Probabilistic Robotics, Sebastian Thrun, 2005 Page 36, 37. <1 x = nf ) =1/ 3 <1 x = f ) =1 p(x = f ) = 0.01 p(x = nf ) = 0.

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1 Probabilistic Robotics, Sebastian Thrun, 005 Page 36, 37 Method : Denote: z n : the sensor measurement in each time x: the state of the sensor f: the state of the sensor is faulty nf: the state of the sensor is not faulty We know: p(z < x nf ) / 3 p(z < x f ) p(x f ) 0.0 p(x nf ) 0.99 when N p(x z ) p(z x) p(x) p(z ) x' f,nf p(z x) p(x) p(z x') p(x') *0.0 p(x f z <) * *0.99 when N pz (, z xpx ) ( ) pz (, z xpx ) ( ) p(x z,z ) pz (, z) pz (, z x') ') x' f, nf *0.0 f z <, z < ) *0.0 + ( ) *0.99 3

2 when N0 pz ( :0 xpx ) ( ) pz ( :0 xpx ) ( ) p(x z :0) pz ( ) pz ( x') ') :0 :0 x' f, nf *0.0 f z:0 < ) *0.0 + ( ) * When Nn Method : Denote: X : the sensor is faulty X : the sensor is right Y: the event that all the N outputs are below m. px ( ) 0.0 PX ( ) 0.99 py ( X) py ( X) ( ) 3 px ( Y) N px (, Y) py ( X) PX ( ) py ( ) PY ( X) P( X) + PY ( X ) P( X ) 0.0 N ( ) * N,,, 0. The posterior probability of a sensor fault is 0.094, , 0.43, ,0.705,0.8804,0.9567,0.985,0.9950,0.9983

3 (a) Suppose Day is a sunny day. What is the probability of the following sequence of days: Day cloudy, Day3 cloudy, Day4 rainy? Denote: s: sunny, c: cloudy, r: rainy state value x : weather of day n by Markov assumption : n, x, x, x) x, x, x), x, x), x, x x) x), x x) ) ), x x) , x, x) x, x), x) 3 x) x) ) ) 3 3 x) pc ( s) 0. x) pcc ( ) 0.4 x) pr ( c) 0. Thus p( x, x, x x ) p( x x ) p( x x ) p( x x ) (b) Write a simulator that can randomly generate sequences of weathers" from this state transition function. [Reference code]: sequencedays 0; s ; c 0; r 0; % the possible value of state T [8 0; 4 4 ; 6 ]'/0; % Markov state transition matrix xp [s c r]'; % the state of the previous day weather {'sunny ', 'cloudy', 'rainy '}; for j:sequencedays xt T * xp; xn [0 0 0]'; % the state of the next day randomprob rand(); cumulativeprob 0; for i:3 cumulativeprob cumulativeprob + xt(i); if (randomprob < cumulativeprob) xn(i) ; break;

4 for i:3 if (xn(i) ) disp(weather{i}); break; [Output (may not exactly the same value)]: sunny, cloudy, sunny, cloudy, sunny, sunny, cloudy, cloudy, sunny, sunny, (c) Use your simulator to determine the stationary distribution of this Markov chain. The stationary distribution measures the probability that a random day will be sunny, cloudy, or rainy. Stationary distribution can be approximated by the frequency of weather far into the future. [Reference code]: E.g. set the statistic sequence days as: sequencedays ; Add the count for statistic in the code: if (xn(i) ) disp(weather{i}); weathercnt(i) weathercnt(i) + ; break; [Output (may not exactly the same value)]: sunny has 6459 days in days, prob cloudy has 8603 days in days, prob rainy has 7439 days in days, prob Thus the stationary distribution of this Markov chain is [0.645, 0.860, 0.074] T. (d) Can you devise a closed-form solution to calculating the stationary distribution based on the state transition matrix above? The stationary distribution means that λ x T x (λ ), in this case, x is the linearly normalized eigen vector of matrix T, when eigen value λ. [Reference code]: [V, D] eig(t) xbar V(:, ) % this st column of V corresponding to eigen value xbar xbar / sum(xbar) [Output]: xbar (e) What is the entropy of the stationary distribution? For stationary Markov chain {Xi} with stationary distribution X and transition matrix P. The entropy is given by: n H( X) lim H( X,..., Xn) lim H( Xi Xi,..., X), by chain rule. n n n n i n p( X ) n n i i i lim H( X X ) p( X )log n

5 [Reference code]: Hx (-) * sum(xbar.* log(xbar)); [Output]: Hx.98 (f) Using Bayes rule, compute the probability table of yesterday's weather given today's weather. (It is okay to provide the probabilities numerically, and it is also okay to rely on results from previous questions in this exercise.) Bayes rule and total probability: i xi ) i ) i xi ) i ) i xi) ) x ) ) i i i i xi If we use stationary probability as the prior: ) [0.649, 0.857, 0.074] T i [Reference code]: Tr zeros(3); for i:3 % weather state: today % weather state: yesterday % calculate p(x_t_ j x_t i) Tr(i, j) T(i, j) * xbar(j) / sum(t(i, :)'.* xbar); for j:3 [Output]: Tr yesterday s weather Today s weather sunny cloudy rainy sunny cloudy rainy If we use uniform probability as the prior: T i ) [,, ] [Reference code]: xbar [ ]' / 3; Tr zeros(3); for i:3 % weather state: today for j:3 % weather state: yesterday % calculate p(x_t_ j x_t i) Tr(i, j) T(i, j) * xbar(j) / sum(t(i, :)'.* xbar); [Output]: Tr

6 Today s weather Yesterday s weather sunny cloudy rainy sunny cloudy rainy (g) Suppose we added seasons to our model. The state transition function above would only apply to the Summer, whereas different ones would apply to Winter, Spring, and Fall. Would this violate the Markov property of this process? Explain your answer. The system still satisfies Markov assumption. It can be viewed as a time-varying Markov Chain (though time-varying, the value of x t+ is still determined by x t only). Alternatively, add season s t as an additional state variable of four values. By ex-ting the weather transition matrix to x, we can still account for all possible transitions from {x t, s t } to {x t+, s t+ }:, s x, s ), s x, s) t+ t+ : t : t t+ t+ t t

7 (a) Suppose Day is sunny (this is known for a fact), and in the subsequent four days our sensor observes cloudy, cloudy, rainy, sunny. What is the probability that Day 5 is indeed sunny as predicted by our sensor? Method : Denote: s: sunny, c: cloudy, r: rainy state value: x : weather of day n, we know x s, x r n 4 sensor data: z n: weather of day n, we know z ~5 ccrs,,, By applying Bayes' rule and Markov assumption: Bel( x ) p( x x, z, z, z, z ) pz ( 5 x5), z, z3, z4, x5) pz ( 5 x5) 5 x, z, z3, z4), z, z, z, z) pz ( x, z, z, z) pz ( x) x, z, z, z) pz ( 5 x5) 5 x, z, z3, z4) x5 s, c, r As we can see in the sensor measurement transition matrix, z 4 rainy, means x 4 rainy ' ' we have x, z, z, z) x r) x) x r) Bel(x 5 s) ' x5 s, c, r p(z s x s) p(x s x r) " % $ p(z 5 s x ' ' 5 ) p(x 5 x 4 r) ' $ ' # x ' 5 s,c,r & from the previous question, we know the transition matrix 0.6*0. 0.6* * *

8 Method : Note that when the weather is rainy, the sensor reports that with absolute certainty. Therefore, we can just analyze the transition from day 4 to day5. We are looking to find p(x 5 s z 5 s) when x 4 r : p(x 5 s z 5 s) p(z 5 s x 5 s) p(x 5 s x 4 r) p(z 5 s) p(z s x s) p(x s x r) " % $ p(z 5 s x ' ' 5 ) p(x 5 x 4 r) ' $ ' # x ' 5 s,c,r & from the previous question, we know the transition matrix 0.6*0. 0.6* * * (b) Once again, suppose Day is known to be sunny. At Days through 4, the sensor measures sunny, sunny, rainy. For each of the Days through 4, what is the most likely weather on that day? Answer the question in two ways: one in which only the data available to the day in question is used, and one in hindsight, where data from future days is also available. For day : For the st case we only know the measures until this day: Method : Bel( x ) p( x x, z ) η pz ( x, x) x) η pz ( x) x) η η /9 / 9, normalizate the sum of prob. to 0 Method :

9 we already know x z s ' x s, c, r s Also the state & measurement transition matrices. Bel( x ) p( x x, z ) pz ( x) x) ' ' pz ( x) x) pz ( x) x) pz ( x) Bel(x s) (0.6*0.8)/(0.6* *0.+0*0) Bel(x c) (0.3*0.)/(0.6* *0.+0*0) 0. Bel(x r) 0 So its most likely weather is sunny with probability 88.89%

10 For day 3: For the st case we only know the measures until this day: Method : x, z ) 3 :3 η pz ( x, x, z) x, z) η pz ( x), x x, z) x η pz ( x) x) x, z) x 8 η pz ( 3 x3) 3 x) x ηpz ( 3 x3).0 η So its most likely weather is sunny with probability 87.8% For day 4: in the case we know the measures of all those days: x, z ) x, z) 4 : So its most likely weather is rainy with probability 00%

11 If we KNOW future measures. For day : For the nd case we know the measures of all those days: we already know x s, x r z s, z s, z r Also the state & measurement transition matrices. Bel( x ) p( x x, z, x ) p( x x, z ), since we already know z, as well as x :4 4 :4 4 4, x, z ) pz ( :4 : 4 :4 :4 :4 :4 3 4 :4 :4 :4 :4 x, x ) p( x, x ) x) pz ( :4 x) pz ( x) pz (, x) pz ( x), z, z) x) x) pz ( x) ) pz ( x) ) x) pz ( x) pz ( 3:4 x) pz ( x) x) pz ( x) 3, z3:4 x), Total prob. pz ( x) η x) pz ( x) x) pz ( x), Bayes rule x3 3 3:4 3 x3 η x) pz ( x) x) pz ( x) pz ( x, z), Bayes rule x3 η x) pz ( x) x) pz ( x) pz ( x), Bayes rule x3 η x) pz ( x) x) pz ( x), z x), Total Prob x3 x4 η x) pz ( x) x) pz ( x) x) pz ( x, x), Bayes rule x3 x4 η x) pz ( x) x) pz ( x) x) pz ( x), Markov assumption x3 x4

12 η x) pz ( x) x) pz ( x) x) pz ( x), Markov assumption x3 x4 assume m x) pz ( x) x when x s, m p( x r x s) p( z r x r) when x c, m p( x r x c) p( z r x r) when x r, m p( x r x r) p( z r x r) η x) pz ( x) x) pz ( x) m x3 assume n x) pz ( x) m x when x s, n p( x s x s) p( z s x s)* c x spz ) ( s x c)* r x spz ) ( s x r)* 0. 0.*0.3* when x c, n p( x s x c) p( z s x s)* c x cp ) ( z3 s x3 c)*0. + r x cpz ) ( s x r)*0. 0.4*0.3* when x r, n p( x s x r) p( z s x s)* c x rpz ) ( s x c)* r x rpz ) ( s x r)*0. 0.6*0.3* η x spz ) ( s x) η η So its most likely weather is sunny with probability 80%

13 For day 3: For the nd case we know the measures of all those days: we already know x s, z s, z s, z r,also we can deduce x r Bel( x ) p( x x, z ) p( x, x, z ) 3 3 :4 3 :4, z:4) pz ( 4 x3) 3, z:3, x) :4 pz ( 4 x3) 3, z:3, x) :4 pz ( 4 x3) pz ( 3 x3, z, x) p( x3, z, x) :4 pz ( x) pz ( x), x, z, x) :4 :4 :4 : x pz ( x) pz ( x) x), z, x) x pz ( x) pz ( x) x) pz ( x), x) x ) 4 x3) pz ( 3 x3) 3 x) pz ( x) x) assume m x) pz ( x) x) x x 3 when x s, m p( x s x s) p( z s x s) p( x s x s) s x cpz ) ( s x cpx ) ( c x s) 3 + s x rpz ) ( s x rpx ) ( r x s) 3 0.8*0.6* *0.3* when x c, m p( x c x s) p( z s x s) p( x s x s) c x cpz ) ( s x cpx ) ( c x s) + c x rpz ) ( s x rpx ) ( r x s) 3 0.*0.6* *0.3* when x r, m p( x r x s) p( z s x s) p( x s x s) r x cpz ) ( s x cpx ) ( c x s) 3 + r x rpz ) ( s x rpx ) ( r x s) *0.3* η 4 r x3) pz ( 3 s x3) 0.00 η η So its most likely weather is cloudy with probability 00%

14 (c) Consider the same situation (Day is sunny, the measurements for Days, 3, and 4 are sunny, sunny, rainy). What is the most likely sequence of weather for Days through 4? What is the probability of this most likely sequence? The probability of the sequence of weather is given by: x, z ) :4 :4 pz ( :4 x, x:4 ), x:4) :4 ) pz ( :4 x, x:4 ) :4 x ) :4 η pz ( 4 x4) pz ( 3 x3) pz ( x)* 4 x3) 3 x) x) Using the sensor model and transition matrix to calculate it, we can get the most likely sequence of weather is s, c, r, which has / ( ) 80% of occurring. There is a 0% probability of c, c, r, and 0% probability of all other possibles.