28.2 The Lower Bound for the Agnostic Case 393
Combining this with Lemma 26.8 we obtain the following bound on the Rademacher
complexity:
R(A)≤
√
2 dlog(em/d)
m
Using Theorem 26.5 we obtain that with probability of at least 1−δ, for every
h∈Hwe have that
LD(h)−LS(h)≤
√
8 dlog(em/d)
m
+
√
2 log(2/δ)
m
Repeating the previous argument for minus the zero-one loss and applying the
union bound we obtain that with probability of at least 1−δ, for everyh∈ H
it holds that
|LD(h)−LS(h)|≤
√
8 dlog(em/d)
m
+
√
2 log(4/δ)
m
≤ 2
√
8 dlog(em/d) + 2 log(4/δ)
m
To ensure that this is smaller thanwe need
m≥
4
^2
·(8dlog(m) + 8dlog(e/d) + 2 log(4/δ)).
Using Lemma A.2, a sufficient condition for the inequality to hold is that
m≥ 4
32 d
^2
·log
(
64 d
^2
)
+
8
^2
·(8dlog(e/d) + 2 log(4/δ)).
28.2 The Lower Bound for the Agnostic Case
Here, we prove that there existsCsuch thatHis agnostic PAC learnable with
sample complexity
mH(,δ)≥C
d+ ln(1/δ)
^2.
We will prove the lower bound in two parts. First, we will show thatm(,δ)≥
0 .5 log(1/(4δ))/^2 , and second we will show that for everyδ≤ 1 /8 we have that
m(,δ)≥ 8 d/^2. These two bounds will conclude the proof.
28.2.1 Showing Thatm(,δ)≥ 0 .5 log(1/(4δ))/
We first show that for any < 1 /
√
2 and anyδ∈(0,1), we have thatm(,δ)≥
0 .5 log(1/(4δ))/^2. To do so, we show that form≤ 0 .5 log(1/(4δ))/^2 ,His not
learnable.
Choose one example that is shattered byH. That is, letcbe an example such