Foundations of Data Science

sleeping experts problem.
Combining Sleeping Experts Algorithm:
Initialize each expert h i with a weight w i = 1. Let ɛ ∈ (0, 1). For each example x, do the
following:
1. [Make prediction] Let H x denote the set of experts h i that make a prediction on x, and
let w x =
∑ w j . Choose h i ∈ H x with probability p ix = w i /w x and predict h i (x).
h j ∈H x
2. [Receive feedback] Given the correct label, for each h i ∈ H x let m ix = 1 if h i (x) was
incorrect, else let m ix = 0.
3. [Update weights] For each h i ∈ H x , update its weight as follows:
( ∑ )
• Let r ix =
h j ∈H x
p jx m jx /(1 + ɛ) − m ix .
• Update w i ← w i (1 + ɛ) r ix
.
Note that ∑ h j ∈H x
p jx m jx represents the algorithm’s probability of making a mistake
on example x. So, h i is rewarded for predicting correctly (m ix = 0) especially
when the algorithm had a high probability of making a mistake, and h i is penalized
for predicting incorrectly (m ix = 1) especially when the algorithm had a low
probability of making a mistake.
For each h i ∉ H x , leave w i alone.
Theorem 6.21 For any set of n sleeping experts h 1 , . . . , h n , and for any sequence of
examples S, the Combining Sleeping Experts Algorithm A satisfies for all i:
E ( mistakes(A, S i ) ) ≤ (1 + ɛ) · mistakes(h i , S i ) + O ( )
log n
ɛ
where S i = {x ∈ S : h i ∈ H x }.
Proof: Consider sleeping expert h i . The weight of h i after the sequence of examples S
is exactly:
[( ∑ ]
∑x∈S
w i = (1 + ɛ) i h j ∈Hx p jxm jx
)/(1+ɛ)−m ix
= (1 + ɛ) E[mistakes(A,S i)]/(1+ɛ)−mistakes(h i ,S i ) .
Let w = ∑ j w j. Clearly w i ≤ w. Therefore, taking logs, we have:
E ( mistakes(A, S i ) ) /(1 + ɛ) − mistakes(h i , S i ) ≤ log 1+ɛ w.
So, using the fact that log 1+ɛ w = O( log W ),
ɛ
E ( mistakes(A, S i ) ) ≤ (1 + ɛ) · mistakes(h i , S i ) + O ( )
log w
ɛ .
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