3

Suppose that $X_1,\ldots,X_n$ are independent sub-Gaussian random variables with $ \max_{1\leq i \leq n} \Vert X_i \Vert_{\psi_2} \leq 1$. I need to prove that:

$$P \left( \max_{1\leq i \leq n} X_i \geq t \sqrt{\log n} \right) \leq 2 \exp (-Ct^2)$$

Approach:

Since $ X_i $ are sub-Gaussian random variables with $ \Vert X_i \Vert_{\psi_2} \leq 1 $, each $ X_i $ satisfies the tail bound: $$ P(|X_i| \geq u) \leq 2 \exp(-cu^2) $$ for some constant $ c > 0 $ and any $ u > 0 $.

We are interested in bounding the probability $ P\left( \max_{1 \leq i \leq n} X_i \geq t \sqrt{\log n} \right) $.

By the union bound: $$ P\left( \max_{1 \leq i \leq n} X_i \geq t \sqrt{\log n} \right) \leq \sum_{i=1}^n P\left(X_i \geq t \sqrt{\log n} \right). $$

For each $ i $, using the sub-Gaussian tail bound with $ u = t \sqrt{\log n} $, we get: $$ P\left(X_i \geq t \sqrt{\log n} \right) \leq \exp(-c t^2 \log n) = n^{-c t^2}. $$

Summing these probabilities, we obtain: $$ P\left( \max_{1 \leq i \leq n} X_i \geq t \sqrt{\log n} \right) \leq n \cdot n^{-c t^2} = n^{1 - c t^2}. $$ For this probability to be small, we need $ 1 - c t^2 \leq -1 $, or $ c t^2 \geq 2 $.

Therefore, choosing $ t $ large enough (e.g., $ t \geq \sqrt{2/c} $), we have: $$ P\left( \max_{1 \leq i \leq n} X_i \geq t \sqrt{\log n} \right) \leq 2 \exp(-Ct^2), $$ where $ C = c / 2 $ is a universal constant.

Where am I going wrong with this? Can anyone help me out here? Thanks!

Maths Freak
  • 773
  • 1
    Probably helpful: https://math.stackexchange.com/questions/1676407/tail-bounds-for-maximum-of-sub-gaussian-random-variables – PhoemueX Oct 30 '24 at 20:24
  • The max of subgaussians is again subgaussian. When you union bound like this, you lose that, so you get the sub-optimal scaling behavior that you have. – Andrew Oct 31 '24 at 01:34
  • @Andrew, but can we always say that maximum of sub gaussians is sub gaussian? Don't we need additional assumptions like $E(X)=0$? – Maths Freak Oct 31 '24 at 15:20
  • I didn't work it out if they're not all centered, but you are already assuming they are centered in your post. – Andrew Oct 31 '24 at 19:32
  • To be clear, I do not know what definition of subgaussian you are using. However, the inequality $P(|X|>t)\leq 2\exp(-ct^2)$ implies that $X$ is centered. – Andrew Oct 31 '24 at 21:32

0 Answers0