\left\{X_j\leq\frac{2zX_i}{X_i-z}|X_i\right\}
f_{X_i}(X_i)dX_i\\
=&I_1(z)+I_2(z)
\end{align}
where
$$
I_1(z)\stackrel{(c)}{=}\int_{0}^{z}f_{X_i}(X_i)dX_i
$$
$$
I_2(z)\stackrel{(d)}{=}\int_{z}^{\infty}F_{X_j}(\frac{2zX_i}{X_i-z})f_{X_i}(X_i)dX_i.
$$
My questions is are
Firs which low of probability is used in step (a)?.
Why hi divide integral in two range, from $0$ to $z$ and from $z$ to $\infty$
Finally please some detail on steps (b), (c) and (d).
Thanks
Best Answer
A bit more detail on (b):
$$\frac{X_iX_j}{2X_i+X_j}\leq z \Leftrightarrow X_iX_j \leq 2zX_i+zX_j$$
and
$$X_iX_j \leq 2zX_i+zX_j \Leftrightarrow X_j(X_i-z) \leq 2zX_i \; .$$
Now, in the last step, you have two possible cases: $X_i>z$ or $X_i < z$. If $X_i < z$ then the inequality is trivially true (because the $X_i$ are positive r.v.) and thus has probability 1. Hence why you end up with (c). The other case gives you (d).