Jonathan Scarlett
@jmscarlett.bsky.social
Associate Professor, National University of Singapore. Working in information theory, machine learning, and statistics.
How about: An MSE of c/n would imply (by Markov) reasonable probability of distinguishing Bernoulli( 1/2+sqrt(10c/n) ) from Bernoulli( 1/2-sqrt(10c/n) ), which is impossible for "small" c because their KL divergence is O(c/n). (So O(c) with n samples -- small KL, hence small TV by Pinsker)
February 2, 2025 at 3:03 PM
How about: An MSE of c/n would imply (by Markov) reasonable probability of distinguishing Bernoulli( 1/2+sqrt(10c/n) ) from Bernoulli( 1/2-sqrt(10c/n) ), which is impossible for "small" c because their KL divergence is O(c/n). (So O(c) with n samples -- small KL, hence small TV by Pinsker)