RFs self study notes (III): clutter model - first determine the number with Poisson distribution, and then use uniform distribution as probability distribution

The clutter model is derived from the binomial distribution , namely $n$ Heavy Bernoulli distribution , When $n$ As we go to infinity , At this time, it can be approximated as Poisson distribution $$C_n^i{p}^i(1-p{)}^{n-i}=\frac{{\lambda }^i{e}^{-\lambda }}{i!}$$
among ,$n$ Indicates the total number ,$i$ Indicates the number of successes ,$p$ Represents the probability of success in Bernoulli distribution ,$\lambda$ It's Poisson's ratio

Number of clutter points

Suppose the field of vision is a two-dimensional plane , Clutter and objects will appear in this two-dimensional plane , Let the area of this two-dimensional plane, that is, the field of view, be $V$, Now I want to know how many clutter will appear at a certain time , How to estimate the number of clutter ？
At first glance, there are few conditions for this problem , What we should be able to think of is ： If the number of clutter and the distribution of clutter itself can conform to some probability distribution models we have learned , For example, binomial distribution 、 Poisson distribution 、 Exponential distribution and so on , Because in this way, it is easier for us to estimate —— Abstract and simplify the problem into the probability model we have learned, so as to establish equations to complete the estimation task , This is the process of modeling ？ therefore , Many of the algorithms we learn are nothing more than a process of modeling , It is a great test of basic math skills .
For this question , Let's model it like this ： View area $V$ Divide evenly into $n$ A small square , There is either a clutter in each small square , Or there is no clutter , That is, in a single small square , The number of clutter conforms to Bernoulli distribution , The clutter itself conforms to the uniform distribution within this square
$$\left(\begin{array}{ccc}Cnt& 1& 0\\ P& p^c& 1-p^c\end{array}\right)$$
among ,$Cnt$ Represents the number of clutter in a small square , The value can only be 1 or 0,P Represents the probability value corresponding to their respective values ：$p^c$ Indicates the probability of success, that is, the probability of a clutter in a small square is $p^c$

There is a Bernoulli distribution in each small square , Then the whole field of vision $n$ The distribution composed of three small squares is $n$ The heavy Bernoulli distribution is the binomial distribution , Using the formula of binomial distribution, we can calculate ： When the number of clutter is $i$ The probability value corresponding to each hour is the probability that the number of clutter in the field of vision is $i$ individual $$P(|C_k|=i)=C_n^i(p^c{)}^i(1-p^c{)}^{n-i}$$ among ,$C_k$ Represents the clutter variable , It's a collection , contain $k$ All clutter points at the moment ,$C_k=[C_k^1,C_k^2,...,C_k^m]$.$|C_k|$ The potential of this set is the number of clutter points

When the field of vision $V$ When the division is very fine, that is $n\to \infty$ when , Binomial distribution becomes Poisson distribution . in other words , The number of clutter in the whole field of view obeys Poisson distribution , Set variables $m_k^c=|C_k|$, It represents the number of clutter , be $m_k^c$ Obey Poisson distribution $$m_k^c\sim Po({\lambda }_c)$$ among ,${\lambda }_c$ It's Poisson's ratio , We need to decide by ourselves

Observe the calculation formula of binomial distribution ：$p^c$ and $n$ It is a parameter that needs to be determined by ourselves , among ,$n$ It reflects the fineness of our division ,$p^c$ It reflects the possibility or intensity of clutter —— The number of times clutter appears in a small square in unit time . And the product of the two $n{p}^c$ It represents the clutter intensity in the whole field of view —— The number of times clutter appears in the whole field of view in unit time . It should be noted that ： The mean value of binomial distribution happens to be $n{p}^c$, It means , We can get the number of clutter per unit time in the field of vision by statistical means , And you get $n{p}^c$ Value .
Observe the calculation formula of Poisson distribution ： Average strength ${\lambda }_c$ It is a parameter that needs to be determined by ourselves . actually ,${\lambda }_c=n{p}^c$, It is consistent with binomial distribution , Including the meaning is the same .

Sum up ： We have completed the number of clutter $m_k^c$ Obey Poisson distribution $Po({\lambda }_c)$, Poisson's ratio ${\lambda }_c$ You need to be sure . In this way, the number of clutter in the field of vision at a certain time can be estimated .

Specific distribution

Up to now , We just get the model of the number of clutter, that is, we just know how many clutter will appear at a certain time , however , What we finally need to know is the distribution characteristics of these clutter points , So that we can distinguish clutter points from real object points . therefore , Next, we need to solve the probability distribution of clutter .
Here we think that the clutter points themselves obey the uniform distribution in the field of view —— The probability of clutter points appearing anywhere in the field of vision is the same $$\begin{gathered} C_k^i\sim unif(V)=f_c(c) \\ \lambda (c)=\{\begin{array}{cc}\lambda & c\in V\\ 0& others\end{array} \end{gathered}$$ among ,$C_k^i$ It means $k$ At the moment $i$ Clutter points ,$\lambda$ It represents the intensity of the whole field of view ( Because it reflects the number , So general $\lambda$>0) Not a probability distribution , It mainly corresponds to the Poisson distribution, which corresponds to the estimated number of clutter . If it is used to represent the probability distribution of clutter points , It needs to be transformed into a form of uniform distribution, that is, the next standardization .

in addition $$\begin{gathered} {\lambda }_c=\int \lambda (c)dc \\ {f}_c(c)=\frac{\lambda (c)}{{\lambda }_c}=\{\begin{array}{cc}\frac{1}{V}& c\in V\\ 0& others\end{array} \end{gathered}$$
This is to standardize it ：$dc$ Integral actually means traversing the entire field of view , If the intensity of each position in the field of vision is the same, that is $\lambda (c)=constan$ Is constant , Then this can be simply expressed as ${\lambda }_c=\lambda (c)V$ That is, the strength is multiplied by the area , The average intensity over the whole field of view is obtained ——${\lambda }_c$ It represents the average intensity of the visual field ,$\lambda (c)$ It represents the strength at a certain position （ When the intensity of each position point is not exactly equal ）. The reason why it is written in the form of integral is that the strength at each position may not be the same ,$\lambda (c)dc$ Represents the number of clutter on a small square , Taking the integral again means ： Add up the clutter number of all squares to get the average clutter number, that is, the average intensity . And yes $\lambda (c)$ After standardization, we get $f_c(c)$, It represents the probability distribution of clutter points —— It was $\lambda (c)$ Represents the distribution of a clutter point , But because it involves the whole field of vision, that is, the unified relationship of many clutter points , To ensure that the probability integral is 1, Therefore, the original distribution needs to be standardized to ensure that the final full integration result is 1.
for instance ： set up $V=4,n=4,\lambda (c)=2$, Then the area of the small square divided is $v=V/n=1$, The average number of clutter on the small square is $v\lambda (c)=2$, Then the expectation of the number of clutter in the whole field of vision is ${\lambda }_c=\int \lambda (c)dc=V\lambda (c)=8$.

Before standardization 1 The distribution of clutter points on the small square No. is $$\lambda (c{)}_1=\{\begin{array}{cc}2& c\in v_1\\ 0& others\end{array}$$ It means that $v_1$ In the area , The number of clutter per unit area per unit time is 2 individual ？. Here we think that the intensity of the four small squares is the same , therefore $\lambda (c{)}_1,\lambda (c{)}_2,\lambda (c{)}_3,$$\lambda (c{)}_4$ The value of is the same , It's just that the targeted areas are different .

After standardization 1 The distribution of clutter points on the small square No. is $$f_c(c{)}_1=\{\begin{array}{cc}\frac{2}{{\lambda }_c}=\frac{1}{4}& c\in v_1\\ 0& others\end{array}$$ This means 1 The strength of the small square $\lambda (c{)}_1$ In overall strength ${\lambda }_c$ The proportion of the , It can reflect how big the clutter points may come from 1 Small square No ？ Because the intensity of the four small squares is the same , That is to say, for an existing clutter point , It is possible that it comes from any position in the field of vision —— Uniform distribution , therefore $$\begin{gathered} f_c(c{)}_1=f_c(c{)}_2=f_c(c{)}_3=f_c(c{)}_4 \\ =\{\begin{array}{cc}\frac{2}{{\lambda }_c}=\frac{1}{4}=\frac{1}{V}& c\in V\\ 0& others\end{array} \end{gathered}$$
It seems that the understanding of standardization is not very good , Let's add later . The question is ： clarify ${\lambda }_c,\lambda (c),V,v,f_c()$ The relationship between ？

The solution of clutter probability distribution is divided into two steps （ Here we need to review the knowledge of classical probability and joint probability distribution ）：
1. Determine the number of clutter points according to Poisson distribution —— When estimating the number, it corresponds to a probability value
2. Calculate the corresponding probability value of each clutter point , Then tired ride
$$\begin{gathered} P(C_k)=P(C_k,m_k^c)=P(m_k^c)P(C_k|m_k^c) \\ =Po(m_k^c;{\lambda }_c)\prod _{i=1}^{m_k^c}{f}_c(c_k^i)=\frac{e^{-{\lambda }_c}}{m_k^c!}\prod _{i=1}^{m_k^c}\lambda (c_k^i) \end{gathered}$$
What did we finally get ？
The question is ： Explain this formula with the relationship between events ？

The sorting of clutter model comes here first , There are still some places that are not clear enough , We need to take a closer look , I'll add later