This post is a follow-up to this previous one, based on what I learned from this second one.
Problem Definition
Consider a polygon with vertices $V_1,\dots,V_n \in \mathbb{R}^2$ and let
\begin{aligned} z&=\underbrace{\sum_{j=1}^n \left[(V_{j+1}-V_j) \frac{L}{L_j}\left(\alpha-\frac{\tau_{j-1}}{L}\right)+V_j\right] 1_{\left[\frac{\tau_{j-1}}{L},\frac{\tau_j}{L}\right)}(\alpha)}_{\triangleq \psi(\alpha)}\\ \alpha&\sim p_\alpha(\alpha)\triangleq\mathcal{U}(\alpha; [0,1])=1_{[0,1]}(\alpha) \end{aligned}
where:
- $V_{n+1}\triangleq V_1$
- $L_j \triangleq \lVert V_{j+1}-V_j\rVert$
- $\tau_{j}\triangleq \sum_{i=1}^j L_j$, $\tau_0\triangleq 0$
- $L\triangleq \tau_n$
- $1_S(x)\triangleq 1$ if $x\in S$ and $1_S(x)\triangleq 0$ otherwise
It turns out that, according to the definition of the map $\psi(\alpha)$ above, $z$ is uniformly distributed over the polygon contour. Now consider the following model:
\begin{aligned} y&=z+v\\ z&=\psi(\alpha) \\ \alpha&\sim \mathcal{U}(\alpha;[0,1])\\ v&\sim\mathcal{N}(v; 0, R) \end{aligned}
where $v$ is zero-mean Gaussian noise with covariance $R$. Under the assumption that $z$ and $v$ are independent, I would like to prove that the following function \begin{equation} \mathcal{L}(y|\psi)=\int_0^1 \mathcal{N}(y-\psi(\alpha);0,R) \, \text{d}\alpha \tag{1} \end{equation} represents the likelihood that a measurement $y$ is sampled from the contour of the polygon $V_1,V_2,\dots,V_n$.
A Simpler Example
Consider the model:
\begin{aligned} y&=z+v\\ z&=\widetilde{\psi}(\alpha) \\ \alpha&\sim \mathcal{U}(\alpha;[0,1]^2)\\ v&\sim\mathcal{N}(v; 0, R) \end{aligned}
where $\widetilde{\psi}(\alpha):[0,1]^2\mapsto P$ is a map from the unit square $[0,1]^2$ to the polygon surface $P$ (i.e., the set of all internal and boundary points of the polygon). In this case, $z$ admits the probability density: \begin{equation} p_z(z)=\frac{p_{\alpha}(\widetilde{\psi}^{-1}(z))}{\left|\textrm{det} \frac{\partial \widetilde{\psi}}{\partial \alpha}(\widetilde{\psi}^{-1}(z))\right|} \end{equation} where $\frac{\partial \widetilde{\psi}}{\partial \alpha}$ is the Jacobian matrix of $\widetilde{\psi}(\alpha)$. Under independence, since $y$ is the sum of $z$ and $v$, the likelihood function can be written as the following convolution:
\begin{aligned} \mathcal{L}(y|\psi)&=(p_v \ast p_z)(y)= \int_P p_v(y-z)\,p_z(z)\,\text{d}z\\ &=\int_P p_v(y-z)\, \frac{p_{\alpha}(\widetilde{\psi}^{-1}(z))}{\left|\textrm{det} \frac{\partial \widetilde{\psi}}{\partial \alpha}(\widetilde{\psi}^{-1}(z))\right|}\text{d}z \end{aligned}
Now consider the change of variable $z\triangleq \widetilde{\psi}(\alpha)$. The convolution becomes: \begin{equation} \begin{aligned} \mathcal{L}(y|\psi) &=\int_P p_v(y-\widetilde{\psi}(\alpha))\, \frac{p_{\alpha}(\widetilde{\psi}^{-1}(\widetilde{\psi}(\alpha)))}{\left|\textrm{det} \frac{\partial \widetilde{\psi}}{\partial \alpha}(\widetilde{\psi}^{-1}(\widetilde{\psi}(\alpha)))\right|}\text{d}\widetilde{\psi}(\alpha)\\ &=\int_{\widetilde{\psi}^{-1}(P)} p_v(y-\widetilde{\psi}(\alpha))\, \frac{p_{\alpha}(\alpha)}{\left|\textrm{det} \frac{\partial \widetilde{\psi}}{\partial \alpha}(\alpha)\right|} \left|\textrm{det} \frac{\partial \widetilde{\psi}}{\partial \alpha}(\alpha)\right|\text{d}\alpha\\ &=\int_{[0,1]^2} p_v(y-\widetilde{\psi}(\alpha))\,p_{\alpha}(\alpha) \,\text{d}\alpha\\ &=\int_{[0,1]^2} \mathcal{N}(y-\widetilde{\psi}(\alpha);0, R)\,\text{d}\alpha \end{aligned} \tag{2} \end{equation}
My Difficulty
The main problem in the derivation of $(1)$ is that $p_z(z)$ does not exist, which implies that the likelihood $\mathcal{L}(y|\psi)$ is not simply the probability density of $y$. I am unsure how to define the likelihood $\mathcal{L}(y|\psi)$ and how to show that $(1)$ follows from this definition. Note that I am attempting to "reverse-engineer" $\mathcal{L}(y|\psi)$ from $(2)$ because I observed that several results for the "surface case" hold in the same form for the "contour case." The difficulty lies in re-deriving the "surface results" for the "contour case," as the standard tools of elementary probability theory cannot be directly applied.
My Attempt
I guess that the workflow should be based on probability measures rather than densities, followed by some kind of trick to derive the likelihood. Following this intuition, consider the following probability measure:
\begin{aligned} \mathbb{P}_y(S)&\triangleq \textrm{Prob}(y\in S)\\ &=\textrm{Prob}(\psi(\alpha)+v\in S)\\ &=\textrm{Prob}(v\in S-\psi(\alpha)) \\ &=\int \textrm{Prob}(v\in S-\psi(\alpha)|\alpha)\,p_\alpha(\alpha) \, \text{d}\alpha\\ &=\int_0^1 \textrm{Prob}(v\in S-\psi(\alpha)|\alpha) \, \text{d}\alpha\\ &=\int_0^1 \int_{S-\psi(\alpha)}\mathcal{N}(v;0,R)\, \text{d}v \, \text{d}\alpha\\ &=\int_0^1 \int_{S}\mathcal{N}(v-\psi(\alpha);0,R)\, \text{d}v \, \text{d}\alpha\\ \end{aligned}
Now, assuming the above derivation is correct, the objective is to determine the relationship between $\mathbb{P}_y(S)$ and $\mathcal{L}(y|\psi)$. In a "non-pathological" case, the desired result should follow from the definition $\mathcal{L}(y|\psi)\triangleq\frac{\text{d}\mathbb{P}_y(S)}{\text{d}y}$, where the derivative is understood in the Lebesgue sense.
Pretending that everything works also in the current "pathological case", I would complete the derivation by writing \begin{equation*} \begin{aligned} \mathcal{L}(y|\psi)&\triangleq \frac{\text{d}\mathbb{P}_y(S)}{\text{d}y}= \int_0^1 \underbrace{\frac{\text{d}}{\text{d}y}\left[\int_S\mathcal{N}(v-\psi(\alpha);0,R)\text{ d}v\right]}_{=\mathcal{N}(y-\psi(\alpha);0, R)}\text{ d}\alpha\\ \end{aligned} \end{equation*}
Questions
(1) Does the definition $\mathcal{L}(y|\psi)\triangleq \frac{\text{d}\mathbb{P}_y(S)}{\text{d}y}$ make sense?
(2) Assuming the above definition makes sense, is the derivation of $(1)$ correct?
(3) If my derivation is incorrect, is it still possible to make sense of $(1)$?
