
Let X and Y be two discrete random variables
defined on the same sample space. The joint
probability mass function p  x, y  is defined by
p  x, y   P  X  x and Y  y 


We must have p  x, y   0 and
  p  x, y   1 .
x
y
Also P  X , Y   A    x , y A p  x, y  .
2

Let X be the deductible on an auto policy and Y
the deductible on a homeowner’s policy for a
particular company. The possible deductibles are
$100 and $250 on auto policies, and $0, $100 and
$200 on homeowners policies. The joint pmf is:
y
x
0
100
200
100
.20
.10
.20
250
.05
.15
.30

Then p(100,100)  P  X  100, Y  100   .10 ,
P(Y  100)  p 100,100   p  250,100 
 p 100, 200   p  250, 200   .75
4

The marginal probability distributions are
defined as
pX  x    y: p x , y 0 p  x, y 
pY  y    x: p x , y 0 p  x, y 
for each possible x
for each possible y
5
y
x
0
100
200
pX  x 
100
.20
.10
.20
.5
250
.05
.15
.30
.5
pY  y 
.25
.25
.50

Let X and Y be two continuous rv’s. The joint
probability density function f(x,y)
is
a
function

satisfying f  x, y   0 and   f  x, y  dxdy  1 .
 

Then P  X , Y   A   f  x, y  dxdy .
A
7

The marginal density functions of X andY are
fX  x 

 f  x, y  dy

fY  y  

 f  x, y  dx

8

A nut company markets cans of deluxe mixed
nuts containing almonds, cashews, and
peanuts. Suppose the net weight of each can
is exactly 1 lb., but the composition can vary.
Let X=weight of almonds and Y=weight of
cashews. Let the joint density be
24 xy 0  x  1,0  y  1, x  y  1
f  x, y   
otherwise
 0
9

This is a valid density (verify), and
P  X , Y    x, y  : 0  x  1,0  y  1, x  y  0.5
1/2 1/2 x

 
0
24 xy dydx  1/16
0
1 y
2
24
xy
dx

12
y
1

y
0  y 1



fY  y    0

0
otherwise

1 x
2
24
xy
dx

12
x
1

x
0  x 1



fX  x   0

0
otherwise

10

Two rv’s are independent if for every pair of x
and y values
p  x, y   p X  x  pY  y  (discrete)
or
f  x, y   f X  x  fY  y  (continuous)
11

For the insurance example,
p 100,100   .10  (.5)(.25)  p X 100  pY 100 
so X and Y are not independent.
12

For the mixed nuts,
f  3 / 4, 3 / 4   0  f X  3 / 4  pY  3 / 4    9 /16 
2
so X and Y are not independent.
To be independent, the density f  x, y  must have
g  x h y
the form
and the region of positive
density must be a rectangle whose sides are
parallel to the coordinate axis.
13

If X , X , , X are discrete rv’s, the joint pmf is
the function
1
2
n
p  x1 , x2 ,

, xn   P  X 1  x1 ,
, X n  xn 
For continuous rv’s, the joint density is the
function f  x , x , , x  such that for n intervals
a , b , ,a , b  ,
1
1
1
n
2
n
n
P  a1  X 1  b1 ,
b1
, an  X n  bn   
a1
bn
 f x ,
1
, xn  dxn
dx1
an
14

The random variables X1 , X 2 , , X n
are
independent if for every subset X i , X i , , X i of
the variables (each pair, each triple, and so
on), the joint pmf or pdf of the subset is equal
to the product of the marginal pmf’s or pdf’s
for all possible values of the variables.
1
2
n
15

Let X and Y be two continuous rv’s with joint
pdf f(x,y). Then for any x such that f X  x   0 ,
the conditional probability density function of
Y given that X=x is fY X  y x   f  x, y  ,   y   .
fX  x

For X and Y discrete, replacing the pdf’s by
pmf’s gives the conditional probability mass
function of Y when X=x.
16

For the mixed nuts example,
fY X  y x  
24 xy
12 x 1  x 
2

2y
1  x 
2
, 0  y  1 x
17