EECS564 Estimation, Filtering, and Detection Exam 2 Week of April 20, 2015

Transcription

1 EECS564 Estimation, Filtering, and Detection Exam Week of April 0, 015 This is an open book takehome exam. You have 48 hours to complete the exam. All work on the exam should be your own. problems have equal weight. If you feel that additional assumptions need to be made to answer any part of a question state your assumption explicitly. Please make sure that your name and student id number are on your exam, and if not using a blue book, make sure all of your pages are stapled in the correct order before handing in. Do not forget to write and sign the honor pledge before you hand in your work. 1 In this problem you will explore the so-called change detection problem for detecting a shift in the mean. Let w k be a white Gaussian noise with variance σ w. Let uk be a unit step function, i.e., uk = 0 for k 0 and uk = 1 for k 0. It is of interest to test the hypotheses : x k = w k, : x k = a k uk τ + w k, k = 1,..., n k = 1,..., n where a k 0 and τ {1,..., n} is the change time assumed fixed and known. a Find the most powerful test of level α for the case that the sequence {a k } k and τ are known and non-random. Be sure to specify an expression for the threshold. Find an expression for the power β. Is your test UMP against unknown positive values of a? Note that in vector form the signal component under has the form s = [0,..., 0, a τ,..., a n ] T where there are τ 1 zeros in this vector. Note that n τ + 1. The likelihood ratio test statistic is n Λ = exp a n kx k σw a k σw uk τ = n Therefore, the MP-LRT is a k x k γ n where γ = a k σ wn 1 1 α. a constant signal. The power is The test is not UMP unless a k = a, i.e., β = 1 N N 1 1 α d where the detectibility index d is given by the positive square root of d = n a k σ w 1

2 b Find the most powerful test of level α for the case that {a k } k are i.i.d. zero mean Gaussian with variance σa and τ is known and non-random. Be sure to specify an expression for the threshold. Find an expression for the power β. Is the test uniformly most powerful against unknown σa? In this case under the observations x k are independent N 0, σw for k τ and independent N 0, σa+σ w for k τ. Hence, the likelihood ratio test statistic is σ n τ+1 Λ = w σ a σa + σw exp σa + σwσ w x k Therefore the MP-LRT is x k Where γ = σwx n τ α. This is UMP as it does not depend on σ a. The power is β = 1 X n τ+1 ρxn τ α where the detectibility index ρ [0, 1] is given by γ ρ = σ w σ a + σ w c Find the most powerful test of level α for the case that a k = a, i.e., a k is constant over time, where a is a zero mean Gaussian r.v. with variance σa. Be sure to specify an expression for the threshold. Find an expression for the power β. Is the test uniformly most powerful against unknown σa This is a bit more involved than parts a-c since the random constant a makes the observations correlated over time. The covariance matrix of the vector of observations x under is obtained by using the vector form of the signal s = [0,..., 0, 1,..., 1] T a, denoted as s = 1 τ a: R 1 = covx = σ a1 τ 1 T τ + σ wi The form of the MP LRT can now be derived using the Woodbury identity for the R 1 1 to express f 1 x. Alternatively, one can derive it directly without using matridentities via the method of conditioning f 1 x = fx afada which, using the results of part a and completion of the square in the exponent of the integrand, gives Λ = f 1x f 0 x = 1 exp 1 n τ σ w /σs n τ σw/σ s x k Therefore, the MP LRT reduces to x k γ

3 where γ = n τ + 1 σ wx α. This is UMP wrt σ a. The power is β = 1 X 1 ρx α where the detectibility index ρ [0, 1] is given by ρ = σ w n τ + 1σ a + σ w d If the change time τ is unknown but a is known as in part a what does the GLRT look like? You do not need to specify the level α threshold or the power of the GLRT. The test statistic is simply the imized version of the LRT statistic found in part a: a n Λ GLR = exp x k τ σw a n τ + 1 σw This cannot be taken any further. Consider the following study of survival statistics among a particular population. A number n of individuals have enrolled in a long term observational study, e.g., a study of life expectancy for heart disease patients or chain smokers. The exact time of death of some of the individuals is reported. The other individuals stopped participating in the study at some known time. For these latter patients the exact time of death is unknown; their survival statistics are said to be censored. The objective is to estimate or test the mean survival time of the entire population. For the i-th individual define the indicator variable w i, where w i = 1 if the time of death is reported and otherwise w i = 0. For an individual with w i = 1 let t i denote their reported time of death. For an individual with w i = 0 let τ i denote they time they stopped participating in the study. Let T i be a random variable that is the time of death of individual i. Assume that the T i s are i.i.d with density ft; λ parameterized by λ, which is related to the mean survival time. Then, for w i = 1 the observation is the real value X i = t i while for w i = 0 the observation is the binary value X i = IT i τ i where IA is the indicator function of event A. Therefore the likelihood function associated with the observation x = [x 1,..., x n ] T is fx; λ = n f w i t i ; λ 1 F τ i ; λ 1 w i where F t is the cumulative density function t 0 fu; λdu. In the following you should assume that T i is exponentially distributed with density ft; λ = λe λt, λ 0 a Find the imum likelihood estimator of λ. Find the CR bound on unbiased estimators of λ. Is your estimator an efficient estimator? First, note that the likelihood function is only proper if w i interpreted as a Bernoulli variable with parameter P w i = 1 = F τ i ; λ, where τ i are non-random exit times those that do not exit the study before death have τ i =. Otherwise, the likelihood function does not integrate to one. 3

4 Define N d = n w i the number of deaths reported and N N d is the rest. The MLE is simply the imizing value over λ of the likelihood function fx; λ = n f w i t i ; λ 1 F τ i ; λ 1 w i = λ N d exp w i t i + 1 w i τ i where we have used the fact that F τ i ; λ = 1 exp λτ i. Define x = n 1 n = n w it i + n 1 w iτ i and w = n 1 n w i = N d /n. Then, a simple analysis of the gradient of the log likelihood function reveals d ln f/dλ = nλ 1 w x and d ln f/dλ = nλ w. Hence the MLE is ˆλ = w x = 1 N d The estimator is not efficient since d ln f/dλ k λ ˆλ λ. The second derivative is λ n w i. This is the conditional Fisher information given w i, giving the conditional CR bound var λ ˆλ λ /N d The unconditional CRB is found by taking the expectation of the second derivative of the log likelihood, which is F λ = λ E λ [w i ] = λ 1 exp λτ i F τ i ; λ = λ Or, defining the indicator function z i = Iτ i = indicating the indices of those individuals who do not exit the study before they die, F λ = λ M + 1 z i 1 exp λτ i where M is the number of these individuals. Note that the Fisher information decreases as more people exit the study, as is expected since this corresponds to less informative observations for λ. b Consider testing the one-sided hypotheses : λ = λ 0 : λ λ 0 where λ 0 0 is fixed. Find the most powerful test of level alpha 1. Is your test uniformly most powerful? 1 Hint: the sum of m i.i.d. standard exponential variables Y i with E[Y i] = 1 is Erlang distributed with parameter m, denoted Er n 4

5 The LRT is Since λ λ o λ Nd exp λ λ o λ o this reduces to H1 η or since the times τ i γ that individuals left the study are known and non-random w i t i γ 1 As t i are i.i.d. exponential λ o random variables under, λ o t i are i.i.d. standard exponential 1 random variables, and we can set the threshold using the fact that under : λ n o w it i is Erlang with N d degrees of freedom, so that γ 1 = λ 1 o Er Nd α. The test is uniformly most powerful since it does not depend on the unknown value of λ. c Now consider the two-sided hypotheses : λ = λ 0 : λ λ 0 Does there exist a UMP test for these hypotheses? Find the level α GLRT for testing vs. There does not exist a UMP since the sign of λ λ o changes over the uncertainty domain of λ unlike in part b. The GLRT is Nd / n Λ GLR = λ o Nd exp which can be rewritten as 1 Nd Λ GLR = C λ n o x exp i N d + λ o λ o where C = N N d d exp N d is a positive constant. Λ GLR is of the form Cu N d expu where u = λ n o, which is a convex function of u. Hence, the GLRT has an decision region of the form γ u γ +. Since n = n w it i + sum n 1 w iτ i, and w i and τ i are non-random and known, this test is equivalent to γ 1 w i τ i λ o w i t i γ + 1 w i τ i. Under the distribution of λ o n w it i is Er Nd which allows us to set the thresholds γ ± to give false alarm probability equal to α. H1 H1 η η 5

6 d Find a 1 α confidence interval for the parameter λ. Using the results of c, if γ and γ + are determined so that the GLRT is of level α, we have P λo γ / λ o γ + / = 1 α which, since λ o is arbitrary, specifies the following 1 α confidence interval for the parameter λ: [ ] γ /, γ + / 3 Here you will consider the multichannel simultaneous signal detection problem. Consider testing the following N hypotheses on the presence of a signal s i in at least one of N channels, where b i is a Bernoulli distributed random variable indicating the presence of signal s i, and w i is noise, all in the i-th channel. : x 1 = s 1 b 1 + w 1... : x N = s N b N + w N Here we assume that s i, b i and w i are mutually independent random variables and all quantities are independent over i. Let ˆb i {0, 1} be the decision function for the i-th channel where ˆb i = 0 and ˆb i = 1 corresponds to deciding b i = 0 and b i = 1 respectively. a What is the decision function ˆb i that corresponds to most powerful likelihood ratio test of level α for testing any one channels for signal, i.e., testing : = w i vs. H i : = s i + w i? Specify the test with threshold for the case that s i and w i are independent zero mean Gaussian random variables with variances σs and σw, respectively. An equivalent pair of hypotheses is: H i0 : b i = 0 vs H i1 : b i = 1. These are simple hypotheses so the MP LRT of level α is the Bayes optimal test, which has the form P X i b i = 1/P X i b i = 0 γ where η = 1 p i /p i, where p i = E[b i ] = P b i = 1. This is equivalent the posterior odds ratio test: E[b i X] Λ B = η E[1 b i X] Thus the Bayes optimal decision function is = { 1 ΛB η 0 Λ B η Under the Gaussian assumption given, σ Λ B = w σs σs + σw exp σs + σwσ w so that the Bayes optimal test is equivalent to x i pi 1 p i x i γ. 6

7 To find the power function and the ROC curves we first find the γ that gives probability of false alarm equal to α, which we find to be: γ = σwx α, where X1 1 1 α denotes the 1 α quantile of the Chi-square distribution with 1 degree of freedom. Therefore, the power function is β = 1 X 1 ρx α with ρ = σ w/σ s + σ w. b With the threshold found in part a what is the probability that at least one test of level α gives a false alarm among all of the N channels? This is the multichannel false alarm rate. Adjust the threshold on your test in part a so that the multichannel false alarm rate is equal to α. What is the new single channel false alarm rate and how does it compare to that of part a? What is the power of your test for correctly detecting signal presence in a given channel with this multichannel test of level α? Evaluate this power for case that s i and w i are independent zero mean Gaussian random variables with variances σs and σw, respectively. The power function associated with ˆb is found by expressing β m = P ˆb i = 1 in terms of α m = P ˆb i = 1. We will make the assumption here that the channels are all identical, i.e., the signals, noises and bernoulli variables in each channel are independent with identical distributions i.i.d., so that P ˆb i = 1, P ˆb i = 1 do not depend on i. In this case, the probability that at least one test of ˆb i gives a positive is P i {ˆb i = 1} H k = 1 P i {ˆb i = 0} H k = 1 N P ˆb i = 0 H k = = 1 P N ˆb i = 0 H k = 1 1 P ˆb i = 1 H k N for H k equal to or to. Hence, defining α as the level of the single channel test, we have that P i {ˆb i = 1} = 1 1 α N Therefore, if we set the left side to α and solve for α we obtain α = 1 1 α 1/N. Therefore, if gα denotes the power function for a single channel test ˆb i then the power function under the constraint P i {ˆb i = 1} = α is β = 1 1 g1 1 α 1/N N For the case that s i and w i are independent zero mean Gaussian random variables with variances σs and σw, respectively, from part a gα = 1 X 1 ρx α. c As an alternative to defining the decision function ˆb i as a MP LRT and adjusting the test for multichannel error protection using part b, here you will consider a different approach to optimal detection that accounts for the multichannel problem directly. Specifically, we define the optimal multichannel set of decision rules {ˆb i } N as those that imize the average number of true positives subject to a constraint on the average proportion of false positives over the N channels: E[ˆb k b k ] subject to E[ˆb k 1 b k ]/N q, 7

8 where q [0, 1] is the mean false positive rate, set by the user. Derive a general form of the optimal decision rule, illustrate it for the case that s i and w i are independent zero mean Gaussian random variables with variances σs and σw, respectively, and evaluate the power function. Using the Lagrange multiplier approach discussed in class to derive the Neyman-Pearson Lemma, we formulate this as the optimization { N } E[ˆb k b k ] + λ q E[ˆb k 1 b k ]/N Collect terms and, for Q a random variable, express E[Q] as E[E[Q X]] where X denotes all the observations: { N } E ˆbk [E[b k X] λe[1 b k X]/N] + λq The sum in the first expectation is imized by assigning ˆb k = 1 if and only if E[b k X] λe[1 b k X]/N 0 or equivalently ˆb k is the decision rule associated with the test E[b k X] E[1 b k X] which has exactly the same form as the single channel test in a and its multi-channel power curve is the same as that derived in b. d As another alternative consider defining the optimal multichannel set of decision rules {ˆb i } N as those that imize the average number of true positives but now subject to a constraint on the average proportion of false positives among all positives found: λ/n E[ˆb k b k ] subject to E[ˆb k 1 b k /M] q, where M = N ˆb i is the total number of positives found by the test. Derive a general form of the optimal decision rule and illustrate it for the case that s i and w i are independent zero mean Gaussian random variables with variances σs and σw, respectively. You do not have to evaluate the power function for this part. Again, using the Lagrange multiplier approach discussed in class to derive the Neyman-Pearson Lemma, we formulate this as the optimization { N } E[ˆb k b k ] + λ q E[ˆb k 1 b k ]/M Collect terms and, again, for Q a random variable, express E[Q] as E[E[Q X]] where X denotes all the observations: { N } E ˆbk [E[b k X] λe[1 b k X]/M] + λq 8

9 Now the fact that M = i ˆb i couples the multichannel tests together. However, if we fix M then the only ˆb k s that should be equal to one are those form which E[b k X] λe[1 b k X]/M 0 or equivalently ˆb k is the decision rule associated with the test E[b k X] E[1 b k X] An equivalent form for this decision rule is λ/m T k = E[1 b k X] E[b k X] = P b k = 0 X P b k = 1 X M/λ This test can be implemented by rank ordering all of the scores also called posterior odds ratios" T k in increasing order T 1... T N and finding the first index M at which T k goes above the straight line T i = i/λ. Only those hypotheses having scores less than than M/λ should be declared as valid discoveries. This is very similar to the Benjamini False Discovery Rate FDR test, and this Bayesian generalization is due to John Storey. 9