Probability that ensemble model is correct based on accuracies of its classifiers

Question

I'm trying to understand what I did wrong when trying to answer this question. The exact question is:

Assume that we have 3 trained prediction models, and each model outputs either -1 or 1. We then tested the accuracies of these models and obtained the following outcomes:

Model	Accuracy
m1	0.60
m2	0.55
m3	0.45

Let M be the ensemble model that outputs a plurality vote of these three models. If we assume that the errors of the models m₁, m₂, and m₃ are independent, what is the probability that M(x) would be correct on a test instance x?

I thought that because this was a plurality vote, and the classification errors are independent of each other, I could simply take the weighted average of the accuracy of the three classifiers:

$ \begin{align*} P(X) &= \sum_{all\ models\ M_j} P(C_i|x,M_j)P(M_j) \\ &= \frac{1}{L} \sum_{all\ models\ M_j} P(C_i|x,M_j) \\ &= \frac{1}{3}(0.60+0.55+0.45)\\ &= 0.53 \end{align*} $

But I was told that this is incorrect (with no context as to why).

Can someone explain why this is incorrect? If this is a plurality vote (which to me assumes that the votes of each classifier are equal), why can I not simply take the weighted average?

Answer 1

It's not the actual data, it's the probabilities. So you should consider all the scenarios of voting.

For the Ensemble to be correct,
Either any two or all the three should be correct

=$[m_1*m_2*(1- m_3) + m_1*(1-m_2)*m_3) + (1-m_1)*m_2*m_3] + [m_1*m_2*m_3]$

= [0.6*0.55*(1-0.45) + 0.6*(1-0.55)*0.45 + (1-0.6)*0.55*0.45] + 0.6*0.55*0.45

= 0.5505

Answer 2

You would get a correct prediction if model's 1, 2 and 3 were

• Correct, Correct, Correct

• Correct, Correct, Wrong

• Correct, Wrong, Correct

• Wrong, Correct, Correct

Likewise, You would get a wrong prediction if they were

• Wrong, Wrong, Correct
• Wrong, Correct, Wrong
• Correct, Wrong, Wrong
• Wrong, Wrong, Wrong

---

So, a total of 8 possibilities.

The likelihood of the first event C, C, C is 0.6 × 0.55 × 0.45

The likelihood of the fifth event W, W, C is 0.4 × 0.45 × 0.45

You want to calculate likelihood of events 1-4 and 5-8 and sum them up separately. You also want to total them all 1-8. The first sum divided the total gives you the answer you are looking for.