One can conclude from this that
f
X,Y
(
x, y
) =
f
X
(
x
)
f
Y
(
y
)
,
or again the joint density factors. Going the other way, one can also see that if the joint density factors,
then one has independence.
Example.
Suppose one has a floor made out of wood planks and one drops a needle onto it. What is the
probability the needle crosses one of the cracks? Suppose the needle is of length
L
and the wood planks are
D
across.
Answer.
Let
X
be the distance from the midpoint of the needle to the nearest crack and let Θ be the angle
the needle makes with the vertical. Then
X
and Θ will be independent.
X
is uniform on [0
, D/
2] and Θ is
uniform on [0
, π/
2]. A little geometry shows that the needle will cross a crack if
L/
2
> X/
cos Θ. We have
f
X,
Θ
=
4
πD
and so we have to integrate this constant over the set where
X < L
cos Θ
/
2 and 0
≤
Θ
≤
π/
2
and 0
≤
X
≤
D/
2. The integral is
π/
2
0
L
cos
θ/
2
0
4
πD
dx dθ
=
2
L
πD
.
If
X
and
Y
are independent, then
P
(
X
+
Y
≤
a
) =
{
x
+
y
≤
a
}
f
X,Y
(
x, y
)
dx dy
=
{
x
+
y
≤
a
}
f
X
(
x
)
f
Y
(
y
)
dx dy
=
∞
∞
a

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 Spring '09
 wen
 Factors, Probability theory, dx dy, joint density factors

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