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I am reading an article related to streaming algorithms named "Turnstile streaming algorithms might as well be linear sketched" by Yi Li, Huy Nguyen and David Woodruff,

At some point they have a random algorithm (uses a tape of random bits) that solved a problem over $\mathbb Z_{|m|}^{n} = \{-m,..,m\}^n$ that succeeds with probability $1-\delta$

They want to reduce the number of bits needed for randomness used by that algorithm using via the following statement:

Theorem 9: Let A be a randomized automaton solving problem P on $\mathbb Z_{|m|}^{n}$ with failure probability at most $\delta$. There is a ranomized automaton B that only needs to pick uniformly at random one of $O(\delta ^{-2}nlogm)$ deterministic instances of A and solves P on $\mathbb Z_{|m|}^{n}$ with failure probability at most $2\delta$

Proof: Let $A_1 ,A_2 ,.., A_{O(n\delta ^{-2}logm)}$ be independent draws of deterministic automata picked by B.

Fix an input $x\in \mathbb Z_{|m|}^{n}$.

Let $p_A(x)$ be the fraction of automata among $A_1 ,A_2 ,.., > A_{O(n\delta ^{-2}logm)}$ that solve problem P correctly on x and $p_B(x)$ be the probability that B solves P on x correctly.

By a Chernoff bound, we have that $Pr\{|p_A(x)-p_B(x)|\geq \delta\} \leq exp(-O(nlogm)) < (2m+1)^{-2n}$.

Taking a union bound over all choices of $x\in \mathbb Z_{|m|}^{n}$, we have $Pr\{|p_A(x)-p_B(x)|\geq \delta \ for\ all\ x\} > 0$.

Therefore, there exists a set of $A_1 ,A_2 ,.., A_{O(n\delta > ^{-2}logm)}$ such that $|p_A(x)-p_B(x)|\leq \delta$ for all $x\in > \mathbb Z_{|m|}^{n}$. The automaton B simply samples uniformly at random from this set of deterministic automata

I am having trouble understanding some parts of the proof:

the random variable $p_A(x)$ should be the amount of $A_i$'s that succeeds divided by the amount of them that B has sampled?

if so than by the low of total probability its easy to see that:

$p_B(x) = \sum_{i}P[B\ choose\ A_i] P[B\ succeeds\ on\ x\ |\ B\ choose' A_i] = \sum_{i}\frac{1}{O(\delta^{-2}nlogm)} P[A_i\ succeeds\ on\ x] = p_A(x)$

where the last move comes from the fact the each instance $A_i$ is deterministic so its either 1 for success or 0 for failure.

So the whole $|p_A(x)-p_B(x)|$ they use is actually 0

I understand from the end of the proof that $|p_A(x)-p_B(x)|$ should some how represent the distance between A's success to B's success but could not see how this happen.

Also I did not understand the use of Chernoff the way they did.

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Here is another way of looking at $p_A$ and $p_B$. We can think of $B$ as a deterministic algorithm which accepts an additional input $r$ representing the randomness. This input is drawn from some distribution $R$. Also, denote by $P(x)$ the correct answer. Then $$ p_B(x) = \Pr_{r \sim R} [B(x,r) = P(x)]. $$ Let $M = O(n\delta^{-2}\log m)$, and sample $r_1,\ldots,r_M \sim R$. The algorithm $A_i$ simply runs $B$ with randomness $r_i$, that is $A_i(x) = B(x,r_i)$. We can define another algorithm $A$ which first choose $i \in \{1,\ldots,M\}$ uniformly at random, and then runs $A_i$. The success probability of $A$ is $$ \begin{align*} p_A(x) &= \Pr_{i \in \{1,\ldots,M\}} [B(x,r_i) = P(x)] \\ &= \Pr_{r \in \{r_1,\ldots,r_M\}} [B(x,r) = P(x)]. \end{align*} $$ Thus in $B$, the randomness is chosen according to $R$, whereas in $A$, the randomness is chosen uniformly from a set $\{r_1,\ldots,r_M\}$. If we choose $r_1,\ldots,r_M \sim R$, then we expect the performance of $A$ to be very similar to that of $B$, and this can be quantified using Chernoff's inequality.

The probability $p_A(x)$ itself depends on the choice of $r_1,\ldots,r_M$. What may confuse you is how we conceive of this two-step random process. We first choose $r_1,\ldots,r_M$, and this gives us the function $p_A(x)$, which is a random function depending on the choice of $r_1,\ldots,r_M$. Note that we don't take probability over the choice of $r_1,\ldots,r_M$ – if we did, we would just get $p_B(x)$.

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  • $\begingroup$ I think I get it, the way B chooses A's M instances is not uniform, but after we got those instances we chose between them at uniform, ill try to write that down with the other parts of the proof. Toda Raba! $\endgroup$ – Matan L May 20 '17 at 10:51

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