Lec #19: JL etc.

So two things that we wanted to check during lecture today, and some notes:

  • If {X_i} is sub-Gaussian with parameter {c_i}, and they are independent, then {\sum a_i X_i} is indeed sub-Gaussian with parameter {\sqrt{ \sum_i a_i^2 c_i^2 }}, as claimed. Basically, just use the MGF definition to show this. And note the analogy to Gaussians, where {c_i} should be the standard deviation, and not the variance.
  • I got the sub-exponential equivalence slightly wrong. The MGF-based definition: there exist parameters {(\nu, b)} such that

    \displaystyle  \mathbb{E}[ e^{t(W - \mathbb{E}[W])} ] \leq exp( \nu^2 t^2/2) \qquad \forall |t| \le 1/b.

    This is equivalent to there existing some other constants {c_1, c_2} such that

    \displaystyle  \Pr[ |W - \mathbb{E}[W]| \geq \lambda ] \leq c_1 e^{- c_2 t}.

    See Theorem~2.2 here for a proof. (Thanks for pointing out that mistake, Tom!)

  • For compressive sensing, there are a million resources online. E.g., here’s Terry Tao on the topic.
  • I should have emphasized: the blood testing example we talked about was the simple case where you were guaranteed that the {x} vector was {1}-sparse. I.e., there is exactly one person with the disease among a population of {D} people. Then the binary search strategy we did used {\log_2 D} linear tests, which is optimal. (To clarify, for each {i}, the {i^{th}} test combines together the samples for all people whose index has a {1} in the {i^{th}} place and tests this combination. Now construct a {\log_2 D}-bit vector with a {1} in the {i^{th}} position precisely if the {i^{th}} test came up positive: this is the index of the person with the infection.

    Note this is a non-adaptive strategy: we can write down all the tests up-front, and can read off the answer from their results. This is opposed to an adaptive strategy that would look at the answers to previous tests to decide which sets to test next.

    What about the case where there are exactly two people infected? The same strategy does not work. In fact, since there are {\binom{n}{2}} answers and each answer gives us {1} bit each time, we must perform at least {\log_2 \binom{n}{2}} tests. Any thoughts on how to do this? An adaptive strategy that uses {2 \log_2 n} tests is easy: can you get a non-adaptive strategy? Using randomness?

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