COMING ATTRACTIONS FOR EECS 564 Exam #3 may be picked up outside 4114 EECS by 5 PM Friday
April 5-9, 1999

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Problem Set #9 Hints:

• The answer to Problem #1 is: (1/a)tanh[½ aT](y(0)+y(T))
• There will be NO CLASS Monday April 12.
• I will be giving a presentation to the National Advisory Board

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Problem Set #6 Hints:

• Problem #1:
  • Note that h(t)*x(t) is known under H₀ and H₁
  • So this is simply binary hypothesis testing in WGN
  • Arithmetic mean=(x+y)/2> harmonic mean=2xy/(x+y) implies d²> 0
  • Well-known inequality: Arithmetic mean > geometric mean > harmonic mean
  • PROOF: (x^½-y^½)²> 0 for first inequality;
  • take the reciprocal of the first inequality and multiply by xy for the second inequality
• Problem #3: R(t) is known for 0< t< T, but h(t)*R(t) is known for all t> 0.
• Answers: a=1.26/T, d²=0.815(optimal d²)
• Problem #4: The signal vector tips do NOT require a 4-D sketch!

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Problem Set #5 Hints:

• Problem #2: Use Mercer's theorem and results from antipodal signaling.
• Problem #3: Do entirely in the time domain.
• Also note (&int x(t)y(t)dt)²=&int&int x(t)y(t)x(s)y(s)dtds.
• Also note sinc*sinc=sinc. "f(t)" here should be "x(t)."
• Problem #6a: Note that Gram-Schmidt can be used to compute the basis.
• But you DON'T have to do it--just know that it can be done.

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Problem Set #4 Hints: On Problem #1, do the problem using the given p_m(M).
Then THINK: what property of that p_m(M) was vital for UMP?
You do NOT have to PROVE your result--just EXPLAIN why it is true.

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On Problem #4, use deterministic Cauchy-Schwarz:
(&int x(t)y(t)dt)²< (&int x(t)²dt)(&int y(t)²dt)

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Problem Set #3 Hints:
1. Problem #1: Your answer should be simple to interpret.
2. Problems #2,4: see back side of "Multichannel Cramer-Rao bound" handout.
3. Problem #3: Follow the hints on the problem set handout.
   • The resulting sufficient statistics have simple, known distributions.
   • Pr[error]=1-erf²(something).
   • For K=0 or A=0 you should get 3/4 (why is this reasonable!).

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In binary hypothesis testing, two different criteria
both lead to the same answer: the likelihood ratio test:

1. The likelihood ratio is LR(R)=p_r|H1(R|H₁) /p_r|H0(R|H₀).
2. The relation > < means (on this web page):
   • If LR(R) > n, choose H₁.
   • If LR(R) < n, choose H₀
   .
3. MIN[Bayes risk=E[cost]] leads to the test LR(R) > < n
   where n=[p₀(C₁₀-C₀₀)]/[p₁ (C₀₁-C₁₁)]; these were defined on Jan. 8.
4. Special case: Min[Pr[error]] (MEP) leads to n=p₀/p₁ (MAP).
5. Neyman-Pearson criterion:
   • P_F=Pr[false alarm]=Pr[choose H₁|H₀ true].
   • P_D=Pr[detection]=Pr[choose H₁|H₁ true].
   • MAX[P_D] subject to constraint P_F < a=level of significance
   • leads to the test LR(R) > < n where n is such that P_F=a.
6. Minimax criterion: MIN_nMAX_p1E[cost]
   • You choose threshold n such that when
   • nature chooses the actual a priori p₁=Pr[H₁]
   • the damage E[cost] is minimized (worst-case scenario)
   • Solution: For MEP, plot P_D+P_F=1 on ROC.
   • The slope of the ROC at the intersection is the threshold n.
7. ROC (Receiver Operating Characteristic): Plot of P_D vs. P_F for all n.