Let $D$ be an $\epsilon$-biased distribution we want to show that $$\text{Supp}(D)\geq \Omega\bigg(\frac{n}{\epsilon^2\log(\frac{1}{\epsilon})}\bigg)$$ I know that there are some proofs for this but I am intrested in the following particular way.

Let $X,Y$ be two distributions over $\{0,1\}^n$. Consider the "convolution" of distributions $Z=X\circledast Y$ to be defined by sampling $x\sim X$ and $y\sim Y$ independently and outputting $x\oplus y$ where $\oplus$ is just bit-wise xor.

Now if we define a distribution $D^{(t)}=D\circledast\cdots\circledast D$ ($t$ times) we can use some Fourier analysis and combinatorics to show that

  1. $D^{(t)}$ is $\epsilon^t$-biased.
  2. $|\text{Supp}(D^{(t)})|\leq\binom{|\text{Supp}(D)|+t}{t}$

We can also show that for every $\delta$-biased distribution $Z$ we have $\lVert Z\rVert_2^2\leq\delta^2+2^{-n}$ when considering the distribution as a vector of probabilities. Now, I was told that I can use the latter to obtain a lower bound on $|\text{Supp}(D^{(t)})|$ and then choose $t$ accordingly to get the desired bound.

However, the only bound I can get for $|\text{Supp}(D^{(t)})|$ is $$|\text{Supp}(D^{(t)})|\geq\sqrt{\frac{1}{\epsilon^{2t}+2^{-n}}}$$ And I don't see how that helps...

  • 1
    $\begingroup$ It's not clear what you're asking. If you're trying to understand an existing result, please cite it. Are you suggesting a novel proof, etc?.. $\endgroup$ – Aryeh Dec 29 '18 at 21:29
  • $\begingroup$ @Aryeh In some lecture notes by Swastik Kopparty (about Boolean functions and derandomization) I have seen an "exercise" consisting of four parts that asks to prove the "famous" lower bound on the support size of $\epsilon$-biased distribution. The first three parts are described above and I managed to handle with them. At the last part the exercise asks to use the first parts in order to get the lower bound, I do not understand how to do so. I do know that there is a proof not following the above line, but I am interested in the above. $\endgroup$ – user621824 Dec 30 '18 at 7:43
  • $\begingroup$ @user621824 OK, first: are you sure about the square root in your RHS? (Note that the $\ell_2$ norm of a distribution over support of size $N$ is lower bounded by $1/\sqrt{N}$; you here have the squared $\ell_2$ norm). Second: set $t$ such that the two terms of the denominator balance. Third: use $\binom{n}{k} \leq (en/k)^k$. This ought to be enough. $\endgroup$ – Clement C. Dec 30 '18 at 9:04
  1. You shouldn't have a square root. Namely, for every $\delta$-biased distribution $Z$ (using your notation), we have $$ \delta^2+2^{-n} \geq \lVert Z\rVert^2_2 \geq \frac{1}{\lvert\operatorname{supp} Z\rvert}\tag{1} $$ since the squared $\ell_2$ norm of a distribution over support of size $N$ is at least that of the uniform distribution on $N$ elements, which is $1/N$.

  2. This gives you, for $t\geq 1$ to choose suitably, $$\lvert\operatorname{supp} D^{(t)}\rvert \geq \frac{1}{\varepsilon^{2t}+2^{-n}} \tag{2} $$ since as you have shown $D^{(t)}$ is $\delta$-biased for $\delta=\varepsilon^t$. A natural choice of $t$, in view of (2), is the one balancing the denominator, i.e., such that $\varepsilon^{2t}=2^{-n}$. This leads to taking $$ t \stackrel{\rm def}{=} \frac{n}{2\log(1/\varepsilon)}\tag{3} $$

  3. Recalling the standard bound $\binom{n}{k}\leq \left(\frac{en}{k}\right)^k$ for $1\leq k\leq n$ along with (2), (3), and what you mentioned as 2. in your question, we thus get $$ \left(\frac{e(\lvert\operatorname{supp} D \rvert+t)}{t}\right)^t \geq \binom{\lvert\operatorname{supp} D \rvert+t}{t} \geq \frac{1}{\varepsilon^{2t}+2^{-n}} = \frac{1}{2\varepsilon^{2t}} \tag{4} $$ so that $$ \frac{e(\lvert\operatorname{supp} D \rvert+t)}{t} \geq \frac{1}{2^{1/t}\varepsilon^{2}} \tag{5} $$

Rearranging and recalling again the choice of $t$ in (3) yields $$ \lvert\operatorname{supp} D \rvert\geq \Omega\!\left(\frac{n}{\varepsilon^{2}\log(1/\varepsilon)} \right) \tag{6} $$ as claimed.

  • $\begingroup$ Sorry for not accepting for a long time. Thank you! $\endgroup$ – user621824 Jan 2 at 16:24
  • $\begingroup$ @user621824 No problem at all. Glad this helped. $\endgroup$ – Clement C. Jan 2 at 16:25

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