Foundations of Data Science

Now consider a very tall tree. If the p is the probability that a root has value +1,
we can iterate the formula for the height one tree and observe that at low temperature
the probability of the root being one converges to some value. At high temperature, the
probability of the root being one is 1 / 2 independent of p. See Figure 9.10. At the phase
transition, the slope of q at p=1/2 is one.
Now the slope of the probability of the root being 1 with respect to the probability of
a leaf being 1 in this height one tree is
∂q
∂p = D ∂C − C ∂D
∂p ∂p
D 2
Since the slope of the function q(p) at p=1/2 when the phase transition occurs is one, we
can solve ∂q = 1 for the value of β where the phase transition occurs. First, we show that
∂p
∣
= 0.
∂D
∂p
∣
p=
1
2
Then
∂q
∣
∂p
∂D
∂p
∣
p=
1
2
∣
D = [ pe 2β + 1 − p ] d
+
[
p + (1 − p) e
2β ] d
∂D
= d [ pe 2β + 1 − p ] d−1 (
∂p e 2β − 1 ) + d [ p + (1 − p) e 2β] d−1 ( )
1 − e
2β
∣
p=
1
2
= d
2 d−1 [
e 2β + 1 ] d−1 (
e 2β − 1 ) + d
2 d−1 [
1 + e
2β ] d−1 (
1 − e
2β ) = 0
= D ∂C − C ∣ ∂D ∣∣∣∣p=
∂p ∂p
D 2 1
2
=
∂C
∂p
D
∣
∣ 1 p=
2
= d [ ]
1
2 e2β + 1 d−1 (
2 e 2β − 1 )
[ 1
]
2 e2β + 1 d [
2 + 1 + 1e2β] = d ( e 2β − 1 )
d
1 + e 2β
2 2
=
d [ pe 2β + 1 − p ] d−1 (
e 2β − 1 )
[pe 2β + 1 − p] d + [p + (1 − p) e 2β ] d ∣ ∣∣∣∣p=
1
2
Setting
And solving for β yields
d ( e 2β − 1 )
1 + e 2β = 1
d ( e 2β − 1 ) = 1 + e 2β
e 2β = d+1
d−1
β = 1 2
ln
d+1
d−1
To complete the argument, we need to show that q is a monotonic function of p. To see
this, write q = 1
1+ B . A is a monotonically increasing function of p and B is monotonically
A
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