Why exponential distribution is used to represent channel gain between two communicating entities? In many many papers the channel fading is represented by a random variable $h$ whose pdf follows exponential distribution. My question is why this is used? By using this pdf we may have higher received power at the destination than the transmitted power at the source. There must be something very crucial that I am missing. I will be very thankful if somebody can explain it how it works? Thanks in advance.
 A: This means the channel gain is Rayleigh distributed, which characterizes the multi-path propagation of signals. In other words, $\sqrt{h}$ is Rayleigh distributed, so $h$ is exponentially distributed. Also, in many cases, the exponential distribution gives nice closed-form expression.
I think you are considering the singular path loss model: $\|x\|^{-\alpha}$, where $\|x\|$ is the distance between the source and the destination $\alpha > 0$ is the path loss exponent. So if $\|x\| < 1$, the signal gets amplified. But in practice, this is not the case. The singular path loss model is usually considered due to its tractability and has its drawback. A better path loss model is to consider $\min\lbrace1, \|x\|^{-\alpha} \rbrace$.
A: Path loss model is just a model, which made many simplification. Those simplification introduces many constrain of the model, for example, where it holds.
In a multipath model, each path actual is still summation of many indiscriminable sub-path. If each sub-path have similar power, it is easy to derived Rayleight path model if we assume there are infinite sub-path. The infinite sub-path assumption introduces the contradiction that received power is larger than transmitted power.
In actual case, the sub-path number is a limit number. When the path distance is large enough, we assume number of the sub-path is large enough then Rayleight model still hold approximately.
