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)=0.01b 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)=13p ( Z i < 1 | X = s e n s o r i s n o t f a u l t y ) = 1 3
So belˉˉˉˉˉˉ(x1)b e l ˉ ( x 1 )equal
belˉˉˉˉˉˉ(x1=sensorfaulty)=1?bel(x0)+0?(1?bel(x0))=bel(x0)b e l ˉ ( 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=0.01ηb 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=0.33ηb 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;
i
4.2 (a) depend on the table
p(Day2=cloudy)=0.2p(Day2=sunny)=0.8p(Day2=rainy)=0p ( 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)=0.056p ( D a y 4 = r a i n y ) = 0.056
p(Day2=cloudy,Day3=cloudy,Day4=rainy)=0.2?0.4?0.2=0.016p ( 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
【probabilistic|probabilistic robotics 读书笔记(一)】(b)
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