0
$\begingroup$

From here:

The notion of pseudoentropy is only useful, however, as a lower bound on the computational entropy in a distribution. Indeed, it can be shown that every distribution on $\{0,1\}^n$ is computationally indistinguishable from a distribution of entropy at most $poly (\log n)$.

How is that any distribution on $n$-cube can be computationally indistinguishable from one with entropy at most $poly(\log n)$?

$\endgroup$
2
$\begingroup$

Take any distribution $D$ on $\{0,1\}^n$. Sample $k(n)$ points $x_1, \cdots, x_{k(n)}$ independently from $D$ and let $\tilde D$ be the uniform distribution that gives a random $x_i$. Then, if $k(n)$ is super-polynomially large e.g. $k=n^{\log n}$, you cannot distinguish $D$ and $\tilde D$ using only $\mathrm{poly}(n)$ samples. Hence $D$ is computationally indistinguishable from a distribution $\tilde D$ with entropy $\mathrm{poly}\log(n)$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.