THE EGS5 CODE SYSTEM

Then to sample from f 1j (E), first let
p = 2 1−j , E ′ = E/p . (2.113)
We then relate the distributions of E and E ′ , as follows. Suppose ˆx, ŷ are random variables with
probability density functions f(x) and g(y) and cumulative distribution functions F (x) and G(y).
Further, suppose that ˆx is related to ŷ by
with h monotonic increasing (↑) or decreasing (↓). Then clearly
ˆx = h(ŷ) (2.114)
F (x) = P r{ ˆx < x} = P r{ h(ŷ) < x}
(
)
= P r{ ŷ < h −1 (x)} if h ↑, P r{ ŷ > h −1 (x)}
(
)
= G(h −1 (x)) if h ↑, 1 − G(h −1 (x)) if h ↓ .
(2.115)
Letting
x = h(y), y = h −1 (x) , (2.116)
we have
F (h(y)) = (G(y) if h ↑, 1 − G(y) if h ↓) . (2.117)
Differentiating with respect to y, we obtain,
f (h(y)) dh(y)
dy
= (g(y) if h ↑, −g(y) if h ↓) . (2.118)
Now
Hence
dh(y)
dy
As an example, we claim that if
g(E ′ ) = 1
ln 2
(∣ ∣ )
∣∣∣ dh(y)
∣∣∣
=
dy ∣ if h ↑, − dh(y)
dy ∣ if h ↓ . (2.119)
f (h(y)) = g(y)/
dh(y)
∣ dy ∣ , (2.120)
( )/ ∣ f(x) = g h −1 ∣∣∣ dh(h −1 (x))
(x)
. (2.121)
dy ∣
( (
1 if E ′ ∈ 0, 1 )
,
2
1 − E ′
E ′
( )) 1
if E ′ ∈
2 , 1
, (2.122)
and
then
E = h(E ′ ) = E ′ p, E ′ = h −1 (E) = E/p,
f(E) = 1 ( 1
ln 2 p if E ∈ (0, p/2), 1 − E/p
E
dh(E ′ )
dE ′ = p , (2.123)
)
if E ∈ (p/2, p)
. (2.124)
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