2.1. Distributions of Two Random Variables 87At times it is convenient to speak of thesupportof a discrete random vec-
tor (X 1 ,X 2 ). These are all the points (x 1 ,x 2 )inthespaceof(X 1 ,X 2 )such
thatp(x 1 ,x 2 )> 0. In the last example the support consists of the six points
{(0,0),(0,1),(1,1),(1,2),(2,2),(2,3)}.
We say a random vector (X 1 ,X 2 )withspaceDis of thecontinuoustype if its
cdfFX 1 ,X 2 (x 1 ,x 2 ) is continuous. For the most part, the continuous random vectors
in this book have cdfs that can be represented as integrals of nonnegative functions.
That is,FX 1 ,X 2 (x 1 ,x 2 ) can be expressed as
FX 1 ,X 2 (x 1 ,x 2 )=∫x 1−∞∫x 2−∞fX 1 ,X 2 (w 1 ,w 2 )dw 1 dw 2 , (2.1.5)for all (x 1 ,x 2 )∈R^2. We call the integrand thejoint probability density func-
tion(pdf) of (X 1 ,X 2 ). Then
∂^2 FX 1 ,X 2 (x 1 ,x 2 )
∂x 1 ∂x 2
=fX 1 ,X 2 (x 1 ,x 2 ),except possibly on events that have probability zero. A pdf is essentially character-
ized by the two properties
(i)fX 1 ,X 2 (x 1 ,x 2 )≥0 and (ii)∫∫
DfX^1 ,X^2 (x^1 ,x^2 )dx^1 dx^2 =1. (2.1.6)
For the reader’s benefit, Section 4.2 of the accompanying resourceMathematical
Comments^1 offers a short review of double integration. For an eventA∈D,we
have
P[(X 1 ,X 2 )∈A]=∫∫AfX 1 ,X 2 (x 1 ,x 2 )dx 1 dx 2.Note that theP[(X 1 ,X 2 )∈A] is just the volume under the surfacez=fX 1 ,X 2 (x 1 ,x 2 )
over the setA.
Remark 2.1.1.As with univariate random variables, we often drop the subscript
(X 1 ,X 2 ) from joint cdfs, pdfs, and pmfs, when it is clear from the context. We also
use notation such asf 12 instead offX 1 ,X 2. Besides (X 1 ,X 2 ), we often use (X, Y)
to express random vectors.
We next present two examples of jointly continuous random variables.Example 2.1.2. Consider a continuous random vector (X, Y) which is uniformly
distributed over the unit circle inR^2. Since the area of the unit circle isπ,thejoint
pdf is
f(x, y)=
{ 1
π −^1 <y<^1 ,−√
1 −y^2 <x<√
1 −y^2
0elsewhere.Probabilities of certain events follow immediately from geometry. For instance, let
Abe the interior of the circle with radius 1/2. ThenP[(X, Y)∈A]=π(1/2)^2 /π=
1 /4. Next, letBbe the ring formed by the concentric circles with the respective
radii of 1/2and
√
2 /2. ThenP[(X, Y)∈B]=π[(√
2 /2)^2 −(1/2)^2 ]/π=1/4. The
regionsAandBhave the same area and hence for this uniform pdf are equilikely.
(^1) Downloadable at the site listed in the Preface.