Online Learning Applications

Theorem 1 Let M be mistakes of RWM and Mi the mistakes of expert i. Then, RWM satisfies E[M ] ≤ (1 + )Mi + log n/ . Recall that to prove this, we kept weights wt(i) on each expert i, and updated them as wt+1(i) = (1 − mt(i)η)wt(i). Here mt was the mistake vector (outcome). RWM played i with probability pt(i) = wt(i)/Wt, where Wt = ∑ iwt(i). This gives Wt+1 = Wt(1 − ηmt · pt). We then used that any expert that makes k mistakes must have weight w0(1− η)k.