Corollary 11.8: under independent priors, GREEDY suffers Bayesian regret $\ge T \cdot \frac{\alpha}{2}(\mu^0_1 - \mu^0_2)\Pr[\mu_2 > 1

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Corollary 11.8: under independent priors, GREEDY suffers Bayesian regret $\ge T \cdot \frac{\alpha}{2}(\mu^0_1 - \mu^0_2)\Pr[\mu_2 > 1 - \alpha]$ · Prove2Me