Final Exam
The final exam will be placed here at 1200 hours on the 30th of April.
Rules:
1. This exam is due back (scanned and sent via email, or placed in the class lockbox) by 11:59:59.00 (AM) on the 1st of May.
2. The only resources you're allowed to use your notes, this class website, either of the two textbooks, and your own wits. No talking with anyone else, nor reading anything else, nor looking at any other website.
3. In the event of questions, the instructor will be in his office at HSH 706 from 1730 to 1830 hours on 30th April. He'll also respond to emails quickly from 1100 to 1200 on the 1st of May.
4. While you're welcome to spend as much of 24 hours as you wish, this exam is designed to be solved in 2 hours.
5. Use this wiki to ask questions, post notices about typos. Please do not dicuss proof techniques/answers/anything else.
Click here to download the question paper. Good luck!
Comments (12)
Mak said
at 1:39 pm on Apr 30, 2009
For Q2, what are the assumptions of X^n nd Y^n? Are they generated from the same iid source?
Mak said
at 1:45 pm on Apr 30, 2009
And for Q3(a), what is the role of \epsilon? By "able to decode correctly", do you mean that ALL Y^n must be decoded correctly, or do you mean we allow some error in terms of \epsilon?
MEI Yuchen said
at 2:36 pm on Apr 30, 2009
In Q2, what does "always" mean? Is it 100%, or allow some errors?
sidjaggi said
at 2:55 pm on Apr 30, 2009
Mak: Q2: Your answer should hold for ARBITRARY X^n, Y^n (regardless of how they're generated).
Q3(a), the role of espilon is to be an arbitrarily small positive constant. by decoding correctly, we mean that an arbitrarily large fraction of Y^n, as some function of \epsilon.
Yuchen: Q2: always = 100% = no errors.
Mak said
at 5:01 pm on Apr 30, 2009
Thanks for the answers.
So for Q3(a), by "needs to transmit at least n(H(Y|X) − \epsilon) bits to Bob", do you mean that there must exist SOME Y^n that requires us to send at least n(H(Y|X) − \epsilon) bits? Or the EXPECTED number of bits to be transmitted is at least n(H(Y|X) − \epsilon)?
sidjaggi said
at 5:09 pm on Apr 30, 2009
(Tong walks into my office to ask some questions)
Tong: asks about Q2: What does the probability of error mean?
Sid: As I'd said in my previous comment, your scheme should work for any X^n, Y^n. There is no randomness in X^n, Y^n -- the only randomness is from whatever scheme you may choose to use.
More precisely, X^n, Y^n are chosen somehow. Then, independently, you choose a scheme, possibly with some random variable R. For each X^n, Y^n, the probability of error over R should be at most the quantity given in the question.
Tong: Q4: Do the elements of Z come from only X or Y, or possibly from some other set too?
The random variable Z is the same as the random variable with probability p, and the same as the random variable Y with probability 1-p. There are no other possibilities for Z since p + 1-p = 1.
Tong: Followup question -- but what if X and Y intersect? The question does not cover this case.
Sid: Well, with probability p, Z equals X, i.e., it equals some value in the domain of X. With probability 1-p it equals Y, i.e, equals some value in the domain of Y. It is quite possible that there are some common values in the domains (the question indeed says this explicitly in the hint), but Z is still well-defined.
Mak said
at 5:28 pm on Apr 30, 2009
No reply for my latest question? = =
Lingxiao XIA said
at 10:17 pm on Apr 30, 2009
hi, do we have to prove the things we have already proved in class? like the probability of an element is from the strongly typical set?
Lingxiao XIA said
at 10:20 pm on Apr 30, 2009
and is the polynomial time complexity a rigid requirement for question 4(b)? do we have to prove it?
Poppin Ke said
at 9:31 am on May 1, 2009
can i submit the paper directly to you, Michael ?
Cho Yiu Ng said
at 10:05 am on May 1, 2009
Liu Ke: You can put it inside the course assignment box on 8/F, SHB. Of course, I recommend you to scan and email the answers to both Prof. Jaggi and me.
Poppin Ke said
at 11:18 am on May 1, 2009
i put it in the assignment box since i could not find the scanner
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