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Adapted from a talk by Ali Malik at the teach-us-anything seminar.
Say you want to distinguish the uniform distribution over $[d]$ from a distribution that’s $\eps$-far from it in TV distance (say, with $90\%$ accuracy) by drawing random samples. How many do you need?
Detecting small support
If you want to do this for any $\eps \le 1/2$, then you need to be able to distinguish between (A) the uniform distribution and (B) a uniform distribution over a random subset of $d/2$ elements. And for that, you need to see some repeated element in the sample (a “collision”): indeed, conditioned on no collision, the sample $(X_1, \ldots, X_n)$ will be a random sample of $n$ distinct points in either case.
What’s the chance of getting a collision among $n$ samples? Well, for both (A) and (B), each of the $\binom{n}{2}$ pairs of samples has a $O(1/d)$ chance to collide, so by a union bound the probability of a collision is $O(n^2/d)$. Therefore, in order to get a collision with constant probability (and thus have a shot at good accuracy), you need $n = \Omega\p{\sqrt{d}}$.
On the other hand, counting the number of collisions $Z$ seems promising when $n$ is, say, $100\sqrt{d}$: for (A), $\E[Z] = \binom{n}{2}/d$, and for (B), $\E[Z] = 2\binom{n}{2}/d$. So as long as $Z$ is reasonably concentrated, we’re good. And it will be! Indeed, the events $X_i=X_j$ are all pairwise independent, so since sums of pairwise-independent Bernoullis vary like they average, we’ll get good concentration by Chebyshev’s once $\binom{n}{2}/d$ is a large enough constant.
General case
If we’re dealing with a general distribution $p=(p_1, \ldots, p_d)$ that’s $\eps$-far from uniform, can we still use the number of collisions?
First, let’s check that $\E[Z]$ is still significantly larger than $\binom{n}{2}/d$. The canonical case of a distribution that’s $\eps$-far from uniform is the distribution where $p_1 = \cdots p_{n/2} = (1+\eps)/d$ and $p_{n/2+1} = \cdots = p_n = (1-\eps)/d$. In that case, the probability that $X_i = X_j$ is
\[
\sum_{r=1}^d \Pr[X_i = X_j = r] = \sum_{r=1}^d p_r^2 = \frac{d}{2}\p{\frac{1+\eps}{d}}^2 + \frac{d}{2}\p{\frac{1-\eps}{d}}^2 = \frac{1+\eps^2}{d},
\]
so we get $\E[Z] = (1+\eps^2)\binom{n}{2}/d$. Nice!
And in fact, the probability of a collision will always be at least $(1+\eps^2)/d$. Indeed, if we express the probabilities $p_i$ by their “difference with uniform” $p_i = 1/d + \delta_i$, then we get
\[
\Pr[X_i = X_j] = \sum_{r=1}^d p_r^2 = \sum_{r=1}^d\p{1/d^2 + 2\delta_r/d + \delta_r^2} = 1/d + 0 + \norm{\delta}_2^2.
\]
And the condition that $p$ is $\eps$-far from uniform in TV distance translates to $\norm{\delta}_1 \geq \eps$, so the minimum value of $\norm{\delta}_2^2$ is reached when $\delta$ is as equally spread out as possible (which corresponds the distribution mentioned earlier):
\[\Pr[X_i = X_j] = 1/d + \norm{\delta}_2^2 \geq 1/d + \p{\norm{\delta}_1/\sqrt{d}}^2 = \frac{1+\norm{\delta}_1^2}{d} \geq \frac{1+\eps^2}{d}.\]
So now we only have to prove concentration. This time, collisions are not necessarily pairwise independent. Consider the events $X_i = X_j$ and $X_k = X_l$.
- If $(i,j)$ and $(k,l)$ don’t share a variable, then those events completely independent.
- But if $(i,j)$ and $(k,l)$ do share a variable (say $i=k$), then if $X_i$ takes a particularly frequent value, then that makes both collisions more likely.
On the other hand, in this second case, the “amount by which they lack independence” is intuitively proportional to probability that we get a 3-way collision. And for the uniform distributions, the probability that any 3-way collision happens is only $O(n^3/d^2)$, which is tiny when $n \approx \sqrt{d}$. So it might be the case that $Z$ is not very well concentrated, but in that case we’re likely to see a 3-way collision, and that would be a very clear sign that the distribution is not uniform!
Therefore, (I think) we can test for independence in this way:
- set $n \ce 10\sqrt{d}/\eps^2$ (this makes $\E[Z] \sim 100/\eps^4$, which should be enough for a $(1+\frac{\eps^2}{3})$-approximation);
- if you see any 3-way collision, say “not independent”;
- otherwise, test if the number of collisions is smaller or larger than $(1+\frac{\eps^2}{2})\binom{n}{2}/d$.
Note: According to Ali, the simple “number of collisions” test does work (without special handling of 3-way collisions), but it requires some effort to prove.