Examples: Joint Densities and Joint Mass Functions

AMS 311

Joe Mitchell

Examples: Joint Densities and Joint Mass Functions

Example 1: X and Y are jointly continuous with joint pdf

f (x, y) =

cx2

+

xy 3

0,

if 0 x 1, 0 y 2 otherwise.

(a). Find c. (b). Find P (X + Y 1). (c). Find marginal pdf's of X and of Y . (d). Are X and Y independent (justify!). (e). Find E(eX cos Y ). (f). Find cov(X, Y ).

We start (as always!) by drawing the support set. (See below, left.)

y

y

2 support set

2

Blue: subset

of support set

1

with y>1-x

x

1

x

1 y=1-x

(a). We find c by setting

1=

-

f (x, y)dydx =

-

1 0

2

(cx2

0

+

xy 3

)dydx

=

2c 3

+

1 3

,

so c = 1.

(b). Draw a picture of the support set (a 1-by-2 rectangle), and intersect it with the set

{(x, y) : x + y 1}, which is the region above the line y = 1 - x. See figure above, right.

To compute the probability, we double integrate the joint density over this subset of the

support set:

P (X + Y 1) =

1 0

2

(x2

1-x

+

xy 3

)dydx

=

65 72

(c). We compute the marginal pdfs:

fX(x) = f (x, y)dy =

-

02(x2

+

xy 3

)dy

=

2x2

+

2x 3

0

if 0 x 1 otherwise

fY (y) = f (x, y)dx =

-

01(x2

+

xy 3

)dx

=

1 3

+

y 6

0

if 0 y 2 otherwise

1

(d). NO, X and Y are NOT independent. The support set is a rectangle, so we need to check if it is true that f (x, y) = fX(x)fY (y), for all (x, y). We easily find counterexamples: f (0.2, 0.3) = fX(0.2)fY (0.3).

(e).

E(eX cos Y ) =

1 0

2

(ex

0

cos

y)(x2

+

xy 3

)dydx

(f). cov(X, Y ) = E(XY ) - E(X)E(Y )

=

1 0

2 0

xy(x2

+

xy 3

)dydx

-

1 0

2 0

x(x2

+

xy 3

)dydx

1 0

2 0

y(x2

+

xy 3

)dydx

Example 2: X and Y are jointly continuous with joint pdf

f (x, y) =

cxy 0,

if 0 x, 0 y, x + y 1 otherwise.

(a). Find c. (b). Find P (Y > X). (c). Find marginal pdf 's of X and of Y . (d). Are X and Y independent (justify!).

We start (as always!) by drawing the support set. (See below, left.)

y

y

1 1

support set

y=x

Blue: subset of support set with y>x

1x y=1-x

0.5 1 x

(a). We find c by setting

so c = 24.

1=

-

f (x, y)dydx =

-

1 0

1-x 0

cxydydx

=

c 24

,

(b). Draw a picture of the support set (a triangle), and intersect it with the set {(x, y) : y x}, which is the region above the line y = x; this yields a triangle whose leftmost x-value is 0 and whose rightmost x-value is 1/2 (which is only seen by drawing the figure!). See figure above, right. To compute the probability, we double integrate the joint density over this subset of the support set:

1/2 1-x

P (Y X) =

24xydydx

0

x

2

(c). We compute the marginal pdfs:

fX(x) = f (x, y)dy =

-

fY (y) = f (x, y)dx =

-

1-x 0

24xydy

=

12x(1

-

x)2

0

1-y 0

24xydx

=

12y(1

-

y)2

0

if 0 x 1 otherwise

if 0 y 1 otherwise

(d). NO, X and Y are NOT independent. The support set is not a rectangle or generalized rectangle, so we know we can find points (x, y) where it fails to be true that f (x, y) = fX(x)fY (y). In particular, f (0.7, 0.7) = 0 = fX(0.7)fY (0.7) > 0.

Example 3: X and Y are jointly continuous with joint pdf

f (x, y) =

cxy 0,

if 0 x 1, 0 y 1 otherwise.

(a). Find c. (b). Find P (|Y - 2X| 0.1). (c). Find marginal pdf 's of X and of Y . (d). Are X and Y independent (justify!).

We start (as always!) by drawing the support set, which is just a unit square in this case.

(See below, left.)

y=2x+0.1

y=2x-0.1

y

y

1 support set

1 Blue: subset

of support set

with -.1 ................
................

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