Leibnitz’ Rule
• How do we find the derivative of an integral like
Z b(λ)
f (x; λ)dx
I(λ) =
a(λ)
when the limits of integration depend upon the
derivation paramter? Leibnitz’ rule states:
Z b(λ)
dI(λ) db(λ)
∂f (x; λ)
da(λ)
=
f (b(λ); λ)−
f (a(λ); λ)+
dx
dλ
dλ
dλ
∂λ
a(λ)
• In our case we have two integrals
Z
Z t
f (z)dz + B ·
E(t) = F ·
−∞
∞
b(z)dz
t
to be derivated wrt the thresholding value t.
. :| Parameter λ corresponds to t
For the first integral we have :
. :) a(λ) → −∞ (constant ⇒ derivative = 0)
. :) b(λ) → t (⇒ derivative = 1)
. :) f (x; λ) → f (x) (independent of t)
For the second integral we have :
. ;) a(λ) → t (⇒ derivative = 1)
. ;) b(λ) → ∞ (constant, ⇒ derivative = 0)
. ;) f (x; λ) → b(x) (independent of t)
• Therefore, we have
dE(t)
= 0 ⇒ F · f (T ) = B · b(T )
dt
• Please note that this result quite generally gives the
minimum error. There are no restrictions on the
distributions b and f !!!
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