  We study the problem of testing identity against a given distribution with a
focus on the high confidence regime. More precisely, given samples from an
unknown distribution $p$ over $n$ elements, an explicitly given distribution
$q$, and parameters $0< \epsilon, \delta < 1$, we wish to distinguish, {\em
with probability at least $1-\delta$}, whether the distributions are identical
versus $\varepsilon$-far in total variation distance. Most prior work focused
on the case that $\delta = \Omega(1)$, for which the sample complexity of
identity testing is known to be $\Theta(\sqrt{n}/\epsilon^2)$. Given such an
algorithm, one can achieve arbitrarily small values of $\delta$ via black-box
amplification, which multiplies the required number of samples by
$\Theta(\log(1/\delta))$.
  We show that black-box amplification is suboptimal for any $\delta = o(1)$,
and give a new identity tester that achieves the optimal sample complexity. Our
new upper and lower bounds show that the optimal sample complexity of identity
testing is \[
  \Theta\left( \frac{1}{\epsilon^2}\left(\sqrt{n \log(1/\delta)} +
\log(1/\delta) \right)\right) \] for any $n, \varepsilon$, and $\delta$. For
the special case of uniformity testing, where the given distribution is the
uniform distribution $U_n$ over the domain, our new tester is surprisingly
simple: to test whether $p = U_n$ versus $d_{\mathrm TV}(p, U_n) \geq
\varepsilon$, we simply threshold $d_{\mathrm TV}(\widehat{p}, U_n)$, where
$\widehat{p}$ is the empirical probability distribution. The fact that this
simple "plug-in" estimator is sample-optimal is surprising, even in the
constant $\delta$ case. Indeed, it was believed that such a tester would not
attain sublinear sample complexity even for constant values of $\varepsilon$
and $\delta$.
