Biomedical Engineering Reference
In-Depth Information
6.3 ONE FUNCTION OF TWO RANDOM VARIABLES
Consider a random variable
z
y
) created from jointly distributed random variables
x
and
y
. In this section, the probability distribution of
z
=
g
(
x
,
y
) is computed using a CDF
technique similar to the one at the start of this chapter. Because we are dealing with regions in
a plane instead of intervals on a line, these problems are not as straightforward and tractable as
before.
With
z
=
g
(
x
,
=
g
(
x
,
y
), we have
F
z
(
γ
)
=
P
(
z
≤
γ
)
=
P
(
g
(
x
,
y
)
≤
γ
)
=
P
((
x
,
y
)
∈
A
(
γ
))
,
(6.15)
where
A
(
γ
)
={
(
x
,
y
):
g
(
x
,
y
)
≤
γ
}
.
(6.16)
The CDF for the RV
z
can then be found by evaluating the integral
F
z
(
γ
)
=
dF
x
,
y
(
α, β
)
.
(6.17)
A
(
γ
)
This result cannot be continued further until a specific
F
x
,
y
and
g
(
x
y
) are considered. Re-
member that in the case of a single random variable, our efforts primarily dealt with algebraic
manipulations. Here, our efforts are concentrated on evaluating
F
z
through integrals, with the
ease of solution critically dependent on
g
(
x
,
y
).
The ease in solution for
F
z
is dependent on transforming
A
(
,
) into proper limits of
integration. Sketching the support region for
f
x
,
y
(the region where
f
x
,
y
γ
=
0, or
F
x
,
y
is not
constant) and the region
A
(
) is often most helpful, even crucial, in the problem solution. Pay
careful attention to the limits of integration to determine the range of integration in which the
integrand is zero because
f
x
,
y
γ
0. Let us consider several examples to illustrate the mechanics
of the CDF technique and also to provide further insight.
=
Example 6.3.1.
Random variables x and y have joint PDF
1
/
4
,
0
<α<
2
,
0
<β<
2
f
x
,
y
(
α, β
)
=
0
,
otherwise
.
Find the CDF for z
=
x
+
y
.
Solution.
We have
A
(
γ
)
={
(
α, β
):
α
+
β
≤
γ
}
. We require the volume under the surface
f
x
,
y
(
α, β
) where
α
≤
γ
−
β
:
∞
γ
−
β
F
z
(
γ
)
=
f
x
,
y
(
α, β
)
d
α
d
β.
−∞
−∞
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