Deriving the distribution of Random variable

Question

I am reading a paper here: http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=7564878

As explained in the following, I cant understand how Eq. 2 is derived by the authors.

$P_I$ is a random variable (Received Power in full-duplex transciever model) and the randomness in $P_I$ is because of random fading power $g = |h|^2$. $P_I$ is given as:

$$P_I = P_T \kappa d^{-\alpha}g$$

$\kappa$, $\alpha$ are constants and $g = |h|^2$. In the following, I cant understand how the authors have come up with Eq. 2.

I will appreciate any help.

Answer 1

Here we have a r.v. $X\sim\text{N}(\mu,\sigma^2)$, and it is required to find the density function for $Y=c\,X^2$, where $c$ is a positive constant.

$$\begin{align}P(Y\le y)&=P(cX^2\le y)\\[2ex] &=P(X^2\le {y\over c})\\[2ex] &=P\left(-\sqrt{y\over c}\le X\le\sqrt{y\over c}\right)\\[2ex] &=F_X\left(\sqrt{y\over c}\right)-F_X\left(-\sqrt{y\over c}\right) \end{align}$$ Differentiating to get the densities ... $$\begin{align}f_Y(y)&={f_X\left(\sqrt{y\over c}\right)\color{blue}{+f_X\left(-\sqrt{y\over c}\right)}\over 2c\sqrt{y\over c}}.\\ \end{align}$$

The portion in blue has been omitted in Equation (2), which is possibly justified by the approximations they are making. In particular, they state that $K$ is "large", which suggests that $v^2$ is only slightly less than $1$, and hence that the variance of the normal distribution is very small. This implies that the distribution of $h$ is very concentrated near the value of $1$, in which case the omitted portion of the density function might be negligible.

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Numerical cross-checks:

Technically, Equation (2) fails to describe a probability density function, because of the mentioned omission. Thus, choosing $\tau>0$ and $0<v<1$, we find, for $f_{P_I}$ given by Equation (2): $$\int_0^\infty f_{P_I}(x)dx \neq 1,$$ although the discrepancy decreases as $v$ increases toward $1.$

Pasting the following into Wolfram Alpha to integrate Equation (2) yields a value less than $1$:

tau=1; v=1/2; integrate (1/(2 sqrt(pi tau x (1-v^2)))) exp(-(sqrt(x/tau)-v)^2/(1-v^2)) dx from x=0 to oo

Pasting the following into Wolfram Alpha to integrate the "corrected" version yields exactly $1$:

tau=1; v=1/2; integrate (1/(2 sqrt(pi tau x (1-v^2)))) (exp(-(sqrt(x/tau)-v)^2/(1-v^2)) +  exp(-(-sqrt(x/tau)-v)^2/(1-v^2)) )  dx from x=0 to oo

Numerical trials suggest that if $0.9\le v < 1$, then Equation (2) integrates to at least $0.998$.