COMING ATTRACTIONS FOR EECS 564
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:
- 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.
- Answers: a=1.26/T, d2=0.815(optimal d2)
- 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:
- Problem #1: Your answer should be simple to interpret.
- 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.
- For K=0 or A=0 you should get 3/4 (why is this reasonable!).
In binary hypothesis testing, two different criteria
both lead to the same answer: the likelihood ratio test:
- The likelihood ratio is LR(R)=pr|H1(R|H1)
The relation > < means (on this web page):
- If LR(R) > n, choose H1.
- If LR(R) < n, choose H0
- MIN[Bayes risk=E[cost]] leads to the test LR(R) > < n
(C01-C11)]; these were defined on Jan. 8.
- Special case: Min[Pr[error]] (MEP) leads to n=p0/p1 (MAP).
- 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.
- ROC (Receiver Operating Characteristic): Plot of PD vs. PF for all n.