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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