Random finite set RFs self-study notes (6): an example of calculation with the formula of prediction step and update step

There are so many derivation formulas written above , I always feel that it is not intuitive and impressive , Personally, I think that looking at specific examples can quickly understand how this thing works , I'll try to describe this example .

Start with the simplest example ： linear equation , Normal distribution , One dimensional scalar .
Some symbols need to be explained ：$p^{-}(x_k)$ It means No $k$ A priori value of time ,$p^{+}(x_k)$ It means No $k$ Posterior value of time . in addition ,$p(x_k|z_{1:k-1})$ It also means $k$ A priori value of time , and $p^{-}(x_k)$ identical ;$p(x_k|z_{1:k})$ It also means $k$ Posterior value of time , and $p^{+}(x_k)$ identical . there $z_{1:k}$ The representation is based on 1 Moment to k All the time $z$ Information .

set up

0 A posteriori probability of time ：$p^{+}(x_0)$—— Subjective guess
1 A priori probability of time ：$p^{-}(x_1)=p(x_1|z_0)=N(x_1;0.5,0.2)$
Motion models ：
$x_k=f(x_{k-1})+q_k=x_{k-1}+q_k$, among ,$q_k\sim N(0,0.35)$
Then you can predict ${\pi }_k(x_k|x_{k-1})=N(x_k;x_{k-1},0.35)$—— Based on $k-1$ Time information to infer $k$ The distribution of time , Isn't this prediction ？
detection probability ：$P^D=0.9$, Here we use constants
Theoretical measurement model ：
$o_k=h(x_k)+r_k=x_k+r_k$, among ,$r_k\sim N(0,0.2)$
Get the theoretical likelihood $g_k(o_k|x_k)=N(o_k;x_k,0.2)$
Clutter intensity ：$$\lambda (c)=\{\begin{array}{cc}0.4& |c|\le 4\\ 0& others\end{array}$$ be ${\lambda }_c=0.4\ast 8=3.2$, be $$f_c(c)=\{\begin{array}{cc}1/8& |c|\le 4\\ 0& others\end{array}$$
Observed value ：$z_1=[-1.3,1.7],z_2=[1.3],z_3=[-0.3,2.3]$ And $m_1=2,m_2=1,m_3=2$


Generally speaking, we should first base on the posterior of subjective conjecture $p^{+}(x_0)$ Yes 1 Make predictions all the time , And then combine 1 Observations at the moment $z_1$ Conduct 1 Update at any time , To get 1 Posterior value of time . But here , The known conditions have been given 1 A priori value of time , therefore , The next thing to do here is to combine $z_1$ Conduct 1 Update at any time , Get the posterior value .

1 Update step formula at time $$\begin{gathered} p(x_k|z_{1:k})=\sum _{{\theta }_{1:k}}p(x_k|z_{1:k},{\theta }_{1:k})p({\theta }_{1:k}|z_{1:k}) \\ =p(x_1|z_1)=\sum _{{\theta }_1}p(x_1|{\theta }_1,z_1)p({\theta }_1|z_1) \\ =p(x_1|\theta =0,z_1)p(\theta =0|z_1) \\ +p(x_1|\theta =1,z_1)p(\theta =1|z_1) \\ +p(x_1|\theta =2,z_1)p(\theta =2|z_1) \end{gathered}$$ Then calculate each item separately ： Probability and weight . It can be seen that , Probability is obviously to use Bayesian formula —— The same as Bayesian filtering ; How to calculate the weight ？

Calculate the probability $$\begin{gathered} p(x_1|\theta =0,z_1)=\frac{p(x_1|z_0)p(\theta =0,z_1|x_1)}{p(\theta =0,z_1)} \\ =\eta p(x_1|z_0)p(\theta =0,z_1|x_1) \end{gathered}$$ there $p(x_1|z_0)$ Is a priori ,$p(\theta =0,z_1|x_1)$ Is likelihood probability —— Likelihood is calculated from the measurement model $$\begin{gathered} p(\theta =0,z_1|x_1)=p(\theta =0,m_1,z_1|x_1) \\ =p(\theta =0,m_1|x_1)p(z_1|x_1,\theta =0,m_1) \end{gathered}$$ and $$\begin{gathered} p(\theta =0,m_1|x_1)=(1-P^D)Po(m_1;{\lambda }_c) \\ p(z_1|x_1,\theta =0,m_1)=\prod _{i=1}^{m_1}{f}_c(z^i) \end{gathered}$$ The former is weight , The latter is probability . What we get in this way is $p(z,m_1,\theta =0|x)$, And you have to multiply by $\eta p(x_1|z_0)$ To get it $p(x_1|\theta =0,z_1)$ The value of is assumed $\theta =0$ A posteriori probability of , But this posterior probability needs to be multiplied by a corresponding weight $p(\theta =0|z_1)$ Talent , As a contribution to the final posterior . namely $$\begin{gathered} p(x_1|\theta =0,z_1)p(\theta =0|z_1) \\ =p(x_1|\theta =0,m_1,z_1)p(\theta =0,m_1|z_1) \\ =\eta p(x_1|z_0)p(\theta =0,z_1|x_1)p(\theta =0,m_1|z_1) \\ =\eta p(x_1|z_0)p(\theta =0,m_1|x_1)p(z_1|x_1,\theta =0,m_1)p(\theta =0,m_1|z_1) \\ =\eta p(x_1|z_0)(1-P^D)Po(m_1;{\lambda }_c)\prod _{i=1}^{m_1}{f}_c(z^i)p(\theta =0,m_1|z_1) \end{gathered}$$ One of the problems we face now is ：$p(\theta =0,m_1|z_1)$ How to find ？ What we know before is how to ask $p(\theta =0,m_1|x_1)$, It's different from this .

alike , We can deduce the probability under other assumptions $$\begin{gathered} p(x_1|\theta =1,z_1)p(\theta =1|z_1) \\ =p(x_1|\theta =1,m_1,z_1)p(\theta =1,m_1|z_1) \\ =\eta p(x_1|z_0)p(\theta =1,z_1|x_1)p(\theta =1,m_1|z_1) \\ =\eta p(x_1|z_0)p(\theta =1,m_1|x_1)p(z_1|x_1,\theta =1,m_1)p(\theta =1,m_1|z_1) \\ =\eta p(x_1|z_0)P^{D}Po(m_1-1;{\lambda }_c)\frac{1}{m_1}{g}_k(z^1|x_1)\frac{{\prod }_{i=1}^{m_1}{f}_c(z^i)}{f_c(z^1)}p(\theta =1,m_1|z_1) \end{gathered}$$
the other one $$\begin{gathered} p(x_1|\theta =2,z_1)p(\theta =2|z_1) \\ =p(x_1|\theta =2,m_1,z_1)p(\theta =2,m_1|z_1) \\ =\eta p(x_1|z_0)p(\theta =2,z_1|x_1)p(\theta =2,m_1|z_1) \\ =\eta p(x_1|z_0)p(\theta =2,m_1|x_1)p(z_1|x_1,\theta =2,m_1)p(\theta =2,m_1|z_1) \\ =\eta p(x_1|z_0)P^{D}Po(m_1-1;{\lambda }_c)\frac{1}{m_1}{g}_k(z^2|x_1)\frac{{\prod }_{i=1}^{m_1}{f}_c(z^i)}{f_c(z^2)}p(\theta =2,m_1|z_1) \end{gathered}$$ It can be seen that , What we have done is to bring in the likelihood probability formula obtained from the previous measurement model , And weight $p(\theta =0,m_1|z_1),p(\theta =1,m_1|z_1),p(\theta =2,m_1|z_1)$ These have not been changed and solved . And a priori probability $p(x_1|z_0)$ Always exist .

Now we have to face a problem ： The weight $p(\theta ,m_1|z_1)$ How to find ？

Learn from the weight processing of particle filter （ Without detection probability and clutter model ）
1. Subjective guess initial value $X_0\sim N(\mu ,{\sigma }^2)$
2. according to $X_0\sim N(\mu ,{\sigma }^2)$ Randomly generate sample points $x_0^1,x_0^2,...,x_0^n$
3. Generate $X_0$ The weight corresponding to the sample point ${\omega }_0^i$—— It can be all $1/n$, It can also be $\frac{f_0(x_0^i)}{\sum f_0(x_0^i)}$. among ,$f_0(x)$ yes $X_0$ The probability density function of is $N(\mu ,{\sigma }^2)$
4. Generate forecast sample points $x_1^i$：$$x_1^{-i}=f_m(x_0^i)+q_1$$ among ,$q_1\sim N(0,Q_1)$ —— It just combines the equation of motion to produce the predicted value .
5. The predicted probability density is obtained according to the predicted sample points ：$$f_1^{-}(x)=\sum _{i=1}^n{\omega }_0^i\delta (x-x_1^{-i})$$ The prediction step just changes the position of the particles, that is $x_0^i\to x_1^i$, But the weight is not changed ${\omega }_0^i$
6. Observed a data $y_1$
7. Bring in the renewal step equation to find the posterior density ：$$\begin{gathered} f_1^{+}(x)=\eta f_r[y_1-h(x)]f_1^{-}(x) \\ =\sum _{i=1}^n\eta f_{r1}[y_1-h(x_1^{-i})]{\omega }_0^i\delta (x-x_1^{-i}) \\ =\sum _{i=1}^n{\omega }_1^i\delta (x-x_1^{-i}) \end{gathered}$$ among ${\omega }_1^i=\eta f_r[y_1-h(x_1^{-i})]{\omega }_0^i,f_{r1}=N(0,R_1)$, That is, the new step changes the weight value ：${\omega }_0^i\to {\omega }_1^i$, But the position of the particles has not changed
8. Reunite power $${\omega }_1^i=\frac{{\omega }_1^i}{{\sum }_{i=1}^n{\omega }_1^i}$$
9. Combine the equation of motion to generate the predicted particles at the next moment $$x_2^{-i}=f_m(x_1^{-i})+q_2$$ among ,$q_2\sim N(0,Q_2)$
10. According to the predicted particle points, we can get the predicted probability density $$f_2^{-}(x)=\sum _{i=1}^n{\omega }_1^i\delta (x-x_2^{-i})$$ The combination of the weight of the previous moment and the particles at this moment —— The prediction step only changes the particle position and the weight remains unchanged
11. Observed a data $y_2$
12. Bring in the renewal step equation to find the posterior density ：$$\begin{gathered} f_2^{+}(x)=\eta f_r[y_1-h(x)]f_1^{-}(x) \\ =\sum _{i=1}^n\eta f_{r2}[y_2-h(x_2^{-i})]{\omega }_1^i\delta (x-x_2^{-i}) \\ =\sum _{i=1}^n{\omega }_2^i\delta (x-x_2^{-i}) \end{gathered}$$ among ${\omega }_2^i=\eta f_r[y_2-h(x_2^{-i})]{\omega }_1^i,f_{r2}=N(0,R_2)$, That is, the new step changes the weight value ：${\omega }_1^i\to {\omega }_2^i$, But the position of the particles has not changed
13. Weight normalization $${\omega }_2^i=\frac{{\omega }_2^i}{{\sum }_{i=1}^n{\omega }_2^i}$$
Iterate so constantly …
Be careful ： Due to normalization , Actually, the one above $\eta$ The coefficient can be omitted directly , namely $${\omega }_2^i=f_r[y_2-h(x_2^{-i})]{\omega }_1^i$$

You can see , The initial weight in particle filter can be set to $1/n$, The subsequent change of particle position is based on the equation of motion $X_k=f_m(X_{k-1})+q_k$ obtain ：$$x_k^i=f_m(x_{k-1}^i)+q_k$$ The change of weight is the update step using Bayesian formula $${\omega }_k^i=\eta f_r[y_k-h(x_k^{-i})]{\omega }_{k-1}^i$$

that ,$SOT$ Weight in $p(\theta ,m_1|z_1)$ Is there a similar formula in the update of ？

The formula $$\begin{gathered} p(x_k|z_{1:k-1})=\sum _{{\theta }_{1:k-1}}{\omega }^{{\theta }_{1:k-1}}{p}_{k|k-1}^{{\theta }_{1:k-1}}(x_k) \\ p(x_k|z_{1:k})=\sum _{{\theta }_{1:k}}{\omega }^{{\theta }_{1:k}}{p}_{k|k}^{{\theta }_{1:k}}(x_k) \end{gathered}$$ At first glance , Isn't this particle filter ？ But take a closer look at , Discovery is the accumulation of all assumptions , Particle filter aims at the accumulation of every sample point , So there is an essential difference between the two , Although very similar in form .
Next is the solution of each term
${\theta }_k=0$ when $$\begin{gathered} {\tilde{\omega }}^{{\theta }_{1:k}}={\omega }^{{\theta }_{1:k}}\int p_{k|k-1}^{{\theta }_{1:k-1}}(x_k|z_{1:k-1})(1-P^D(x_k))d{x}_k \\ {p}_{k|k}^{{\theta }_{1:k}}(x_k|z_{1:k})=\eta p_{k|k-1}^{{\theta }_{1:k-1}}(x_k|z_{1:k-1})(1-P^D(x_k)) \end{gathered}$$ These are what the prediction step and the update step do —— That is, the calculation process of Bayesian filtering

${\theta }_k\in [1,2,3,...,m]$ when $$\begin{gathered} {\tilde{\omega }}^{{\theta }_{1:k}}=\frac{{\omega }^{{\theta }_{1:k}}}{\lambda (z^{\theta })}\int p_{k|k-1}^{{\theta }_{1:k-1}}(x_k|z_{1:k-1})P^D(x_k)g_k(z_k^{{\theta }_k}|x_k)d{x}_k \\ {p}_{k|k}^{{\theta }_{1:k}}(x_k|z_{1:k})=\eta p_{k|k-1}^{{\theta }_{1:k-1}}(x_k|z_{1:k-1})P^D(x_k)g_k(z_k^{{\theta }_k}|x_k) \end{gathered}$$
Under each assumption, it is a calculation process of bell filter , Then add all the hypothetical calculation results together

It's a little complicated , I feel confused for a while , Let's start with this , I'll add later （ The derivation of this prediction step formula should have ）