- 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 #1:
- Note that h(t)*x(t) is
*known*under H_{0}and H_{1} - So this is simply binary hypothesis testing in WGN
- Arithmetic mean=(x+y)/2> harmonic mean=2xy/(x+y) implies d
^{2}> 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

- Note that h(t)*x(t) is
- 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, d
^{2}=0.815(optimal d^{2}) - Problem #4: The signal vector
*tips*do NOT require a 4-D sketch!

- 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.

Then THINK: what property of that p

You do NOT have to PROVE your result--just EXPLAIN why it is true.

On Problem #4, use

(&int x(t)y(t)dt)

- 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.
- Pr[error]=1-erf
^{2}(something). - For K=0 or A=0 you should get 3/4 (why is this reasonable!).

both lead to the same answer:

- The likelihood ratio is LR(R)=p
_{r|H1}(R|H_{1}) /p_{r|H0}(R|H_{0}). - The relation > < means (on this web page):
- If LR(R) > n, choose H
_{1}. - If LR(R) < n, choose H
_{0}

- If LR(R) > n, choose H
- MIN[Bayes risk=E[cost]] leads to the test LR(R) > < n

where n=[p_{0}(C_{10}-C_{00})]/[p_{1}(C_{01}-C_{11})]; these were defined on Jan. 8. *Special case:*Min[Pr[error]] (MEP) leads to n=p_{0}/p_{1}(MAP).- Neyman-Pearson criterion:
- P
_{F}=Pr[false alarm]=Pr[choose H_{1}|H_{0}true]. - P
_{D}=Pr[detection]=Pr[choose H_{1}|H_{1}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.

- P
- Minimax criterion: MIN
*You*choose threshold n such that when*nature*chooses the actual a priori p_{1}=Pr[H_{1}]- 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.

_{n}MAX_{p1}E[cost]- ROC (Receiver Operating Characteristic): Plot of P
_{D}vs. P_{F}for all n.