I had a question about a general statement regarding finding a computationally indistinguishable distribution given any distribution, observed (in the third paragraph of Section 11, page 31) here. This is the statement:

For any distribution $D$ over ${0,1}^{n}$, there exists a distribution $D’$ such that $D$ and $D’$ are $epsilon$ indistinguishable with respect to any classical distinguisher of size $n^{k}$.

Let me restate the proof.

Let $D$ be any distribution over ${0,1}^{n}$. Then choose $w$ elements

independently with replacement from $D$, and let $D′$ be the uniform

distribution over the resulting multiset (so in particular, $H(D′) leq

log_{2} n$). Certainly $D′$ can be sampled by a circuit of size $mathcal{O}(nw)$: just

hardwire the elements. Now, by a Chernoff bound, for any fixed circuit

$C$, clearly $D$ and $D′$ are $epsilon$-indistinguishable with respect to $C$, with

probability at least $1 − e^{-epsilon^{2} w}$ over the choice of $D′$. But there

are “only” $n^{mathcal{O}(n^{k})}$ Boolean circuits of size at most $n^k$. So by the union

bound, by simply choosing $w = Omegaleft(frac{n^{k} log n}{epsilon^2}right)$, we can ensure that $D$ and $D′$ are $ε$-indistinguishable with respect to all circuits

of size at most $n^{k}$, with high probability over $D′$.

I do not understand how the Chernoff bound is applied. How do we know the action of the circuit $C$? I also don’t understand why $w = Omegaleft(frac{n^{k} log n}{epsilon^2}right)$ in the union bound. Since we need to “protect against” $n^{mathcal{O}(n^{k})}$ Boolean circuits, shouldn’t $w$ be something like $w = Omegaleft(frac{n^{mathcal{O}(n^k)} log n}{epsilon^2}right)$?