CMPT 407 / 710 - Complexity Theory : Lecture 13 Valentine

As in the case of BPP, we can reduce the error probability of any MA protocol to be less than an inverse exponential in the input size. Here’s how. Let L ∈ MA be any language. Let R(x, y, z) be a polytime relation for L such that, for every x ∈ L, there is a y with Prz[R(x, y, z) = 1] ≥ 3/4; and for every x 6∈ L, for every y, it holds that Prz[R(x, y, z) = 1] ≤ 1/4. Consider a new protocol where, upon receiving a string y, Arthur randomly and independently chooses k strings z1, . . . , zk, and accepts iff R(x, y, zi) = 1 for more than half of these k strings. We use Chernoff bounds to analyze the correctness of the described protocol. Suppose first that x ∈ L. Then Merlin can send Arthur a string y such that Prz[R(x, y, z) = 1] ≥ 3/4. Every string z1, 1 ≤ i ≤ k, randomly chosen by Arthur has probability at least 3/4 of satisfying R. The expected number of zi’s that satisfy R is μ ≥ 3 4k. Let Xi, 1 ≤ i ≤ k, be a random variable that is 1 if zi satisfies R, and 0 otherwise. Let X = ∑k i=1Xi. As we just argued, the expectation of X is μ. Using Chernoff bounds, we get that Pr[X ≤ k/2] ≤ Pr[|X − μ| > k/4] < 2e−k/48. Thus, Arthur will accept with probability exponentially close to 1. On the other hand, suppose that x 6∈ L. Then, whatever y is sent to Arthur, Prz[R(x, y, z) = 1] ≤ 1/4. Let Xi be random variables as defined above, and let X = ∑k i=1Xi. Then the expectation of X is ≤ 1 4 k. The probability that Arthur accepts in this case is Pr[X > k/2], which, by Chernoff bounds, is at most 2e−k/48. Thus, in this case, Arthur will accept with probability exponentially close to 0.