PDF of distances in the circular disk

Question

I am trying to understand a research article. I cannot understand how the authors have come across the pdf (see the eq. in the third Image) . I have share the relevant details below

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Following are the two hypothesis that authors want to test on.

The PDF of the distance between the MUE and the MBS is given as following figure

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Answer 1

For $\mathbb{H}_{1}$, the assumption is that M-UE is uniformly distributed in region II. Letting $l$ be the distance, and fixing some $L$, we have:

$$P(l \leq L) = \frac{\text{Area in Region II at most }L\text{ away from the M-BS}}{\text{Total area in region II}} = \frac{\pi L^{2} - \pi(d - r)^{2}}{\pi(d + r)^{2} - \pi(d - r)^{2}} = \frac{L^{2} - (d - r)^{2}}{4dr}$$

Assuming $d - r \leq L \leq d + r$. This is the CDF of $L$, and the PDF can be found by differentiating w/r/t $L$. The same method works for $\mathbb{H}_0$, note that the area of regions I and III combined is $\pi[(d - r)^{2} - \varepsilon^{2}] + \pi[R^{2} - (d + r)^{2}] = \pi(R^{2} - 4dr - \varepsilon^{2})$, which is where that denominator comes from.

Answer 2

Let's start with $\mathbb H_1$, we assume M-UE is uniformly distributed on region II. Let us call $U$ the region II and let $a(U)$ be its area. Let $g$ be PDF for the position of M-UE. We know that: $$\forall (x,y) \in U:\ g(x,y) = C$$ and $$\int_U g = C \cdot a(U) = 1.$$ Hence $C = 1/a(U)$. Also, a quick calculation shows that $a(U) = \pi(d+r)^2 - \pi(d-r)^2 = 4\pi dr$. Now, we need to get to $l$ somehow, so we will change the coordinate system from $(x,y)$ to $(l,\theta)$. Let's say $(x,y)=\varphi(l,\theta)$. The exact formula is $\varphi_x(l,\theta) = l\cdot\cos\theta$ and $\varphi_y(l,\theta) = l\cdot\sin\theta$. Let $J$ be the determinant of the Jacobian matrix of $\varphi$. One can easily compute that $J(l,\theta) = l$ Lets say $\hat g$ is same PDF as $g$ but with $(l,\theta)$ as coordinates, then $$\hat g(l,\theta) = g(\varphi(l,\theta))\cdot J(l,\theta)= \frac{1}{a(U)}\cdot l.$$ The first equality was simply the substitution rule for measures (and PDFs as well, since PDFs are measures).

We know the PDF for coordinates $(l,\theta)$, now we ask yourself "what is the PDF for $l$"? Well, let $f$ be the corresponding PDF, then $$f(l) = \int_{[0,2\pi]}\hat g(l,\theta)\ d\theta = \frac{2\pi l}{a(U)} = \frac{2l}{4dr}.$$

This is exactly what the formula $f_l$ from the paper tells us for the case $\mathbb H_1$.

One can repeat the same text for $\mathbb H_0$, the only real difference will be that $U$ is now union of regions I and III, and its area $a(U)$ is $\pi R^2 - \pi (d+r)^2 + \pi(d-r)^2 - \pi\varepsilon^2 = \pi (R^2 - 4dr -\varepsilon^2)$ ($\varepsilon$ comes from the fact that authors assume M-UE can't be too close to M-BS, so the ball of the size $\varepsilon$ is excluded). This will also change the final result to $f(l) = \frac{2l}{R^2 - 4dr -\varepsilon^2}$, which you can also recognize in the final formula.