RFs self study notes (4): actual measurement model - the mixture of OK and CK, and then calculate the likelihood probability

It involves the knowledge of joint probability distribution , Granulate the formula to facilitate computer calculation

The theoretical measurement model was introduced earlier $O_k$ And clutter model $C_k$, Here is the actual measurement model $Z_k$, among ,$Z_k=(O_k,C_k)$, Express $O_k,C_k$ Random arrangement means that the order is random . Think back to Bayesian filtering , What is the function of establishing measurement model ？ Measure the likelihood probability of the model , Then we can use Bayesian formula to solve the posterior probability . So-called , Update step with Bayesian formula and measurement model .

Because the number of clutter is uncertain , This leads to the number of final measured values $|Z_k|$ It's also uncertain , Set variables $m=|Z_k|$, Represents the number of element points contained in the measurement set . In addition, the concept of hypothesis needs to be introduced $$\theta =\{\begin{array}{cc}i>0& if-Z^i-is-an-object-detection\\ 0& if-undetect-object\end{array}$$ for instance ： Suppose now $Z=[0.1,5.0,3.2,6.0]$ Yes 4 Measurements , this 4 All the measured values may be clutter, that is, no object is detected this time , It may also be that one of them is an object and the other three are clutter —— This is the assumption we started .

The first 0 The first assumption is $\theta =0$ when ： Indicates that no object has been detected , That is to say, this time $Z$ The values in are all clutter . Then we can calculate the possibility of this situation according to the clutter model and the measurement model ： If the detection probability of the sensor $P^D=1$, That is to say, it is impossible to miss the inspection , So obviously this is the first 0 This assumption is very unreliable , The probability that this assumption can be established is 0. If the detection of the sensor is successful $P^D=0.9$, That is to say, it may be missed , So this is 0 A hypothesis can happen , You need to calculate the number 0 The probability that a situation represented by a hypothesis can occur , At this time, after adding this assumption, this problem “ reduce to ” It's like this ： In the known clutter model and measurement model, as well as the measured value this time $Z$ after , The measured values this time are all clutter, that is, they all come from the clutter model , Find the probability of this happening ？ First, it is judged that there is no object detected ：$1-P^D$; Then the number of clutter conforms to Poisson distribution , Calculate the probability value corresponding to the number of clutter ; Finally, ask $Z_k^i$ It's all clutter , Find the corresponding probability value , Then you can get tired . Calculation formula ：$$\begin{gathered} (1-P^D)Po(m;{\lambda }_c)\prod _{i=1}^m{f}_c(Z^i)= \\ (1-P^D)\frac{e^{-{\lambda }_c}}{m!}\prod _{i=1}^m\lambda (Z^i)= \\ (1-0.9)\frac{e^{-3}}{4!}\lambda (0.1)\lambda (5.0)\lambda (3.2)\lambda (6.0) \end{gathered}$$ here , We take the average Poisson's ratio as 3,$Z$ The number of elements in is obviously 4, The other is to bring $Z_k^i$ Into the intensity function . We can also determine an intensity function arbitrarily here $$f_c(c)=\{\begin{array}{cc}\frac{1}{4}& c\in [3,7]\\ 0& others\end{array}$$ and $\lambda (c)={\lambda }_c{f}_c(c),$ Then it can be evaluated ：$0.1\ast \frac{e^{-3}}{4!}\ast (3/4{)}^3$
Something doesn't feel right ！！！—— This $\lambda (0.1)$ What is the value of ？ yes 0 still 1？（ What is uniform distribution ？？ Probability density and probability distribution should be able to distinguish ！）

The first 1 The first assumption is $\theta =1$ when ： Express $Z^{\theta }=Z^1$ It's an object , The other three are clutter . Obviously, this situation exists , We also need to calculate the probability of this happening . problem “ reduce to ”： In the known clutter model and measurement model, as well as the measured value this time $Z$ after , Among the measured values this time $Z^{\theta }=Z^1$ It is the object that comes from the measurement model , The rest are clutter, that is, they all come from the clutter model , Find the probability of this happening ？ The calculated probability is ：$$\begin{gathered} P^{D}Po(m-1;{\lambda }_c)\frac{1}{m}{g}_k(z^{\theta }|x)\frac{{\prod }_{i=1}^m{f}_c(z^i)}{f_c(z^{\theta })} \\ =P^D\frac{g_k(z^{\theta }|x)}{\lambda (z^{\theta })}\frac{e^{-{\lambda }_c}}{m!}\prod _{i=1}^m\lambda (z^i) \\ =0.9\ast \frac{g_k(z^1|x)}{\lambda (z^1)}\frac{e^{-3}}{4!}\lambda (z^1)\lambda (z^2)\lambda (z^3)\lambda (z^4) \end{gathered}$$ there $g_k()$ Is the probability distribution of the theoretical measurement model , Reflect the measurement accuracy of the sensor , When simplifying problems , It can be said that $g_k()$ It's a normal distribution , The reference quantity is $x$.

The first 2 The first assumption is $\theta =2$ when ： Express $Z^{\theta }=Z^2$ It's an object , The other three are clutter .

The first 3 The first assumption is $\theta =3$ when ： Express $Z^{\theta }=Z^3$ It's an object , The other three are clutter .

The first 4 The first assumption is $\theta =4$ when ： Express $Z^{\theta }=Z^4$ It's an object , The other three are clutter .

common 5 A hypothesis , It covers all possible situations

Put this 5 The probabilities of each case can be added ？？ The final result is a multimodal distribution ？？

Derivation and calculation of likelihood probability
$m$ Express $Z$ The number of elements in ,$\theta$ Indicates the assumptions made
$$p(Z|X)=p(Z,m|X)=\sum _{\theta =0}^{m}p(Z,m,\theta |X)$$ because $Z$ After setting ,$m$ Then it was decided that $m$ It's actually contained in $Z$ Information in , So there can be $p(Z|X)=p(Z,m|X)$, here $p(Z,m|X)$ It's about $(Z,m)$ Two dimensional joint probability distribution , But these two variables are not independent , It's about containing relationships . and $p(Z,m,\theta |X)$ It's about $(Z,m,\theta )$ Three dimensional joint probability distribution , here $Z$ and $\theta$ Then there is no containment relationship , It is closer to the general two-dimensional joint distribution . And yes $\theta$ The sum of variables is equal to $\theta$ Find integral , Is equivalent to $\theta$ It's gone , The result is about $Z$ Edge density of .

 Supplement the formula of joint probability distribution 

p(A,B) = p(B|A)p(A)

p(a,b|c) = p(a,b,c)/p(c) = [p(a,b,c)/p(b,c)] * [p(b,c)/p(c)]
         = p(a|b,c)p(b|c)

set up $A$ event ：$Z$;$B$ event ：$m$;$C$ event ：$\theta$;$D$ event ：$X$; among $B$ Events are contained in $A$ Incident

be $$p(A,B,C|D)=\frac{p(A,B,C,D)}{p(D)}$$ because $$p(A|B,C,D)=\frac{p(A,B,C,D)}{p(B,C,D)}$$ Bring it into the above formula $$\begin{gathered} p(A,B,C|D)=\frac{p(A,B,C,D)}{p(D)} \\ =\frac{p(A|B,C,D)p(B,C,D)}{p(D)} \\ =p(A|B,C,D)p(B,C|D) \end{gathered}$$ Or get $$p(Z|X)=\sum _{\theta =0}^{m}p(Z|m,\theta ,X)p(\theta ,m|X)$$ Now briefly explain the meaning of each item

$p(Z|m,\theta ,X)$： Number of known elements $m$、 Specific to a certain assumption $\theta$、 Truth value $X$ when , Find out how much probability the sensor will measure $Z$—— With $\theta =1$ For example ： It is measured as $z^1$ It's an object and comes from a theoretical measurement model , Other $z^2,z^3,z^4$ They are all clutter points and come from the clutter model . This is under hypothetical conditions （ Adding this assumption simplifies the original problem ） Ask again $Z$, It represents a simplified problem .
$p(m,\theta |X)$： according to $X$ determine $m,\theta$ Value —— That is, how much probability there will be $m$ Element and is the $\theta$ A hypothesis , Similar to the reliability of assumptions . according to $X$ To infer the number of measurement results and which measurement value will be an object , The reliability of this inference .

Understanding the derivation of this part and the significance of these two terms can better understand the actual measurement model , So as to obtain the actual likelihood probability .

Next, calculate these two terms respectively , Then multiply it to find the , Finally, add up to get the final result .

When $\theta =0$ when , Indicates that the object has not been detected ,$m$ Clutter detection .
At this time ,$p(m,\theta |X)$ The corresponding inference is ： There are $0$ An object , Yes $m$ A clutter . The probability of the former is $1-P^D$, The probability of the latter is $Po(m;{\lambda }_c)$. This item is mainly used to determine assumptions —— That is to distinguish which item is an object , Which items are clutter , So it's actually about the determination of the number . Therefore, it just corresponds to the Bernoulli detection part of the measurement model and the Poisson distribution part of the clutter model , These two parts are related to the number . The distribution characteristics of a specific point are $g_k()$ and $f_c()$, This is a $p(Z|m,\theta ,X)$ This one is responsible . Then the formula is ：$$p(m,\theta =0|X)=(1-P^D)Po(m;{\lambda }_c)$$
and $p(Z|m,\theta ,X)$： When $Z$ The elements in it are all clutter $m$ A clutter , Find the probability of this happening —— Simplified problems after adding assumptions . The formula is ：$$p(Z|m,\theta =0,X)=\prod _{i=1}^m{f}_c(z^i)$$
When $\theta =1$ when , Express $z^1$ It's an object , other $m-1$ All measurement points are clutter .
here ,$p(m,\theta =1|X)$ The corresponding inference is ： There are $1$ An object is $z^1$, Yes $m-1$ A clutter . The probability of the former is $P^D$ That is, the object is successfully detected , The probability of the latter is $Po(m-1;{\lambda }_c)\frac{1}{m}$—— Poisson distribution indicates that there are $m-1$ A clutter , but $\frac{1}{m}$ What does it mean ？ Express $P(\theta =1)=1/m?$ Calculation formula ：$$p(m,\theta =1|X)=P^{D}Po(m-1;{\lambda }_c)\frac{1}{m}$$
and $p(Z|m,\theta =1,X)$： When $Z$ There are only $z^1$ It's an object , Other elements are clutter $m-1$ A clutter , Find the probability of this happening —— Simplified problems after adding assumptions . Its calculation formula ：$$p(Z|m,\theta =1,X)=g_k(z^1|x)\frac{{\prod }_{i=1}^m{f}_c(z^i)}{f_c(z^1)}$$ Why... Here $z$ As a result of $g_k()$ The argument to the function , Should not be $o$ Do you ？—— because $z$ By $(o,c)$ The result of mixed arrangement , So naturally there is one $z^j$ Namely $o$, So this $z^j$ Just follow this $o$ Conform to the theoretical measurement model $g_k()$ 了 . The rest $z$ The element is determined to be a clutter point , From the clutter model , Nature is in line with $f_c()$ 了 .
Empathy , The calculation formula of other assumptions can be deduced, namely $\theta \in (1,2,3,...,m)$, Then add up the results corresponding to all assumptions to get $p(z|x)$, Expression for $$\begin{gathered} p(z|x)=\sum _{\theta =0}^{m}p(z,m,\theta |x) \\ =[(1-P^D)+P^D\sum _{\theta =1}^m\frac{g_k(z^{\theta }|x)}{\lambda (z^{\theta })}]\frac{e^{-{\lambda }_c}}{m!}\prod _{i=1}^m\lambda (z^i) \end{gathered}$$ This is the calculation formula of likelihood probability obtained from the actual measurement model , Later, it can be used in Bayesian formula , To complete the update step .