Homework 5

Hi all: Homework 5 has been posted on the webpage. It’s based on Chernoff bounds, some streaming, and a proof of a matrix Chernoff bound, and is due on March 20th. (In 2 weeks or so.)

Questions and comments below!

P.S. If you haven’t see the Lowner ordering before, solve exercise 4 before you start on problem 5.

Advertisements
This entry was posted in Uncategorized. Bookmark the permalink.

18 Responses to Homework 5

  1. Adam says:

    Is there a reason problem 5 is in parentheses?

    Like

  2. Adam says:

    Okay thanks! I also was wondering: in problem 4.b., we define S_n as sum(X1, X2, …, Xn). Is the same definition used on the left and right side of the equation? Or is S_n redefined as sum(X1, X2, …, X_(n-1)) on the right?

    If both sides use the same definition, it seems like weird things happen…the left side side is not a function of X_n because the Expected Value “integrates away” the X_n term, but that doesn’t happen on the right side. So the right side would still be a function of X_n.

    Like

  3. Adam says:

    Also I’m a little confused on 5.(a).I feel like I’ve found a contradiction which usually means I misunderstood something

    Suppose R=1/4. Suppose the matrices are 1×1.

    Suppose Y_i= {R} = {1/4}

    Suppose E[Y_i] = U_i = {1}

    X_i = {1/4-1} = {(1/4-1)/(1/4)} = {3}

    X_i * X_i = {9}

    The biggest eigenvalue of X_i*X_i is 9, so the 2-norm of X_i is 3, which is more than 1.

    Did I do something wrong here? Is there some way to ensure R>1/2?

    Like

    • I see the confusion: when we say \| Y_i \|_2 \leq R we mean all outcomes of Y_i have spectral value at most R. Much like saying Y_i \in [0,1]. In your example where E[Y_i] = 1, then there must be some other outcomes where Y_i is large.

      Like

  4. Adam says:

    Thanks! Hopefully my last question for a while:

    In part 1.b., I have a solution that works for N = 1/1000000 \epsilon^2 d but does not work for N=99999 \epsilon^2 d. Is that acceptable?

    I think that is a correct interpretation of the Omega notation but wanted to double check before writing the whole thing out.

    Like

  5. Jason says:

    What is the definition of 2-universal hash family in problem 2? From what I look up, it satisfies Pr(h(x) != h(y)) <= 1/N for all x != y. But that isn't sufficient, because h(0) could always be 0, and the tester would always return High. Should we assume that Pr(h(x) = y) = 1/N for all y?

    Like

    • Jason says:

      sorry, for the last sentence, I meant:
      But that isn’t sufficient, because h(0) could always be 0, and the tester would always return High if 0 is in the stream. Should we assume in addition that Pr(h(x) = y) = 1/N for all y?

      Like

      • Jason says:

        oops, sorry, I realized that it’s also defined in the lecture notes, and that we can indeed assume that.

        Like

  6. Love these comments that answer themselves! 🙂 Indeed, the definition is in Lecture 20 (section 2.1.1) and folds in the idea of being uniform on single elements, and being pairwise independent.

    Like

  7. David says:

    Just a note to others, when doing Exercise 4 on this homework, be sure to apply functions as defined here: http://www.cs.cmu.edu/~anupamg/advanced/lectures/lecture21.pdf#page=7

    f(A) does not denote, for instance, entrywise application.

    Like

  8. Jenny says:

    For problem 4c, it looks if you directly use 4(a) and 4(b) you don’t get a factor of d anywhere….

    Like

  9. Adam says:

    Is there an analog of 4.c. for expectation?

    Something like:
    Let x be a real number given from some probability distribution such that x is always in the range [a,b], and E[f(x)]<= g(x)
    Similarly let A be given by some probability distribution such that its eigenvalues are always in the range [a,b].
    From this we conclude E[f(A)]<= g(A)

    I don't recall seeing this theorem anywhere yet but it feels very natural. The problem is that E[f(A)] isn't actually a function of A, which makes weird things happen.

    Like

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s