In a manner parallel to that for the singlevariable case, we obtain a mapping of
probability mass from the basic space to the plane. Since W1(Q)W1(Q) is an event
for each reasonable set Q on the plane, we may assign to Q the probability mass
P
X
Y
(
Q
)
=
P
[
W

1
(
Q
)
]
=
P
[
(
X
,
Y
)

1
(
Q
)
]
P
X
Y
(
Q
)
=
P
[
W

1
(
Q
)
]
=
P
[
(
X
,
Y
)

1
(
Q
)
]
(4)Because of the preservation of set operations by inverse mappings as in the singlevariable case,
the mass assignment determines
PXYPXY as a probability measure on the subsets of the plane R^{2}. The argument
parallels that for the singlevariable case. The result is the probability distribution
induced by W=(X,Y)W=(X,Y). To determine the probability that the
vectorvalued function W=(X,Y)W=(X,Y) takes on a (vector) value in region Q, we simply
determine how much induced probability mass is in that region.
To determine P(1≤X≤3,Y>0)P(1≤X≤3,Y>0), we determine the region for which the
first coordinate value (which we call t) is between one and three and the second
coordinate value (which we call u) is greater than zero. This corresponds to
the set Q of points on the plane with 1≤t≤31≤t≤3 and u>0u>0. Gometrically,
this is the strip on the plane bounded by (but not including) the horizontal axis and by the
vertical lines t=1t=1 and t=3t=3 (included). The problem is to determine how much
probability mass lies in that strip. How this is acheived depends upon the nature of the
distribution and how it is described.
As in the singlevariable case, we have a distribution function.
Definition
The joint distribution function FXYFXY for W=(X,Y)W=(X,Y) is
given by
F
X
Y
(
t
,
u
)
=
P
(
X
≤
t
,
Y
≤
u
)
∀
(
t
,
u
)
∈
R
2
F
X
Y
(
t
,
u
)
=
P
(
X
≤
t
,
Y
≤
u
)
∀
(
t
,
u
)
∈
R
2
(5)This means that FXY(t,u)FXY(t,u) is equal to the probability mass in the region QtuQtu
on the plane such that the first coordinate is less than or equal to t and the second
coordinate is less than or equal to u. Formally, we may write
F
X
Y
(
t
,
u
)
=
P
[
(
X
,
Y
)
∈
Q
t
u
]
,
where
Q
t
u
=
{
(
r
,
s
)
:
r
≤
t
,
s
≤
u
}
F
X
Y
(
t
,
u
)
=
P
[
(
X
,
Y
)
∈
Q
t
u
]
,
where
Q
t
u
=
{
(
r
,
s
)
:
r
≤
t
,
s
≤
u
}
(6)Now for a given point (a,b)(a,b), the region QabQab is the set of points (t,u)(t,u) on
the plane which are on or to the left of the vertical line through (t,0)(t,0)and
on or below the horizontal line through (0,u)(0,u) (see Figure 1 for specific point t=a,u=bt=a,u=b).
We refer to such regions as semiinfinite intervals on the plane.
The theoretical result quoted in the real variable case extends to ensure that a distribution on the
plane is determined uniquely by consistent assignments to the semiinfinite intervals
QtuQtu. Thus, the induced distribution is determined completely by the joint
distribution function.
Distribution function for a discrete random vector
The induced distribution consists of point masses. At point (ti,uj)(ti,uj) in the range of
W=(X,Y)W=(X,Y) there is probability mass pij=P[W=(ti,uj)]=P(X=ti,Y=uj)pij=P[W=(ti,uj)]=P(X=ti,Y=uj). As in the general case, to determine P[(X,Y)∈Q]P[(X,Y)∈Q] we determine how much probability
mass is in the region. In the discrete case (or in any case where there are point mass
concentrations) one must be careful to note whether or not the boundaries are
included in the region, should there be mass concentrations on the boundary.
The probability distribution is quite simple. Mass 3/10 at (0,2), 6/10 at (1,1), and
1/10 at (2,0). This distribution is plotted in Figure 2. To determine (and visualize)
the joint distribution function, think of moving the point (t,u)(t,u) on the plane. The
region QtuQtu is a giant “sheet” with corner at (t,u)(t,u). The value of FXY(t,u)FXY(t,u)
is the amount of probability covered by the sheet. This value is constant over any grid
cell, including the lefthand and lower boundariies, and is the value taken on at the lower
lefthand corner of the cell. Thus, if (t,u)(t,u) is in any of the
three squares on the lower left hand part of the diagram, no probability mass is covered
by the sheet with corner in the cell. If (t,u)(t,u) is on or in the square having probability
6/10 at the lower lefthand corner, then the sheet covers that probability, and the value of
FXY(t,u)=6/10FXY(t,u)=6/10. The situation in the other cells may be checked out by this procedure.
Distribution function for a mixed distribution
The pair {X,Y}{X,Y} produces a mixed distribution as follows (see Figure 3)
Point masses 1/10 at points (0,0), (1,0), (1,1), (0,1)
Mass 6/10 spread uniformly over the unit square with these vertices
The joint distribution function is zero in the second, third, and fourth quadrants.
 If the point (t,u)(t,u) is in the square or on the left and lower boundaries, the
sheet covers the point mass at (0,0) plus 0.6 times the area covered within the square.
Thus in this region
FXY(t,u)=110(1+6tu)FXY(t,u)=110(1+6tu)
(7)  If the pont (t,u)(t,u) is above the square (including its upper boundary) but to the
left of the line t=1t=1, the sheet covers two point masses plus the portion of the
mass in the square to the left of the vertical line through (t,u)(t,u). In this case
FXY(t,u)=110(2+6t)FXY(t,u)=110(2+6t)
(8)  If the point (t,u)(t,u) is to the right of the square (including its boundary) with
0≤u<10≤u<1, the sheet covers two point masses and the portion of the mass in the
square below the horizontal line through (t,u)(t,u), to give
FXY(t,u)=110(2+6u)FXY(t,u)=110(2+6u)
(9)  If (t,u)(t,u) is above and to the right of the square (i.e., both 1≤t1≤t and 1≤u1≤u).
then all probability mass is covered and FXY(t,u)=1FXY(t,u)=1 in this region.