Graphics Reference
In-Depth Information
Just as in the discrete case, if
S
is a continuum probability space with associ-
ated density
p
and
Y
:
S
→
T
,
(30.22)
we can use
Y
to define a probability density on
T
(see Figure 30.7). We'll restrict
our attention to the special case where
Y
is an invertible function, although we'll
apply the results to cases where
Y
is “almost invertible,” like the latitude-longitude
parameterization of the sphere, where the noninvertibility is restricted to a set
whose “size” is zero. In the case of the spherical parameterization, if we ignore all
points of the international dateline, the parameterization is 1-1, and the dateline
itself has size zero (where “size,” for a surface, means “area”).
Our definition closely follows the discrete case:
p
Y
(
t
)=
p
(
Y
−
1
(
t
))
.
(30.23)
Returning to the example program, the variable
U
has as its pdf the function
p
U
:[
0, 1
]
→
R
:
r
→
1.
(30.24)
This evidently integrates to 1 on the whole interval.
For any particular number,
like
r
=
0.3,
the probability that
U
=
r
is
r
r
p
U
(
r
)
dr
=
r
r
1
dr
=
0. Intuitively, if we're picking a random number
between 0 and 1, the probability of picking any
particular
number ends up
being zero. Despite this, we do pick
some
number. This is a situation where the
probability of a union of disjoint events (the events being “pick 0.134,” “pick
π/
13,” and infinitely many other similar events) is
not
the sum of the individ-
ual probabilities. Thus, when we shift from finite sets to infinite ones, some of
our intuition about probability turns out to be mistaken.
The pdf for
W
is not so obvious. Evidently, values near 0 occur less often as
outputs of
W
than do those near 1, but what's the exact pattern? The answer is
obviously
not
that it's the square root of the pdf for
U
. We seek a function
p
W
with the property that
=
b
a
Pr
{
a
≤
W
≤
b
}
p
W
(
r
)
dr
.
(30.25)
In the left-hand side of this equation is the event that
a
≤
W
≤
b
; let's rewrite
this:
{
≤
≤
}
=
{
∈
[
0, 1
]:
a
≤
W
(
s
)
≤
}
;
a
W
b
s
b
(30.26)
≤
√
s
=
{
∈
[
0, 1
]:
a
≤
}
;
s
b
(30.27)
[
0, 1
]:
a
2
b
2
=
{
s
∈
≤
s
≤
}
.
(30.28)
The probability of that last event is
b
2
a
2
. (Why?) So we need to find a function
−
p
W
with the property that for any
a
,
b
∈
[
0, 1
]
,
b
p
W
(
r
)
dr
=
b
2
a
2
.
−
(30.29)
a
A little calculus shows that
p
W
(
r
)=
2
r
(see Exercise 30.4).
