0

It is a common technique to obtain tail bounds from the MGF. But now suppose I have the other way round. Suppose it is given,

$$\mathbb P [X_i \geq t] \leq e^{-t^2/2\sigma^2}$$

then what can you say about,

$$\mathbb P [\sum_{i=1}^n X_i \geq t]$$

Clearly the ideal way would be to recovering the MGFs. But using standard CDF formulation,

$$\int \mathbb P[e^{\lambda X_i } \geq u] du$$

doesn't seem to work well i.e., I was not able to recover the MGF $\log \mathbb E \exp (\lambda X) \leq \lambda^2 \sigma^2/2$ which one intuitively expects. Any thoughts or ideas which I am missing?

DuttaA
  • 283
  • Proposition 2.5.2 in Vershynin's book is exactly what you are looking for: https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html – Clement C. Nov 12 '22 at 02:08
  • @ClementC. aware of it. But it is the other way round. – DuttaA Nov 12 '22 at 02:16
  • https://math.stackexchange.com/questions/4264714/tail-bound-implies-sub-gaussian might be what I am looking for. I'll update as sson as I work it out. – DuttaA Nov 12 '22 at 02:18
  • How is this the other way around? The theorem states that all conditions are equivalent... so they all imply each other. You can extract the argument you want (MGF bound from tail bound, assuming the mean is zero) from that proof. – Clement C. Nov 12 '22 at 06:13
  • @ClementC. 1.) I think your MGF turns out to be worse (i.e. you give worst bound by some constant factors). 2.) In the book they sweep everything under some universal constant $C,K$. I am expecting something more accurate. – DuttaA Nov 12 '22 at 11:45
  • The point about the constant is fair, but the second... as I said, you can extract the proof from there. The proof Vershynin gives can be made "more accurate, he just swept these under the rug as universal constants for simplicity, but just going through the proof will give actual values. My comment was basically siggesting you try to adapt that proof.. – Clement C. Nov 12 '22 at 23:24
  • @ClementC. I actually have gone through the proofs and have tried to adapt it. Unfortunately I am not particularly a math person and Vershynin's book sometimes uses mathematical inequalities (which might be common to math people) unfamilar to me. I do agree with your assessment, it is just I am much too cowardly to do it, since I beleive someone more familiar with the probability concepts will come up with tighter bounds, – DuttaA Nov 12 '22 at 23:32

0 Answers0