7. Two Random Variables
In many experiments, the observations are expressible not
as a single quantity, but as a family of quantities. For
example to record the height and weight of each person in
a community or the number of people and the total income
in a family, we need two numbers.
Let X and Y denote two random variables (r.v) based on a
probability model (, F, P). Then
Px1  X ( )  x2   FX ( x2 )  FX ( x1 )  
x2
x1
and
P y1  Y ( )  y2   FY ( y2 )  FY ( y1 )  
y2
y1
f X ( x)dx,
fY ( y )dy.
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What about the probability that the pair of r.vs (X,Y) belongs
to an arbitrary region D? In other words, how does one
estimate, for example, P( x1  X ( )  x2 )  ( y1  Y ( )  y2 )  ?
Towards this, we define the joint probability distribution
function of X and Y to be
FXY ( x, y )  P( X ( )  x )  (Y ( )  y )
 P( X  x, Y  y )  0,
(7-1)
where x and y are arbitrary real numbers.
Properties
(i) FXY (, y)  FXY ( x,)  0, FXY (,)  1.
(7-2)
since  X ( )  ,Y ( )  y    X ( )   , we get
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FXY (, y )  P X ( )    0.
Similarly  X ( )  ,Y ( )    ,
we get FXY (, )  P()  1.
(ii)
Px1  X ( )  x2 , Y ( )  y   FXY ( x2 , y )  FXY ( x1, y ).
(7-3)
P X ( )  x, y1  Y ( )  y2   FXY ( x, y2 )  FXY ( x, y1 ).
(7-4)
To prove (7-3), we note that for x2  x1,
X ( )  x2 ,Y ( )  y   X ( )  x1,Y ( )  y   x1  X ( )  x2 ,Y ( )  y 
and the mutually exclusive property of the events on the
right side gives
P X ( )  x2 ,Y ( )  y   P X ( )  x1,Y ( )  y   Px1  X ( )  x2 ,Y ( )  y 
which proves (7-3). Similarly (7-4) follows.
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(iii)
Px1  X ( )  x2 , y1  Y ( )  y2   FXY ( x2 , y2 )  FXY ( x2 , y1 )
 FXY ( x1, y2 )  FXY ( x1, y1 ).
(7-5)
This is the probability that (X,Y) belongs to the rectangle R0
in Fig. 7.1. To prove (7-5), we can make use of the
following identity involving mutually exclusive events on
the right side.
x1  X ( )  x2 ,Y ( )  y2   x1  X ( )  x2 ,Y ( )  y1   x1  X ( )  x2 , y1  Y ( )  y2 .
Y
y2
R0
y1
X
x2
x1
Fig. 7.1
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This gives
Px1  X ( )  x2 ,Y ( )  y2   Px1  X ( )  x2 ,Y ( )  y1   Px1  X ( )  x2 , y1  Y ( )  y2 
and the desired result in (7-5) follows by making use of (73) with y  y2 and y1 respectively.
Joint probability density function (Joint p.d.f)
By definition, the joint p.d.f of X and Y is given by
 2 FXY ( x, y )
f XY ( x, y ) 
.
x y
(7-6)
and hence we obtain the useful formula
FXY ( x, y )  
x


y

(7-7)
f XY (u, v ) dudv.
Using (7-2), we also get




 
f XY ( x, y ) dxdy  1.
(7-8)
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To find the probability that (X,Y) belongs to an arbitrary
region D, we can make use of (7-5) and (7-7). From (7-5)
and (7-7)
P  x  X ( )  x  x, y  Y ( )  y  y   FXY ( x  x, y  y )
 FXY ( x, y  y )  FXY ( x  x, y )  FXY ( x, y )

x  x
x

y  y
y
f XY (u, v )dudv  f XY ( x, y )xy.
(7-9)
Thus the probability that (X,Y) belongs to a differential
rectangle x y equals f XY ( x, y)  xy, and repeating this
procedure over the union of no overlapping differential
rectangles in D, we get the useful result
Y
D
y
x
X
Fig. 7.2
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P( X , Y )  D   

( x , y )D
f XY ( x, y )dxdy.
(7-10)
(iv) Marginal Statistics
In the context of several r.vs, the statistics of each individual
ones are called marginal statistics. Thus FX (x) is the
marginal probability distribution function of X, and f X (x) is
the marginal p.d.f of X. It is interesting to note that all
marginals can be obtained from the joint p.d.f. In fact
FX ( x)  FXY ( x,), FY ( y )  FXY (, y ).
(7-11)
Also
f X ( x)  


f XY ( x, y )dy, fY ( y )  


f XY ( x, y )dx.
(7-12)
To prove (7-11), we can make use of the identity
( X  x)  ( X  x )  (Y  )
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so that FX ( x)  P X  x   P X  x,Y    FXY ( x,).
To prove (7-12), we can make use of (7-7) and (7-11),
which gives
FX ( x)  FXY ( x,)  
x




f XY (u, y ) dudy
(7-13)
and taking derivative with respect to x in (7-13), we get
f X ( x)  


f XY ( x, y )dy.
(7-14)
At this point, it is useful to know the formula for
differentiation under integrals. Let
H ( x)  
b( x )
a( x)
h( x, y )dy.
(7-15)
Then its derivative with respect to x is given by
b ( x ) h ( x , y )
dH ( x ) db( x )
da ( x )

h ( x , b) 
h ( x, a )  
dy.
a ( x ) x
dx
dx
dx
Obvious use of (7-16) in (7-13) gives (7-14).
(7-16)
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
If X and Y are discrete r.vs, then pij  P( X  xi ,Y  y j ) represents
their joint p.d.f, and their respective marginal p.d.fs are
given by
P( X  xi )   P( X  xi , Y  y j )   pij
and
j
(7-17)
j
P(Y  y j )   P( X  xi , Y  y j )   pij
i
(7-18)
i
Assuming that P( X  xi ,Y  y j ) is written out in the form of a
rectangular array, to obtain P( X  xi ), from (7-17), one need to
add up all entries in the i-th row.
p
p
p
 p
 p
It used to be a practice for insurance
p
p
 p
 p
companies routinely to scribble out






these sum values in the left and top  p p p  p  p
p
p
 p
 p
margins, thus suggesting the name
Fig. 7.3 9
marginal densities! (Fig 7.3).
PILLAI
ij
i
ij
11
12
1j
1n
21
22
2j
2n
i1
i2
ij
in
m1
m2
mj
mn
j
From (7-11) and (7-12), the joint P.D.F and/or the joint p.d.f
represent complete information about the r.vs, and their
marginal p.d.fs can be evaluated from the joint p.d.f. However,
given marginals, (most often) it will not be possible to
compute the joint p.d.f. Consider the following example:
Y
Example 7.1: Given
1
constant, 0  x  y  1,
f XY ( x, y )  
0,
otherwise .

y
(7-19)
X
0
1
Fig. 7.4
Obtain the marginal p.d.fs f X (x) and fY ( y ).
Solution: It is given that the joint p.d.f f XY ( x, y) is a constant in
the shaded region in Fig. 7.4. We can use (7-8) to determine
that constant c. From (7-8)

 



2
cy
f XY ( x, y )dxdy     c  dx dy   cydy 
y 0  x 0
y 0

2
1
y
1
1

0
c
 1. (7-20)
2
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Thus c = 2. Moreover from (7-14)
f X ( x)  


f XY ( x, y )dy  
1
2dy  2(1  x), 0  x  1,
(7-21)
2dx  2 y,
(7-22)
yx
and similarly
fY ( y )  


f XY ( x, y )dx  
y
x 0
0  y  1.
Clearly, in this case given f X (x) and fY ( y) as in (7-21)-(7-22),
it will not be possible to obtain the original joint p.d.f in (719).
Example 7.2: X and Y are said to be jointly normal (Gaussian)
distributed, if their joint p.d.f has the following form:
f XY ( x, y ) 
1
2 X  Y 1  
2
e
1  ( x   X )2 2  ( x   X )( y  Y ) ( y  Y ) 2



 XY
2 (1  2 )   X2
 Y2




,
   x  ,    y  , |  | 1.
(7-23)
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By direct integration, using (7-14) and completing the
square in (7-23), it can be shown that
f X ( x)  


f XY ( x, y )dy 
1
2
2
X
e( x  X )
2
2
/ 2 X
~
N (  X , X2 ),
(7-24)
/ 2 Y2
~
N ( Y , Y2 ),
(7-25)
and similarly
fY ( y )  


f XY ( x, y )dx 
1
2 Y2
e( y Y )
2
Following the above notation, we will denote (7-23)
as N (X , Y , X2 , Y2 ,  ). Once again, knowing the marginals
in (7-24) and (7-25) alone doesn’t tell us everything about
the joint p.d.f in (7-23).
As we show below, the only situation where the marginal
p.d.fs can be used to recover the joint p.d.f is when the
random variables are statistically independent.
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Independence of r.vs
Definition: The random variables X and Y are said to be
statistically independent if the events X ( )  A and {Y ( )  B}
are independent events for any two Borel sets A and B in x
and y axes respectively. Applying the above definition to
the events X ( )  x and Y ( )  y, we conclude that, if
the r.vs X and Y are independent, then
P( X ( )  x)  (Y ( )  y )  P( X ( )  x) P(Y ( )  y )
(7-26)
i.e.,
FXY ( x, y )  FX ( x) FY ( y )
(7-27)
or equivalently, if X and Y are independent, then we must
have
f XY ( x, y )  f X ( x) fY ( y ).
(7-28)
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If X and Y are discrete-type r.vs then their independence
implies
P( X  xi ,Y  y j )  P( X  xi )P(Y  y j ) for all i, j.
(7-29)
Equations (7-26)-(7-29) give us the procedure to test for
independence. Given f XY ( x, y), obtain the marginal p.d.fs f X (x)
and fY ( y) and examine whether (7-28) or (7-29) is valid. If
so, the r.vs are independent, otherwise they are dependent.
Returning back to Example 7.1, from (7-19)-(7-22), we
observe by direct verification that f XY ( x, y)  f X ( x) fY ( y). Hence
X and Y are dependent r.vs in that case. It is easy to see that
such is the case in the case of Example 7.2 also, unless   0.
In other words, two jointly Gaussian r.vs as in (7-23) are
independent if and only if the fifth parameter   0.
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Example 7.3: Given
 xy 2e y , 0  y  , 0  x  1,
f XY ( x, y )  
otherwise.
 0,
(7-30)
Determine whether X and Y are independent.
Solution:


f X ( x )   f XY ( x, y )dy  x  y 2e  y dy
0
0

y 

 x   2 ye
 2  ye y dy   2 x,
0
0


Similarly
fY ( y )  
1
0
y2 y
f XY ( x, y )dx 
e ,
2
0  x  1.
0  y  .
(7-31)
(7-32)
In this case
f XY ( x, y )  f X ( x) fY ( y ),
and hence X and Y are independent random variables.
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