Final answer:
The modified go-with-weighted majority algorithm attenuates the weight of each expert by a factor of (1 - ε) instead of 1/2. This algorithm's performance can be shown to be bounded by (1 - ε/2)ᴹ, where M is the algorithm's loss. The proof can be done using mathematical induction.
Step-by-step explanation:
The given question is related to the performance of a modified go-with-weighted majority algorithm. In the modified algorithm, the weight of each expert is attenuated by a factor of (1 - ε) instead of 1/2, where ε is a value between 0 and 1/2.
The goal is to show that the sum of the weights Wₜ after round t is less than or equal to (1 - ε/2)ᴹ, where M is the algorithm's loss.
To prove this, we can use mathematical induction. We start with the base case t=1 and show that the inequality holds. Then we assume it holds for t=k and prove it for t=k+1.
Let's assume that Wₖ ≤ (1 - ε/2)ᴹ holds for some k, and show that Wₖ₊₁ ≤ (1 - ε/2)ᴹ holds as well. Using the modified attenuation factor, we have:
wₖ₊₁⁽ᶦ⁾ = (1 - ε)wₖ⁽ᶦ⁾ if expert i makes a correct prediction,
wₖ₊₁⁽ᶦ⁾ = (1 - ε)wₖ⁽ᶦ⁾/2 if expert i makes a mistake,
Wₖ₊₁ = ∑ᵢᴺ = 1wₖ₊₁⁽ᶦ⁾ = ∑ᵢᴺ = 1(1 - ε)wₖ⁽ᶦ⁾ if all experts make correct predictions,
Wₖ₊₁ = ∑ᵢᴺ = 1(1 - ε)wₖ⁽ᶦ⁾/2 if some experts make mistakes.
Since the algorithm's loss M is the total number of mistakes, we have M ≤ N/2. By substituting the above equations into Wₖ₊₁, we obtain:
Wₖ₊₁ = (1 - ε)Wₖ if all experts make correct predictions,
Wₖ₊₁ = (1 - ε/2)Wₖ if some experts make mistakes.
From the induction hypothesis, we know that Wₖ ≤ (1 - ε/2)ᴹ. If all experts make correct predictions, we have Wₖ₊₁ = (1 - ε)Wₖ ≤ (1 - ε)(1 - ε/2)ᴹ = (1 - ε - ε/2 + ε²/2)ᴹ. Since ε ∈ (0, 1/2), we have ε²/2 < ε/2, so Wₖ₊₁ ≤ (1 - ε/2)ᴹ.
If some experts make mistakes, we have Wₖ₊₁ = (1 - ε/2)Wₖ ≤ (1 - ε/2)(1 - ε/2)ᴹ = (1 - ε/2)²ᴹ = (1 - ε/2)ᴹ₊₁.
Therefore, the inequality Wₜ ≤ (1 - ε/2)ᴹ holds for all rounds t, and the modified algorithm's performance is guaranteed to be bounded by (1 - ε/2)ᴹ.