Joint Distribution - Example
Lecture 17: Joint Distributions
Statistics 104
Colin Rundel
March 26, 2012
Section 5.1 Joint Distributions of Discrete RVs
Joint Distribution - Example, cont.
Let B be the number of Black socks and W the number of White socks drawn, then the joint distribution of B and W is given by:
W
012
0
1 8 6 15 66 66 66 66
B
1
12 66
24 66
0
36 66
2
15 66
0
0
15 66
28 32 6 66 66 66 66 66
1/66 8/66 6/66 12/66
P(B = b, W = w ) = 24/66
0/66 15/66 0/66 0/66
If b=0,w=0 If b=0,w=1 If b=0,w=2 If b=1,w=0 If b=1,w=1 If b=1,w=2 If b=2,w=0 If b=2,w=1 If b=2,w=2
64
2
P(B = b, W = w ) =
b
w
2-b-w 12
, for 0 b, w 2 and b + w 2
2
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 2 / 32
Section 5.1 Joint Distributions of Discrete RVs
Joint Distribution - Example
Draw two socks at random, without replacement, from a drawer full of twelve colored socks:
6 black, 4 white, 2 purple Let B be the number of Black socks, W the number of White socks drawn, then the distributions of B and W are given by:
0
1
2
P(B=k)
6 12
5 11
=
15 66
2
6 12
6 11
=
36 66
6 12
5 11
=
15 66
P(W=k)
8 12
7 11
=
28 66
2
4 12
8 11
=
32 66
4 12
3 11
=
6 66
Note - B HyperGeo(12, 6, 2) =
Statistics 104 (Colin Rundel)
66 k 2-k
12 2
48
and W HyperGeo(12, 4, 2) =
k 2-k 12
2
Lecture 17
March 26, 2012
1 / 32
Section 5.1 Joint Distributions of Discrete RVs
Marginal Distributions
Note that the column and row sums are the distributions of B and W respectively.
P(B = b) = P(B = b, W = 0) + P(B = b, W = 1) + P(B = b, W = 2) P(W = w ) = P(B = 0, W = w ) + P(B = 1, W = w ) + P(B = 2, W = w )
These are the marginal distributions of B and W . In general,
P(X = x) = P(X = x, Y = y ) = P(X = x|Y = y )P(Y = y )
y
y
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 3 / 32
Section 5.1 Joint Distributions of Discrete RVs
Conditional Distribution
Conditional distributions are defined as we have seen previously with
P(X = x, Y = y ) joint pmf
P(X = x|Y = y ) =
=
P(Y = y )
marginal pmf
Therefore the pmf for white socks given no black socks were drawn is
1
P (W
=
w |B
=
0)
=
P(W = w , B = P(B = 0)
0)
=
66
8
66
6
66
15 66
=
1 15
15 66
=
8 15
15 66
=
6 15
if W = 0 if W = 1 if W = 2
Joint CDF
Section 5.1 Joint Distributions of Continuous RVs
F (x, y ) = P[X x, Y y ] = P[(X , Y ) lies south-west of the point (x, y )]
Y (x,y)
q
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 4 / 32
Joint CDF, cont.
Section 5.1 Joint Distributions of Continuous RVs
The joint Cumulative distribution function follows the same rules as the univariate CDF,
Univariate definition:
lim F (x) = 0
x -
x
F (x) = P(X x) =
f (z)dz
-
lim F (x) = 1
x
x y F (x) F (y )
Bivariate definition:
y
x
F (x, y ) = P(X x, Y y ) =
f (x, y ) dx dy
- -
lim F (x, y ) = 0
x,y -
lim F (x, y ) = 1
x,y
x x ,y y F (x, y ) F (x , y )
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 6 / 32
Statistics 104 (Colin Rundel)
X
Lecture 17
March 26, 2012 5 / 32
Section 5.1 Joint Distributions of Continuous RVs
Marginal Distributions
We can define marginal distributions based on the CDF by setting one of the values to infinity:
x
F (x, ) = P(X x, Y ) =
f (x, y ) dy dy
- -
= P(X x) = FX (x)
y
F (, y ) = P(X , Y y ) =
f (x, y ) dx dy
- -
= P(Y y ) = FY (y )
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 7 / 32
Joint pdf
Section 5.1 Joint Distributions of Continuous RVs
Similar to the CDF the probability density function follows the same general rules except in two dimensions,
Univariate definition:
f (x) 0 for all x
f (x)
=
d dx
F
(x
)
-
f
(x )dx
=
1
Bivariate definition:
f (x, y ) 0 for all (x, y )
f (x, y ) =
F (x, y )
x y
f (x, y ) dx dy = 1
- -
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 8 / 32
Section 5.1 Joint Distributions of Continuous RVs
Probability and Expectation
Univariate definition: Bivariate definition:
P(X A) = f (x) dx
A
E [g (X )] =
g (x) ? f (x) dx
-
P(X A, Y B) =
f (x, y ) dx dy
AB
E [g (X , Y )] =
g (x, y ) ? f (x, y ) dx dy
- -
Marginal pdfs
Section 5.1 Joint Distributions of Continuous RVs
Marginal probability density functions are defined in terms of "integrating out" one of the random variables.
fX (x) = fY (x) =
f (x, y ) dy
-
f (x, y ) dx
-
Previously we defined independence in terms of E (XY ) = E (X )E (Y ) X and Y are independent. This is equivalent in the joint case of f (x, y ) = fX (x)fY (y ) X and Y are independent.
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 9 / 32
Section 5.1 Joint Distributions of Continuous RVs
Example 1 - Joint Uniforms
Let X , Y Unif(0, 1), it is straight forward to see graphically that
0 xy
F (x, y ) = x
y 1
if x (-, 0) or y (-, 0) if x (0, 1), y (0, 1) if x (0, 1), y (1, ) if x (1, ), y (0, 1) if x (1, ), y (1, )
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 10 / 32
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 11 / 32
Example 1, cont.
Section 5.1 Joint Distributions of Continuous RVs
Based on the CDF we can calculate the pdf using the 2nd partial derivative with regard to x and y.
f (x, y ) =
F (x, y )
x y
0 1
=0
0 0
if x (-, 0) or y (-, 0) if x (0, 1), y (0, 1) if x (0, 1), y (1, ) if x (1, ), y (0, 1) if x (1, ), y (1, )
1 if x (0, 1), y (0, 1) =
0 otherwise
Example 1, cont.
Section 5.1 Joint Distributions of Continuous RVs
Based on the pdf we can calculate the marginal densities:
1 if x (0, 1), y (0, 1) f (x, y ) =
0 otherwise
fX (x) = =
f (x, y ) dy
-
1 0
1
dy
if x (0, 1)
0 dy otherwise
1 if x (0, 1) =
0 otherwise
1 if y (0, 1) fY (y ) = 0 otherwise
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 12 / 32
Example 1, cont.
Section 5.1 Joint Distributions of Continuous RVs
Expectation is also straight forward:
E (X ) =
xf (x, y ) dx dy
- -
11
1
=
x dx dy =
00
0
x2 1 dy
20
11
11
=
dy = y = 1/2
02
20
E (Y ) =
yf (x, y ) dx dy
- -
11
1
=
y dx dy =
xy |10 dy
00
0
1
y2 1
= y dy =
= 1/2
0
20
WShtaictishticss1h0o4 u(CldolinnRountdebl) e surprising... Lecture 17
March 26, 2012 13 / 32
Example 1, cont.
Section 5.1 Joint Distributions of Continuous RVs
E (XY ) =
xyf (x, y ) dx dy
- -
11
1
=
xy dx dy =
00
0
x2y 1 dy
2 x=0
1y
y2 1
=
dy =
= 1/4
02
40
Note that E (XY ) = E (X )E (Y ), what does this tell us about X and Y ?
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 14 / 32
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 15 / 32
Section 5.1 Joint Distributions of Continuous RVs
Example 1, another way
If we did not feel comfortable coming up with the graphical arguments for F (x, y ) we can also use the fact that the pdf is constant on (0, 1) ? (0, 1) to derive the same distribution / density.
f (x, y ) = c
1=
f (x, y ) dx dy
- -
11
=
c dx dy
00
1
1
=
cx|10 dy = c dy
0
0
= cy |10 = c
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 16 / 32
Example 2, cont.
Section 5.1 Joint Distributions of Continuous RVs
Since the joint density is constant then
2 f (x, y ) = c = , for 0 x + y 3
9 based on the area of the triangle, but we need to be careful to define on what range. We can define the range in two ways since X and Y depend on each other, so we can define the range of X in terms of Y or Y in terms of X .
f (x, y ) =
2 9
if y (0, 3), x (0, 3 - y )
0 otherwise
=
2 9
if x (0, 3), y (0, 3 - x)
0 otherwise
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 18 / 32
Example 2
Section 5.1 Joint Distributions of Continuous RVs
Let X and Y be drawn uniformly from the triangle below
4
3
Y
2
1
0
0
1
2
3
4
Statistics 104 (Colin Rundel)
X Lecture 17
March 26, 2012 17 / 32
Find the joint pdf, cdf, and marginals. Section 5.1 Joint Distributions of Continuous RVs
Example 2, cont.
Depending on which range definition you choose it makes life easier when evaluating the marginal densities.
fX (x) =
f (x, y ) dy
-
3-x 2
=
dy
09
2 = (3 - x) for x (0, 3)
9
fY (y ) =
f (x, y ) dy
-
3-y 2
=
dx
09
2 = (3 - y ) for y (0, 3)
9
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 19 / 32
Example 2, cont.
Section 5.1 Joint Distributions of Continuous RVs
Finding the CDF with calculus is hard in this case, still a pain with graphical approaches but easier...
y
x
F (x, y ) =
f (x, y ) dx dy
- -
0
2 9
xy
2
9
xy
-
(y -(3-x))(x-(3-y )) 2
=2
9
3x
-
x2 2
2
9
3y
-
y2 2
1
if x (-, 0) or y (-, 0) if x (0, 3), y (0, 3), x + y (0, 3) if x (0, 3), y (0, 3), x + y (3, 6) if x (0, 3), y (3, ) if x (3, ), y (0, 3) if x (3, ), y (3, )
Example 3
Section 5.1 Joint Distributions of Continuous RVs
Let f (x, y ) = cx2y for x2 y 1.
Find: a) c b) P[X Y ] c) fX (x) and fY (y )
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 20 / 32
Example 3 - Range
Section 5.1 Joint Distributions of Continuous RVs
1
Y
-1 Statistics 104 (Colin Rundel)
0 X
Lecture 17
1 March 26, 2012 22 / 32
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 21 / 32
Example 3.a
Section 5.1 Joint Distributions of Continuous RVs
Let 0
f (x, y
y) 1,
-=cyx
2y
xforx2 y
y
1,
we
can
rewrite
the
bounds
as
1=
f (x, y ) dx dy
- -
1 y
=
cx2y dx dy
0 -1
1
x 3 y
=
cy
dy
0
3 x=-y
1
=
cy 5/2/3 + cy 5/2/3 dy
0
4cy 7/2 1 =
21
y =0
4 =c
21
c = 21/4
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 23 / 32
Example 3 - pdf
Section 5.1 Joint Distributions of Continuous RVs
0.20
0.15
f(x,y) 0.10
0.05
0.00
1.0
0.8
1.0
0.6
0.5
0.4
y
0.2
0.0 -1.0
-0.5
0.0 x
Statistics 104 (Colin Rundel)
Lecture 17
0.15 0.10 0.05 0.00
March 26, 2012 24 / 32
Example 3.b, cont.
Section 5.1 Joint Distributions of Continuous RVs
1
P(X Y ) =
|x| 21 x2y dy dx
-1 x2 4
= 2 1 x 21 x2y dy dx 0 x2 4
42 1 =
40
x2y2 x dx
2 x2
42 1 x 4 x 6
=
- dx
40 2 2
42 x 5 x 7 1
=
-
4 10 14 0
42 1 1
=
-
4 10 14
42 2
=
= 0.3
4 70
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 26 / 32
Example 3.b
Section 5.1 Joint Distributions of Continuous RVs
We need to integrate over the region where x2 y 1 and x y which is indicated in red below
1
1
Y
Y
-1
0
1
-1
0
1
X
X
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 25 / 32
Example 3.c
Section 5.1 Joint Distributions of Continuous RVs
fX (x) =
1 21 x2y dy x2 4
21 =
4
x2y2 1 2 x2
21 =
x2 - x6
, for x (-1, 1)
8
fY (y ) =
y 21 x2y dx -y 4
21 =
4
x 3y y 3 -y
21 y 5/2
=
2
4
3
= 7 y 5/2, for y (0, 1) 2
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 27 / 32
Example 3.c, cont.
Section 5.1 Joint Distributions of Continuous RVs
It is always a good idea to check that the marginals are proper densities.
1
fX (x)dx =
-1
1 21 -1 8
x2 - x6 dx
21 x 3 x 7 1
=
-
8 3 7 -1
21 1 1
=
- =1
43 7
1
fY (y )dy =
0
1 7 y 5/2dy 02
= 7 2 y 7/2 1 = 1
27
0
Example 4
Section 5.1 Joint Distributions of Continuous RVs
Let Y be the rate of calls at a help desk, and X the number of calls between 2 pm and 4 pm one day; Let's say that:
f (x, y ) = (2y )x e-3y x!
for y > 0, x = 0, 1, 2, . . ..
Find: a) P(X = 0) b) P(Y > 2) c) P[X = x] for all x
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 28 / 32
Example 4.a
Section 5.1 Joint Distributions of Continuous RVs
f (x, y ) = (2y )x e-3y , for y > 0, x = 0, 1, 2, . . . x!
P(X = 0) = f (0, y )dy
0
=
e-3y dy
0
= - 1 e-3y
3
0
= 1/3
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 30 / 32
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 29 / 32
Example 4.b
Section 5.1 Joint Distributions of Continuous RVs
f (x, y ) = (2y )x e-3y , for y > 0, x = 0, 1, 2, . . . x!
P(Y > 2) =
f (x, y )dy
2 x=0
= (2y )x e-3y dy 2 x=0 x !
=
e -3y
(2y )x
dy
2
x=0 x !
=
e -3y
(2y ) (2y )2
1+
+
+ ? ? ? dy
2
x =0
1
2
=
e-3y e2y dy =
e-y dy
2
2
=
-e -y
2
= e-2
= 0.13534
Statistics 104 (Colin Rundel)
Lecture 17
March 26, 2012 31 / 32
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