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

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

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: (x½-y½)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.
• Problem #4: The signal vector tips do NOT require a 4-D sketch!

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.

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

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

Problem Set #3 Hints:
2. Problems #2,4: see back side of "Multichannel Cramer-Rao bound" handout.
• 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!).

In binary hypothesis testing, two different criteria
the likelihood ratio test:
1. The likelihood ratio is LR(R)=pr|H1(R|H1) /pr|H0(R|H0).
• The relation > < means (on this web page):
• If LR(R) > n, choose H1.
• If LR(R) < n, choose H0
.
2. 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.
3. Special case: Min[Pr[error]] (MEP) leads to n=p0/p1 (MAP).
• Neyman-Pearson criterion:
• PF=Pr[false alarm]=Pr[choose H1|H0 true].
• PD=Pr[detection]=Pr[choose H1|H1 true].
• MAX[PD] subject to constraint PF < a=level of significance
• leads to the test LR(R) > < n where n is such that PF=a.
• Minimax criterion: MINnMAXp1E[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 PD+PF=1 on ROC.
• The slope of the ROC at the intersection is the threshold n.
4. ROC (Receiver Operating Characteristic): Plot of PD vs. PF for all n.