probabilistic robotics 读书笔记（一）

Ⅱ recursive state estimation

1 basic concepts in probability

2 robot environment interaction

3 bayes filters

4 exercises

4.1

The prior probability for a sensor to be faulty

bel(X0=sensorfaulty)=0.01 b e l ( X 0 = s e n s o r f a u l t y ) = 0.01

The probability of measurement when sensor is not faulty

p(Zi<1|X=sensorisnotfaulty)=13 p ( Z i < 1 | X = s e n s o r i s n o t f a u l t y ) = 1 3

So bel¯¯¯¯¯¯(x1) b e l ¯ ( x 1 ) equal

bel¯¯¯¯¯¯(x1=sensorfaulty)=1∗bel(x0)+0∗(1−bel(x0))=bel(x0) b e l ¯ ( x 1 = s e n s o r f a u l t y ) = 1 ∗ b e l ( x 0 ) + 0 ∗ ( 1 − b e l ( x 0 ) ) = b e l ( x 0 )

bel1(x1=sensorfaulty)=η∗1∗0.01=0.01η b e l 1 ( x 1 = s e n s o r f a u l t y ) = η ∗ 1 ∗ 0.01 = 0.01 η

bel1(x1=sensorisnotfaulty)=η∗13∗0.99=0.33η b e l 1 ( x 1 = s e n s o r i s n o t f a u l t y ) = η ∗ 1 3 ∗ 0.99 = 0.33 η

calculate the bel recursively by the code

 int main()
    {
        double bel0,p,bel_1,bel,a,n;
        printf("please input the belief and probability bel0,p\n");
        scanf("%lf %lf",&bel0,&p);
        bel=bel0;
    n=10;
        for(int i=0;i1*bel;
           a=1/(bel_1+p-p*bel_1);
           bel=a*1*bel_1;
        }
        printf("the belief that sensor is faulty is %lf",bel);
        return bel; 
    }

4.2

(a) depend on the table

p(Day2=cloudy)=0.2p(Day2=sunny)=0.8p(Day2=rainy)=0 p ( D a y 2 = c l o u d y ) = 0.2 p ( D a y 2 = s u n n y ) = 0.8 p ( D a y 2 = r a i n y ) = 0

p(Day3=cloudy)=p(Day3=cloudy|Day2=cloudy)∗p(Day2=cloudy)+p(Day3=cloudy|Day2=sunny)∗p(Day2=sunny)+p(Day3=cloudy|Day2=rainy)∗p(Day2=rainy)=0.4∗0.2+0.2∗0.8+0.6∗0=0.24(2)(3)(4)p(Day3=rainy)=0.04p(Day3=sunny)=0.72 (2) p ( D a y 3 = c l o u d y ) = p ( D a y 3 = c l o u d y | D a y 2 = c l o u d y ) ∗ p ( D a y 2 = c l o u d y ) + p ( D a y 3 = c l o u d y | D a y 2 = s u n n y ) ∗ p ( D a y 2 = s u n n y ) + p ( D a y 3 = c l o u d y | D a y 2 = r a i n y ) ∗ p ( D a y 2 = r a i n y ) (3) = 0.4 ∗ 0.2 + 0.2 ∗ 0.8 + 0.6 ∗ 0 (4) = 0.24 p ( D a y 3 = r a i n y ) = 0.04 p ( D a y 3 = s u n n y ) = 0.72

p(Day4=rainy)=0.056 p ( D a y 4 = r a i n y ) = 0.056

p(Day2=cloudy,Day3=cloudy,Day4=rainy)=0.2∗0.4∗0.2=0.016 p ( D a y 2 = c l o u d y , D a y 3 = c l o u d y , D a y 4 = r a i n y ) = 0.2 ∗ 0.4 ∗ 0.2 = 0.016

(b)