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Chapter 5
Joint Probability Distributions and Random Samples
 5.1 - Jointly Distributed Random Variables
 5.2 - Expected Values, Covariance, and
Correlation
 5.3 - Statistics and Their Distributions
 5.4 - The Distribution of the Sample Mean
 5.5 - The Distribution of a Linear Combination
Suppose Z is a third discrete random
variable that depends on X and Y, i.e.,
there exists a joint pmf z  p ( x, y ) for
every ordered pair (x, y)…
Y
Suppose X and Y are two
discrete random variables with
pmfs pX(x) and pY(y) respectively:
x
pX(x)
y
pY(y)
x1
pX (x1)
y1
pY(y1)
x2
pX (x2)
y2
pY(y2)














xr
pX (xr)
yc
pY(yc)
1
p X ( xi )  0
r
p
i 1
X
( xi )  1
1
pY ( y j )  0
c
p
j 1
Y
(yj) 1
y1
y2
…
x1
p(x1, y1)
p(x1, y2)
...
p(x1, yc) pX (x1)
x2
p(x2, y1)
p(x2, y2)
...
p(x2, yc) pX (x2)
...
...
...
p(xr, y1)
p(xr, y2)
...
pY (y1)
pY (y2)
X
xr
c
 p( x , y )  1
i 1 j 1
...
p(xr, yc) pX (xr)
pY (yc)
p( xi , y j )  0
r
yc
i
j
1
Suppose Z is a third discrete random
variable that depends on X and Y, i.e.,
there exists a joint pmf z  p ( x, y ) for
every ordered pair (x, y)…
Y
Suppose X and Y are two
discrete random variables with
pmfs pX(x) and pY(y) respectively:
x
pX(x)
y
pY(y)
x1
pX (x1)
y1
pY(y1)
x2
pX (x2)
y2
pY(y2)














yc
pY(yc)
xr
pX (xr)
1
p X ( xi )  0
r
p
i 1
X
( xi )  1
1
pY ( y j )  0
c
p
j 1
Y
(yj) 1
y1
y2
…
x1
p(x1, y1)
p(x1, y2)
...
p(x1, yc) pX (x1)
x2
p(x2, y1)
p(xp2,(yx2i ), y j ) 
...0
p(x2, yc) pX (x2)
X
r
...
xr
c
yc

... p( x , y...)  1 ...
i 1 j 1
p(xr, y1)
p(xr, y2)
pY (y1)
pY (y1)
i
j
...
p(xr, yc)
pX (xr)
pY (yc)
1
Suppose Z is a third discrete random
variable that depends on X and Y, i.e.,
there exists a joint pmf z  p ( x, y ) for
every ordered pair (x, y)…
Y
Suppose X and Y are two
discrete random variables with
pmfs pX(x) and pY(y) respectively:
x
pX(x)
y
pY(y)
x1
pX (x1)
y1
pY(y1)
x2
pX (x2)
y2
pY(y2)














yc
pY(yc)
xr
pX (xr)
1
p X ( xi )  0
r
p
i 1
X
( xi )  1
1
pY ( y j )  0
c
p
j 1
Y
(yj) 1
y1
y2
…
x1
p(x1, y1)
p(x1, y2)
...
p(x1, yc) pX (x1)
x2
p(x2, y1)
p(xp2,(yx2i ), y j ) 
...0
p(x2, yc) pX (x2)
X
r
...
xr
c
yc

... p( x , y...)  1 ...
i 1 j 1
p(xr, y1)
p(xr, y2)
pY (y1)
pY (y1)
i
j
...
p(xr, yc)
pX (xr)
pY (yc)
1
marginal pmf of X
Suppose X and Y are two
discrete random variables with
pmfs pX(x) and pY(y) respectively:
x
pX(x)
y
pY(y)
x1
pX (x1)
y1
pY(y1)
x2
pX (x2)
y2
pY(y2)














yc
pY(yc)
xr
pX (xr)
1
p X ( xi )  0
r
p
i 1
X
( xi )  1
1
pY ( y j )  0
c
p
j 1
Y
(yj) 1
c
p X ( x)   p( xi , y j )
z  p ( x, y )
j 1
Y
y1
y2
…
x1
p(x1, y1)
p(x1, y2)
...
p(x1, yc) pX (x1)
x2
p(x2, y1)
p(xp2,(yx2i ), y j ) 
...0
p(x2, yc) pX (x2)
X
r
...
xr
c
yc

... p( x , y...)  1 ...
i
i 1 j 1
p(xr, y1)
p(xr, y2)
pY (y1)
pY (y1)
j
...
marginal pmf of Y
r
pY ( y )   p( xi , y j )
i 1
p(xr, yc)
pX (xr)
pY (yc)
1
p
p
p
p
p
p( xi , y j )  0
r
c
 p( x , y )  1
Joint Probability
Mass Function
i
i 1 j 1
j
marginal pmf of X
c
p X ( x )   p ( x, y j )
pmf z  p ( x, y )
p
Y
j 1
X
( x)  1
x
y1
y2
…
x1
p(x1, y1)
p(x1, y2)
...
p(x1, yc) pX (x1)
x2
p(x2, y1)
p(x2, y2)
...
p(x2, yc) pX (x2)
X
xr
yc
p( xi , y j )  p X ( xi ) pY ( y j )
i  1, 2, , r j  1, 2, , c
...
...
...
...
p(xr, y1)
p(xr, y2)
...
p(xr, yc)
pX (xr)
pY (y1)
pY (y2)
…
pY (yc)
1
marginal pmf of Y
r
pY ( y )   p( xi , y )
p
Y
y
i 1
( y)  1
Def: X and Y are
statistically independent if
i.e., each cell probability
is equal to the product of
its marginal probabilities.
p( x, y)  pX ( x) pY ( y)
Joint Probability
Mass Function
pmf z  p ( x, y )
Y
y1
y2
…
yc
x1
p(x1, y1)
p(x1, y2)
...
p(x1, yc)
x2
p(x2, y1)
p(x2, y2)
...
p(x2, yc)
...
...
...
...
p(xr, y1)
p(xr, y2)
...
p(xr, yc)
X
xr
Joint Probability
Mass Function
pmf z  p ( x, y )
Z
In principle, one can construct a
probability histogram much as
before, with the height of each
rectangle centered at the point (x, y)
equal to the pmf z = p(x, y).
What happens as the
partition of the X and Y
axes becomes arbitrarily
small (i.e., the number of
rows and columns  ∞)?
Recall…
Time
intervals
intervals
= 1.0
= 5.0
2.0
1.0
secs
secs
Time
0.5
“Density”
Interval widths can be made
arbitrarily small, i.e, the scale at
which X is measured can be made
arbitrarily fine, since it is continuous.
f ( x)
p ( x)
(height) (area)
|
x
x
(width)
pmf
P( X  x)  p( x)  f ( x) x

As x  0 and # rectangles  ∞,
this “Riemann sum” approaches the
area under the density curve f(x),
expressed as a definite integral.
pdf
  f ( x) 
dxx  1


Total Area
b
P(a  X  b)   f ( x))
dxx
b
a
a
Similarly…
.
14
p( xi , y j )  0
r
c
 p( x , y )  1
Joint Probability
Mass Function
i
i 1 j 1
j
pmf z  p ( x, y )
Y
x1
x2
X
xr
y1
y2
…
p(x1, y1)
p(x1, y2)
...
p(x2, y1)
...
p(x2, y2)
yc
p(x1, yc) pX (x1)
p(x2, yc) pX (x2)
...
...
...
p(xr, y1)
p(xr, y2)
...
p(xr, yc)
pX (xr)
pY (y1)
pY (y2)
…
pY (yc)
1
r
pY ( y )   p( xi , y )
p
Y
y
i 1
( y)  1
c
p X ( x )   p ( x, y j )
p
...
marginal pmf of Y
marginal pmf of X
x
j 1
X
( x)  1
r
pf( x( ix,,yyj))  00
c
 1 1
 
 f (px(,xy,)ydy) dx

Joint
Joint Probability
Probability
Density
Function
Mass Function

i
 
i 1 j 1
j
ppmf
df zz  fp((xx,, yy))
Y
x1
x2
X
y1
y2
…
p(x1, y1)
p(x1, y2)
...
p(x2, y1)
...
p(x2, y2)
yc
p(x1, yc) pX (x1)
marginal
of X
X
marginal pmf
pdf of
p(x2, yc) pX (x2)
fpX ( x( x) )f p((xx, ,yy) dy
)
c

j 1
X

xr
 f p( x()xdx
) 11
...
...
...
...
p(xr, y1)
p(xr, y2)
...
p(xr, yc)
pX (xr)
pY (y1)
pY (y2)
…
pY (yc)
1
marginal pmf
ofYY
pdf of
r
fYp( y()y)  f p
( x(,xy,)ydx
)
Y


i 1
i
 f p( y()ydy) 11
 Y Y
y

x
X
X
j
a
x
A
X
y
P ( X , Y )  A  ?
b
d
c
P( a  X  b
Y
b
d
p ( x, y )
c  Y  d )  lim  f ( x, y )  y  x
x 0
y 0 x  a y c

b
x a

d
y c
f ( x, y ) dy dx
P ( X , Y )  A   f ( x, y) dA
Z
Joint Probability
Density Function
A
Volume under
density f(x, y)
over A.
pdf z  f ( x, y )
f ( x, y )  0
for all ( x, y )  2 ,
 f ( x, y) dy dx  1
“area element”
dA dy
2
dx
f ( x, y )
Y
0

( x, y )
X
Area A
2
dA  dy dx
or
dA  dx dy
Example: Uniform Distribution
Joint Probability
Density Function
Z
Recall for one r.v. X…
1 3, 0  x  3
f ( x)  
otherwise
0,
pdf z  f ( x, y )
f ( x, y )  0
for all ( x, y )  2 ,
 f ( x, y) dy dx  1
13
1
3
0
2
0
Y
2
X
Example: Uniform Distribution
1
Z
joint pdf f ( x, y ) 
6
( x, y) | 0  x  3, 0  y  2
Joint Probability
Density Function
pdf z  f ( x, y )
1
6
f ( x, y )  0
for all ( x, y )  2 ,
 f ( x, y) dy dx  1
2
0
3
X
2
Y
2
(0 else)
Joint Probability
Density Function
pdf z  f ( x, y )
Example: Uniform Distribution
1
joint pdf f ( x, y ) 
0
6
( x, y) | 0  x  3, 0  y  2 (0 else)

Confirm pdf...
 f ( x, y) dydx  1?
2
y2
y0
|
x
1
xx0? y 0?  6  dy dx
x 3?
y  2?
3
2

1
6
 

1
6
  y

1
6

1
6
x 0 y 0
1 dy dx
3
2
x 0
0

3
x 0
1
6
2
3
dx
2 dx
 2x 0
3
 16 (6  0)  1

Example:
Joint Probability
Density Function
pdf z  f ( x, y )
1x  y
jointpdf
pdf f f((xx, ,yy))
0
6 15
( x, y) | 0  x  3, 0  y  2 (0 else)

Confirm pdf...
 f ( x, y) dydx  1?
2
y   x,
z0
Example:
x y
0
15
( x, y) | 0  x  3, 0  y  2 (0 else)
joint pdf f ( x, y) 
Joint Probability
Density Function
pdf z  f ( x, y )

 f ( x, y) dydx  1?
Confirm pdf...
2
 x y
x0 y 0  15  dy dx
x 3
y 2

1
15

1
15
3
2
x 0
y 0
 
( x  y ) dy dx

3

1
15

3

1
15
 x  2 x   151 (9  6  0)  1
0
2
 x y  y  dx
y 0
x 0
x 0
2
1
2
2
(2 x  2) dx
3

Example:
x y
15
( x, y) | 0  x  3, 0  y  2 (0 else)
joint pdf f ( x, y) 
Joint Probability
Density Function
pdf z  f ( x, y )
P ( X , Y )  A  ?

x2
x 0
y 1
A
x2

1
15

1
15
2
 
x 0

2
1
y 0
( x  y ) dy dx
1
 x y  y  dx
y 0
x 0
2
1
2
2
2
1

x
(1)

(1)
 0 dx
2
x0 
2
1
1
 15   x  12  dx 
x 0
5

A  ( x, y) | 0  x  2, 0  y  1
 x y
y 0  15  dy dx
y 1
1
15
Example:
x y
15
( x, y) | 0  x  3, 0  y  2 (0 else)
joint pdf f ( x, y) 
Joint Probability
Density Function
pdf z  f ( x, y )
P ( X , Y )  A  ?
 x x y y 
dxdx
 dy
x0? y ?0  1515 dy

x 3?
2 2
y x
9
y  ?92 x 2
 151 
3
 151 
3

2 x2
9
x 0 y 0
( x  y ) dy dx
2 x2
9
 x y  12 y  dx
y 0
x 0
A
|
x
A  ( x, y ) | 0  x  3, 0  y  x
2
9
2

2
3

1
15
2 2
1 2 2 2


x
(
x
)

2 ( 9 x )  0  dx
x0  9

1
15
 
3
x 0
2
9
x3  812 x4  dx  19 50
 0.38
Example:
P ( X , Y )  A   f ( x, y) dA
A
6
1
 2 
dy dx
 x0.5 y 0 1  xy
A
1
1
2x

1

ln
2


 dx
 2 x 0.5  x
6
1
2x
1

 1


ln
1

xy
 dy dx
 2 x 0.5  x
y 0

6 ln 2

 1 1 
 ln    0 dx
2 x  0.5 

 x 2 

6 ln 2

6 (ln 2) 2
6
6
1
1
 1

 2    ln  21   dx
 x 0.5  x

6
1
2
2
2
1
x0.5 x dx
1
ln | x |
1
1
2
 0.2921
26
p( xi , y j )  0
r
c
 p( x , y )  1
Joint Probability
Mass Function
i
i 1 j 1
j
marginal pmf of X
c
p X ( x )   p ( x, y j )
pmf z  p ( x, y )
p
Y
j 1
X
( x)  1
x
y1
y2
…
x1
p(x1, y1)
p(x1, y2)
...
p(x1, yc) pX (x1)
x2
p(x2, y1)
p(x2, y2)
...
p(x2, yc) pX (x2)
...
...
...
...
p(xr, y1)
p(xr, y2)
...
p(xr, yc)
pX (xr)
pY (y1)
pY (y2)
…
pY (yc)
1
X
xr
marginal pmf of Y
r
pY ( y )   p( xi , y )
p
Y
y
i 1
( y)  1
yc
p( xi , y j )  0
r
c
 p( x , y )  1
Joint Probability
Mass Function
i
i 1 j 1
j
marginal pmf of X
c
p X ( x )   p ( x, y j )
pmf z  p ( x, y )
p
Y
j 1
X
( x)  1
x
y1
y2
…
x1
p(x1, y1)
p(x1, y2)
...
p(x1, yc) pX (x1)
x2
p(x2, y1)
p(x2, y2)
...
p(x2, yc) pX (x2)
...
...
...
...
p(xr, y1)
p(xr, y2)
...
p(xr, yc)
pX (xr)
pY (y1)
pY (y2)
…
pY (yc)
1
X
xr
yc
marginal pmf of Y
r
pY ( y )   p( xi , y )
p
Y
y
i 1
( y)  1
Extend this to the
continuous scenario….
Z
Joint Probability
Density Function
z  f ( x, y*)
pdf z  f ( x, y )
fY ( y*)
marginal pdf of X
y  y*
marginal pdf of Y
fYfY( (yy*))  


X
ff ((x, y*)
) dx
dx
Fix
y  y*
y  y*
Y
f X ( x)  


f ( x, y ) dy
Example (revisted):
x y
15
( x, y) | 0  x  3, 0  y  2 (0 else)
joint pdf f ( x, y) 
Joint Probability
Density Function
pdf z  f ( x, y )
marginal pdf of Y
fY ( y )  


f ( x, y ) dx
3
 x y
1 1 2



x

x
y

 
dx 15  2


x

0
x 0
 15 
1
10 (3  2 y )
x 3
marginal pdf of X
f X ( x)  


f ( x, y ) dy  
 x  y   1  x y  1 y2 2 
dy 15 
2
 y 0
y  0  15 


y 2
2
15
( x  1)
Example (revisted):
x y
15
( x, y) | 0  x  3, 0  y  2 (0 else)
joint pdf f ( x, y) 
Joint Probability
Density Function
pdf z  f ( x, y )




fY ( y) dy 

1
10
2
1
10
0
(3  2 y ) dy
2
3 y  y   1
0
2
Check?
marginal pdf of Y
fY ( y)  101 (3  2 y)  0
marginal pdf of X
f X ( x)  152 ( x  1)



f X ( x) dx 
Exercise

=1


Example (revisted):
x y
15
( x, y) | 0  x  3, 0  y  2 (0 else)
joint pdf f ( x, y) 
Joint Probability
Density Function
pdf z  f ( x, y )
Calculate cdf FY ( y )  P(Y  y )
y


 fY (ty))dtdy 
1
10
0


y2
1
10
(3  2ty))dt
dy
3tyt y  1011  3 y  y 2 
00
2
2 2y
Note: As y increases from 0 to 2,
this increases continuously and
monotonically from 0 to 1.
marginal pdf of Y
fY ( y)  101 (3  2 y)
marginal pdf of X
f X ( x)  152 ( x  1)



f X ( x) dx 
Exercise
=1
Example (revisted):
x y
15
( x, y) | 0  x  3, 0  y  2 (0 else)
joint pdf f ( x, y) 
Joint Probability
Density Function
pdf z  f ( x, y )
cdf FY ( y)  P(Y  y) 
1
10
(3 y  y 2 )
P(Y  1)  F
?Y (1)  4 10  2 5
marginal pdf of Y
fY ( y)  101 (3  2 y)
marginal pdf of X
f X ( x)  152 ( x  1)



f X ( x) dx 
Exercise
=1
Example (revisted):
x y
15
( x, y) | 0  x  3, 0  y  2 (0 else)
joint pdf f ( x, y) 
Joint Probability
Density Function
pdf z  f ( x, y )
cdf FY ( y)  P(Y  y) 
1
10
(3 y  y 2 )
P(Y  1)  F
?Y (1)  4 10  2 5
Exercise:
cdf FX ( x)  P( X  x) 
1
15
( x 2  2 x)
P( X  2)  FX (2)  8 15
marginal pdf of Y
fY ( y)  101 (3  2 y)
marginal pdf of X
f X ( x)  152 ( x  1)



f X ( x) dx 
Exercise
=1
Example (revisted):
x y
15
( x, y) | 0  x  3, 0  y  2 (0 else)
joint pdf f ( x, y) 
Joint Probability
Density Function
pdf z  f ( x, y )
P(Y  1)  F
?Y (1)  4 10  2 5
P( X  2)  FX (2)  8 15
P (Both)  P( X  2
y 1
A
P  ( X , Y )  A  1 5 (see slide 24)
x2
marginal pdf of Y
fY ( y)  101 (3  2 y)
marginal pdf of X
f X ( x)  152 ( x  1)
Y  1)  ?



f X ( x) dx 
Exercise
=1
Joint Probability
Density Function
marginal cdf of Y
FY ( y )  P(Y  y )  

X
pdf z  f ( x, y )




f ( x, y ) dy dx  1
f(x2, y1x) f(xy 
2, y2)
Y
...
...
f(x2, yc)
P ...
( X , Y )  ...
A 
...
A
f(xr, y1)
f(x1, yc)
f(xr, y2)
...f ( x, yf(x) dA, y )
A
r
marginal pdf of X
f X ( x)  


f ( x, y ) dy
marginal cdf of X
x
FX ( x)  P( X  x)   f X (t ) dt

marginal pdf of Y
f(x1,fy(2x) , y ) ...
0
f(x1, y1)
y
fY (t ) dt
Integrate fY
from   to y
fY ( y )  


f ( x, y ) dx
c
1
Integrate fY
from   to +
Integrate f X
from   to +
Integrate f X
from   to x
Example
x2 y3
joint pdf f ( x, y) 
36
( x, y) | 0  x  3, 0  y  2
Joint Probability
Density Function
pdf z  f ( x, y )
Confirm pdf...
(0 else)
 f ( x, y) dydx  1
Exercise
2
marginal pdf of Y
fY ( y )  


marginal pdf of X
f X ( x)  


f ( x, y ) dy
f ( x, y ) dx
2
4
2


x y
x
x
y
x

dy   y 3 dy 

y 0 36
36 y 0
36  4 
9
y 0
2
2 3
2
2
2
Example
x2 y3
joint pdf f ( x, y) 
36
( x, y) | 0  x  3, 0  y  2
Joint Probability
Density Function
pdf z  f ( x, y )
Confirm pdf...
(0 else)
 f ( x, y) dydx  1
Exercise
2
NOTE: For "standard rectangles"
f ( x, y)  f X ( x) fY ( y)

X and Y are
statistically independent!
marginal pdf of Y
fY ( y )  


marginal pdf of X
f X ( x)  


x2
f ( x, y ) dy  9
y3
f ( x, y ) dx  4
Exercise
Example
x2 y3
joint pdf f ( x, y) 
36
( x, y) | 0  x  3, 0  y  2
Joint Probability
Density Function
pdf z  f ( x, y )
(0 else)
P ( X , Y )  A  ?
NOTE: For "standard rectangles"
f ( x, y)  f X ( x) fY ( y)

y 1
X and Y are
statistically independent!
marginal pdf of Y
x0
A
x2


Exercise
P ( X , Y )  A  P( X  2
y0
marginal pdf of X
f X ( x)  
fY ( y )  
yy33

f ( x, y ) dx  44


xx2 2
f ( x, y ) dy  99
 222 xx112 2 x 21 
yy3 3y 31 y3 
dy
dydy
dx
dx dx
   dx
dy 


xx
y9
0 9
x0 0 9
y

0
y

0

 364 4 4

Y  1)
Example
x2 y3
joint pdf f ( x, y) 
36
( x, y) | 0  x  3, 0  y  2
Joint Probability
Density Function
pdf z  f ( x, y )
(0 else)
P ( X , Y )  A  ?
NOTE: For "standard rectangles"
f ( x, y)  f X ( x) fY ( y)

y 1
X and Y are
statistically independent!
marginal pdf of Y
x0
A
x2


Exercise
P ( X , Y )  A  P( X  2
y0
marginal pdf of X
f X ( x)  
fY ( y )  
yy33

f ( x, y ) dx  44


x2
f ( x, y ) dy  9
 22 2 x1x12 2 x 21 y
y3 3y 31 y3 
dy
dydy
dx
dx dx
     dx
dy 


xxx0 09
y9
0 9
y

0
y

0

 364 4
4 
Y  1)
Example
x2 y3
joint pdf f ( x, y) 
36
( x, y) | 0  x  3, 0  y  2
Joint Probability
Density Function
pdf z  f ( x, y )
(0 else)
P ( X , Y )  A  ?
NOTE: For "standard rectangles"
f ( x, y)  f X ( x) fY ( y)

y 1
X and Y are
statistically independent!
marginal pdf of Y
x0
A
x2
fY ( y )  


y3
f ( x, y ) dx  4
Exercise
P ( X , Y )  A  P( X  2
y0
marginal pdf of X
f X ( x)  


2
x

f ( x, y ) dy
9
 P( X  2)
P(Y  1)
Y  1)
Example
x2 y3
joint pdf f ( x, y) 
36
( x, y) | 0  x  3, 0  y  2
Joint Probability
Density Function
pdf z  f ( x, y )
(0 else)
P ( X , Y )  A  ?
NOTE: For "standard rectangles"
f ( x, y)  f X ( x) fY ( y)

y 1
X and Y are
statistically independent!
marginal pdf of Y
x0
A
x2
fY ( y )  


y3
f ( x, y ) dx  4
Exercise
P ( X , Y )  A  P( X  2) P(Y  1) 
y0
t2
8
FX (2)  P( X  2)  
dt 
0 9
27
3
1t
1
FY (1)  P(Y  1)   dt 
0 4
16
2
marginal pdf of X
f X ( x)  


x2
f ( x, y ) dy  9
1
54
Example
x2 y3
joint pdf f ( x, y) 
36
( x, y) | 0  x  3, 0  y  2
Joint Probability
Density Function
pdf z  f ( x, y )
(0 else)
P ( X , Y )  A  ?
NOTE: For "standard rectangles"
f ( x, y)  f X ( x) fY ( y)

yx 2
x0
X and Y are
statistically independent!
marginal pdf of Y
A
x2
fY ( y )  


y3
f ( x, y ) dx  4
Exercise
P ( X , Y )  A  P( X  2
2) P(YY 1)
1) 
y0
2
marginal pdf of X
f X ( x)  


x2
f ( x, y ) dy  9
2t
x
2
FX (2)
P2x( X
x
1 x
2 y
x4222)y33 0 9
8
dt


1
27
dydx
dx 


 36
y 1dydx
t3
1
0
x 0 36
y 0 y4
FY (1)  P(Y  1) 
 00 dt  16128
4
16
  
1
54
Example
x2 y3
joint pdf f ( x, y) 
36
( x, y) | 0  x  3, 0  y  2
Joint Probability
Density Function
pdf z  f ( x, y )
(0 else)
P ( X , Y )  A  ?
NOTE: For "standard rectangles"
f ( x, y)  f X ( x) fY ( y)
X and Y are not independent if
don’t have “standard” rectangles!

X and Y are
statistically independent!
marginal pdf of Y
fY ( y )  


y3
f ( x, y ) dx  4
P ( X , Y )  A  P( X  2
f X ( x)  


x2
f ( x, y ) dy  9
Y  1)
1
x y 
1

 4  dx  16128 
x 0 36
54
 0
1
marginal pdf of X
x 2
Exercise
2
4
Example
x2 y3
joint pdf f ( x, y ) 
36
( x, y) | 0  x  3, 0  y  mx
Joint Probability
Density Function
pdf z  f ( x, y )
y  m x, where
4
84
m
 1.009133
3

andY Yare
are
XXand
statistically dependent
independent
statistically
! !
marginal pdf of Y
fY ( y )  


f X ( x)  


(0 else)
NOTE: For "standard rectangles"
f ( x, y) 
 f XX ( x) fYY ( y)
X and Y are not independent if
don’t have “standard” rectangles!
marginal pdf of X
Exercise
y3
f ( x, y ) dx  4
Exercise
mx
2
7 x6
x
x y 
x x y
3
dy   y dy 



f ( x, y ) dy  9y 0 36
y

0
36
36  4 
36
y 0
2
2
2 3
2
mx
2
2
4
P  ( X 1 , , X n )  A 
P ( X , Y )  Af( x1 , f (,xx, ny))dA
dA
Joint Probability
Density Function
 
A
pdf z  f ( x, y )
“Hypervolume”
Volume under
under density f(x,
f y)
over Aover
.
A.
dA  dx1 dx2 dxn
or any permutation
Definition of statistical
independence of X and Y
f ( x, y)  f X ( x) fY ( y)
can be extended to any
number of variables.
A
X1
Area A
X2
X3
X4
XY n