YES, trivially. Even $E(\|X_0-X_1\|^2)$ is already bounded by 4x the variance. (or even 2x ?)
Am I missing something? How is $n$ related to $d$?
YES, trivially. Even $E(\|X_0-X_1\|^2)$ is already bounded by 4x the variance. (or even 2x ?)
Am I missing something? How is $n$ related to $d$?