= Chi-squared testing lower bound
{title2=$\max\{\alpha,\beta\}\geq\tfrac12-\tfrac14\sqrt{\chi^2(P\Vert Q)}$}
The <Cauchy-Schwarz inequality> gives $\|P-Q\|_{\mathrm{TV}}=\frac12\mathbb E_Q|dP/dQ-1|\leq\frac12\sqrt{\chi^2(P\Vert Q)}$. Every <statistical hypothesis testing> rule has sum of its <Type I error> and <Type II error> at least $1-\|P-Q\|_{\mathrm{TV}}$, proving the displayed bound. A <mixture model> of alternatives yields a lower bound on the worst alternative error because a maximum dominates an average.
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