Two different models of Rician fading

Question

The amplitude of the signal in Rician fading case is characterized by the Rician distribution (let's call this model 1). However, the received power in Rician fading is sometimes modelled as (model 2): \begin{equation} \label{model2} P = q + (1-q)p, \end{equation} where $p$ is exponential of mean $1$ ,and $q \in [0,1]$ is the part of the energy received in the line-of-sight.

Probability density function for $P$ can be derived as a convolution of $ p \mapsto \delta(p-q)$ and $p \mapsto u(p)\frac{e^{-\frac{ p}{1-q}}}{1-q}:$ $$\text{PDF}_{\text{model 2}}(p) = \frac{e^{-\frac{ p- q}{1-q}}}{1-q}u(p-q), $$ where $u(\cdot)$ is a step-function. (This can be seen from the fact that Laplace transform of $P$ is a product of Laplace transforms of these functions.)

In the plot that follows, for model 1 we acquire PDF for received power by simulating the Rician distribution for amplitude and taking the second exponent. We use Rician parameters s.t. $K= 1$, $\Omega = 1$.

The distribution of received power is clearly different for $q \neq 0$ in model 1 and model 2. Case $q=0$ is, of course, the Rayleigh-fading case.

My question is: How do these two models relate? It seems that model 2 neglects the change that signals received from non-line-of-sight paths would add destructively to the line-of-sight signal. Signal power never goes under $q$ (which is the line-of-sight element). In model 1 the signal-power can go to zero.