Recursive Estimation

Transcription

1 Recursive Estimation Raffaello D Andrea Spring 08 Problem Set 3: Extracting Estimates from Probability Distributions Last updated: April 9, 08 Notes: Notation: Unless otherwise noted, x, y, and z denote random variables, p x denotes the probability density function of x, and p x y denotes the conditional probability density function of x conditioned on y. Note that shorthand as introduced in Lecture and and longhand notation is used. The expected value of x and its variance is denoted by E [x] and Var [x], and Pr Z denotes the probability that the event Z occurs. A normally distributed random variable x with mean µ and variance σ is denoted by x N µ, σ. Please report any errors found in this problem set to the teaching assistants hofermat@ethz.ch or csferrazza@ethz.ch.

2 Problem Set Problem Let x and w be scalar CRVs, where x, w are independent: px, w px pw, and w [, ] is uniformly distributed. Let z x + 3w. a Calculate pz x using the change of variables. b Let x. Verify that Pr z [, 4] x Pr w [0, ] x. Problem Let w R and x R 4 be CRVs, where x and w are independent: px, w px pw. The two elements of the measurement noise vector w are w, w, and the PDF of w is uniform: p w w p w,w w, w The measurement model is where z Hx + Gw H /4 for w and w 0 otherwise. [ ] [ ], G Calculate the conditional joint PDF p z x using the change of variables.. Problem 3 Find the maximum likelihood ML estimate ˆx ML : arg max pz x when x z i x + w i, i, where x, w i are mutually independent, with w i N 0, σ i and zi, w i, x R. Problem 4 Solve Problem 3 if the joint PDF of w is p w w p w,w w, w /4 for w and w 0 otherwise.

3 Problem 5 Let x R n be an unknown, constant parameter. Furthermore, let w R m, n m, represent the measurement noise with its joint PDF defined by the multivariate Gaussian distribution p w w π m / det Σ exp / wt Σ w with mean E [w] 0 and variance E [ ww T ] : Σ Σ T independent. Let the measurement be R m m. Assume that x and w are z Hx + w where the matrix H R m n has full column rank. a Calculate the ML estimate ˆx ML : arg max pz x for a diagonal variance x σ 0 0 Σ 0 σ σm Note that in this case, the elements w i of w are mutually independent. b Calculate the ML estimate for non-diagonal Σ, which implies that the elements w i of w are correlated. Hint: The following matrix calculus is useful for this problem see, for example, the Matrix Cookbook for reference: Recall the definition of a Jacobian matrix from the lecture: Let gx : R n R m be a function that maps the vector x R n to another vector gx R m. The Jacobian matrix is defined as gx x : g x x g x x. g mx x g x x g x x.. g mx x g x x n g x x n. g mx x n Rm n where x j is the j-th element of x and g i x the i-th element of gx. Derivative of a quadratic form: Chain rule: x xt Ax x T A + A T R n, A R n n. u gx, x R n, u R n x ut Au u T A + A T u x, u x gx x Rn n. 3

4 Problem 6 Find the maximum a posteriori MAP estimate ˆx MAP : arg max pz x px when x z x + w, x, z, w R where x and w are independent, x is exponentially distributed, 0 for x < 0 p x x e x for x 0 and w N 0,. Problem 7 In the derivation of the recursive least squares algorithm, we used the following result: trace ABA T AB if B B T A where trace is the sum of the diagonal elements of a matrix. Prove that the above holds for the two-dimensional case, i.e. A, B R. Problem 8 In the derivation of the recursive least squares algorithm, we also used the following result trace AB A B T. Prove that the above holds for A, B R. Problem 9 Consider the recursive least squares algorithm presented in lecture, where the estimation error at step k was shown to be ek I KkHk ek Kkwk. Furthermore, E [wk] 0, E [ek] 0, and x, w,...} are mutually independent. Show that the measurement error variance P k Var [ek] E [ eke T k ] is given by T P k I KkHk Pk I KkHk + KkRkK T k where Rk Var [wk]. Problem 0 Use the results from Problems 7 and 8 to derive the gain matrix Kk for the recursive least squares algorithm that minimizes the mean squared error MMSE by solving 0 T Kk trace I KkHk Pk I KkHk + KkRkK T k. 4

5 Problem Apply the recursive least squares algorithm when where zk x + wk, k,,... E [x] 0, Var [x], and E [wk] 0, Var [wk]. The random variables x, w, w,... } are mutually independent and zk, wk, x R. a Express the gain matrix Kk and variance P k as explicit functions of k. b Simulate the algorithm for normally distributed continuous random variables wk and x with the above properties. c Simulate the algorithm for x and wk being discrete random variables that take the values and with equal probability. Note that x and wk satisfy. Simulate for values of k up to Problem Consider the random variables and measurement equation from Problem c, i.e. zk x + wk, k,,... 3 for mutually independent x and wk that take the values and with equal probability. a What is the MMSE estimate of x at time k given all past measurements z, z,..., zk? Remember that the MMSE estimate is given by [ ˆx MMSE k : arg min E ˆx x ] z, z,..., zk. ˆx x z,z,...,zk b Now consider the MMSE estimate that is restricted to the sample space of x, i.e. [ ˆx MMSE k : arg min E ˆx x ] z, z,..., zk ˆx X x z,z,...,zk where X denotes the sample space of x. Calculate the estimate of x at time k given all past measurements z, z,..., zk. 5

6 Sample solutions Problem a The change of variables formula for conditional PDFs when z gx, w is p z x z x p w xh z, x x g. w x, h z, x where w h z, x is the unique solution to z g x, w. We solve z g x, w for w to find h z, x w z x 3 h z, x and apply the change of variables: p z x z x p w xh z, x x g w x, h z, x p w x 3 z x x 3 3 p w z x. 3 With the given uniform PDF of w, the PDF of pz x can then be stated explicitely: 6 for x 3 z x + 3 p z x z x 0 otherwise b We begin by computing Pr z [, 4] x : Pr z [, 4] x 4 p z x z d z The probability Pr w [0, ] x is Pr w [0, ] x Pr w [0, ] 4 which is indeed identical to Pr z [, 4] x. 0 6 d z. p w w d w 0 d w Problem Analogous to the lecture notes, the multivariate change of variables formula for conditional PDFs is p z x z x p w x h z, x x detg where w h z, x is the unique solution to z H x + G w. The function h z, x is given by [ ] [ ] w G z H x 0 z x [ ] z x 3 z x 3 6

7 where we used x x, x, x 3, x 4 and z z, z. Applying the change of variables, it follows that p z x z x p w x h z, x x detg p w x G z H x x 3 3 p w z x, 3 z x 3. Using the PDF p w, this can then be expressed as 6 for x z x + and x z x 3 3 p z x z x 0 otherwise. Problem 3 For a normally-distributed random variable with w i N 0, σ i, we have pw i exp w i σi. Using the change of variables, and by the conditional independence of z, z given x, we can write pz, z x pz x pz x exp z x σ + z x σ. Differentiating with respect to x, and setting to 0, leads to pz, z x x 0 xˆx ML z ˆx ML z ˆx ML + σ σ 0 ˆx ML z σ + z σ σ + σ which is a weighted sum of the measurements. Note that for σ 0 : ˆx ML z, for σ 0 : ˆx ML z. Problem 4 Using the total probability theorem, we find that p wi w i for w i [, ] 0 otherwise. and pw, w pw pw. Using the change of variables, p zi x z i x if z i x 0 otherwise. We need to solve for p z,z x z, z x p z x z x p z x z x. Consider the different cases: z z >, for which the likelihoods do not overlap, see the following figure: 7

8 p z x z x z p z x z x x z Since no value for x can explain both, z and z, given the model z i x + w i, we find p z,z x z, z x 0. z z, where the likelihoods overlap, which is shown in the following figure: pz, z x pz x pz x x 4 x z z z z We find p z,z x z, z x 4 for x [ z, z + ] [ z, z + ] 0 otherwise where A B denotes the intersection of the sets A and B. Therefore the estimate is ˆx ML [ z, z + ] [ z, z + ]. All values in this range are equally likely, and any value in the range is therefore a valid maximum likelihood estimate ˆx ML. Problem 5 a We apply the multivariate change of coordinates: p z x z x p w x h z, x x detg p w x z H x x p w z H x π m / det Σ exp / z H xt Σ z H x We proceed similarly as in class, for z, w R m and x R n : H H H. H m, H i [ ] h i h i... h in, where hij R. Using p z x z x exp m z i H i x i σ i. 8

9 differentiating with respect to x j, and setting to zero yields m z i H iˆx ML h ij 0, i σ i [ ] h j h mj }} column of H From which we have and H T W z H ˆx ML 0 σ ˆx ML H T W H H T W z. σ... 0 σm j,,..., n } } W 0 z H ˆx ML 0, j,..., n. Note that this can also be interpreted as a weighted least squares, wx z Hx ˆx arg min wx T W wx x where W is a weighting of the measurements. The result in both cases is the same! b We find the maximizing ˆx ML by setting the derivative of pz x with respect to x to zero. With the given facts about matrix calculus, we find 0 x p z x z x x exp z H xt Σ z H x where we use the fact that Σ T Σ T Σ. z H xt Σ + Σ T z H x x z H x T Σ H Finally, z H x T Σ H 0 z T Σ H x T H T Σ H and after we transpose the left and right hand sides, we find H T Σ z H T Σ H x. It follows that ˆx ML H T Σ H H T Σ z. 9

10 Problem 6 We calculate the conditional probability density function, p z x z x exp z x. π Both the measurement likelihood p z x z x and the prior p x x are shown in the following figure p x x p z x z x x z We find p z x z x p x x 0 for x < 0 π exp x exp z x for x 0. We calculate the MAP estimate ˆx MAP that maximizes the above function for a given z. We note that p z x z x p x x > 0 for x 0, and the maximizing value is therefore ˆx MAP 0. Since the function is discontinuous at x 0, we have to consider two cases:. The maximum is at the boundary, ˆx MAP 0. It follows that p z x z ˆx MAP MAP px ˆx exp π z. 4. The maximum is where the derivative of the likelihood p z x z x p x x with respect to x is zero, i.e. 0 p z x z x p x x x from which follows ˆx MAP z xˆx MAP + z ˆx MAP at the maximum. Substituting back into the likelihood, we obtain p z x z ˆx MAP MAP px ˆx exp z + exp π exp z +. 5 π px ˆx MAP We note that p z x z ˆx MAP is therefore ˆx MAP 0 for z < z for z. pz x z ˆx MAP px ˆx MAP for z, and the ML estimate 0

11 Problem 7 We begin by evaluating the individual elements of AB and ABA T with A ij denoting the i, jth element of matrix A element in ith row and jth column: AB ij a i b j + a i b j ABA T ij a ib + a i b a j + a i b + a i b a j From this we obtain the trace as trace ABA T ABA T + ABA T and the derivatives trace ABA T a b + a b a + a b + a b a + a b + a b a + a b + a b a a b + a b +a a }} b AB b trace ABA T a b + a b + a b AB a trace ABA T a b + a b + a b AB a trace ABA T a a b + a b + a b AB We therefore proved that trace ABA T A holds for A, B R. AB Problem 8 We proceed similarly to the previous problem: that is trace AB AB + AB a b + a b + a b + a b trace AB a b B B T trace AB a b B B T trace AB a b B B T trace AB a b B B T trace AB A B T for A, B R.

12 Problem 9 By definition, P k E ek [ eke T k ]. Substituting the estimation error ek I KkHk ek Kkwk in the definition, we obtain [ ] T P k E I KkHk ek Kkwk I KkHk ek Kkwk ek,wk [ T E I KkHk ek e T k I KkHk + Kkwkw T kk T k ek,wk ] T I KkHk ek w T kk T k Kkwke T k I KkHk T I KkHk Pk I KkHk + KkRkK T k [ I KkHk ek w T k ] K T k Kk E ek,wk E ek,wk [ wke T k ] T I KkHk. Because ek and wk are independent, [ E wke T k ] [ E [wk] E e T k ]. ek,wk wk ek [ Since E [wk] 0, E [wk] E e T k ] 0, and it follows that wk ek T P k I KkHk Pk I KkHk + KkRkK T k. Problem 0 First, we expand T I KkHk Pk I KkHk + KkRkK T k Pk KkHkPk Pk H T kk T k+kk HkPk H T k+rk K T k. We use the results from problems 7 and 8 and evaluate the summands individually, since tracea + B tracea + traceb. We find Kk trace Pk 0 Kk trace KkHkPk Pk H T k, note that Pk P T k Kk trace Pk H T kk T k Pk H T k, note that trace A trace A T Kk trace Kk HkPk H T k + Rk K T k Kk HkPk H T k + Rk.

13 Note that R T k Rk and HkPk H T k T HkPk H T k. Finally, Kk HkPk H T k + Rk Pk H T k. Kk Pk H T k HkPk H T k + Rk Problem a We initialize P 0, ˆx0 0, Hk and Rk k,,..., and get: Kk Pk Pk + 6 P k Kk Pk + Kk Pk + Pk + Pk + Pk Pk + Pk. From 7 and P 0, we have P, P 3,..., and we can show by induction see below that P k + k, Kk + k Proof by induction: Assume from 6. P k + k. 8 Start with P 0 +0, which satisfies 8. Now we show for k + : P k + + k + k + P k + P k +k + by assumption 8 +k q.e.d. + k + b The Matlab code is available on the class web page. from 7 Recursive least squares works, and provides a reasonable estimate after several thousand iterations. The recursive least squares estimator is the optimal estimator for Gaussian noise. c The Matlab code is available on the class web page. Recursive least squares works, but we can do much better with the proposed nonlinear estimator derived in Problem. 7 3

14 Problem a The MMSE estimate is ˆx MMSE k arg min ˆx x, ˆx x p x z,...,zk x z, z,..., zk. 9 Let us first consider the MMSE estimate at time k. The only possible combinations are: x w z We can now construct the conditional PDF of x when conditioned on z: x z p x z x z / 0 / Therefore, if z or z, we know x precisely: x and x, respectively. If z 0, it is equally likely that x or x. The MMSE estimate is for z : for z : for z 0: ˆx MMSE since p x z 0 ˆx MMSE since p z z 0 ˆx MMSE arg min ˆx arg min ˆx arg min ˆx ˆx + 0. p x z 0 ˆx + p x z 0 ˆx + ˆx + ˆx + Note that the same result could have been obtained by evaluating ˆx MMSE k E [x 0] x z x, xp x z x 0 + At time k, we could construct a table for the conditional PDF of x when conditioned on z and z as above. However, note that the MMSE estimate is given for all k if one measures either or at time k, for z, ˆx MMSE k k,,... for z, ˆx MMSE k k,,

15 On the other hand, a measurement z 0 does not provide any information, i.e. x and x stay equally likely. As soon as a subsequent measurement zk, k, has a value of or, x is known precisely which is also reflected by the MMSE estimate. As long as we keep measuring 0, we do not gain additional information and ˆx MMSE k 0 if zl 0 l,,..., k. Matlab code for this nonlinear MMSE estimator is available on the class webpage. b The solution is identical to part a, except that the estimate is the minimizing value from the sample space of x: ˆx MMSE k arg min ˆx x p ˆx X x z,...,zk x z,..., zk. 0 x, The same argument holds that, if one measurement is either - or, we know x precisely. However, for z 0, the estimate is ˆx MMSE arg min p x z 0 ˆx + p x z 0 ˆx + ˆx X arg min ˆx X ±. ˆx + ˆx + Note that X, } and therefore, we cannot differentiate and set to zero the right hand side of Equation 0 to find the minimizing value. This implies that the estimate is not necessarily equal to E [x z], which is the case for measuring z 0. In this case E x z [x 0] is equal 0, whereas the MMMSE estimator yields ± for ˆxMMSE. The remainder of the solution is identical to the above part a in that the estimate remains ˆx MMSE ± until a measurement is either - or. 5