Let
(
X
,
Y
)
∼
Uniform
(
D
)
(X,Y)∼Uniform(D), where
D
:
=
{
3
X
>
Y
,
0
<
x
<
1
,
y
>
0
}
D:={3X>Y,0<x<1,y>0}

1 point
Find the joint density of
X
X and
Y
.
Y.

f
X
Y
(
x
,
y
)
=
{
3
2
for
0
<
x
<
1
,
0
<
y
<
3
x
0
otherwise
f
XY
​
(x,y)=
⎩
⎨
⎧
​

2
3
​

0
​

for 0<x<1,0<y<3x
otherwise
​

f
X
Y
(
x
,
y
)
=
{
2
3
for
0
<
x
<
1
,
0
<
y
<
3
x
0
otherwise
f
XY
​
(x,y)=
⎩
⎨
⎧
​

3
2
​

0
​

for 0<x<1,0<y<3x
otherwise
​

f
X
Y
(
x
,
y
)
=
{
2
3
for
0
<
x
<
1
,
y
>
3
x
0
otherwise
f
XY
​
(x,y)=
⎩
⎨
⎧
​

3
2
​

0
​

for 0<x<1,y>3x
otherwise
​

f
X
Y
(
x
,
y
)
=
{
2
3
for
0
<
x
<
1
,
y
>
0
0
otherwise
f
XY
​
(x,y)=
⎩
⎨
⎧
​

3
2
​

0
​

for 0<x<1,y>0
otherwise
​

To find the joint density \( f_{XY}(x, y) \) of \( (X, Y) \) given the specified region \( D \), we follow these steps:

1. **Identify the region \( D \)**:
   The region \( D \) is given by:
   \[
   D = \{(x, y) \mid 3x > y, 0 < x < 1, y > 0\}
   \]

   This implies that \( y \) must be less than \( 3x \) for \( 0 < x < 1 \).

2. **Determine the area of region \( D \)**:
   Since \( X \) and \( Y \) are uniformly distributed over the region \( D \), we need to compute the area of \( D \) to find the joint density. The region is bounded by the lines \( y = 0 \), \( y = 3x \), \( x = 0 \), and \( x = 1 \).

   The area of this triangular region can be calculated as:
   \[
   \text{Area}(D) = \int_{0}^{1} \int_{0}^{3x} dy \, dx
   \]
   Evaluate the inner integral first:
   \[
   \int_{0}^{3x} dy = 3x
   \]
   Now, integrate with respect to \( x \):
   \[
   \int_{0}^{1} 3x \, dx = 3 \int_{0}^{1} x \, dx = 3 \left[ \frac{x^2}{2} \right]_{0}^{1} = 3 \cdot \frac{1}{2} = \frac{3}{2}
   \]

   Therefore, the area of \( D \) is \( \frac{3}{2} \).

3. **Find the joint density**:
   Since \( (X, Y) \) is uniformly distributed over \( D \), the joint density function \( f_{XY}(x, y) \) is constant within \( D \) and zero outside of \( D \). The constant value is the reciprocal of the area of \( D \).

   Thus, the joint density \( f_{XY}(x, y) \) is:
   \[
   f_{XY}(x, y) = \begin{cases}
   \frac{2}{3} & \text{if } 0 < x < 1 \text{ and } 0 < y < 3x \\
   0 & \text{otherwise}
   \end{cases}
   \]

Therefore, the correct joint density function is:
\[
f_{XY}(x, y) = \begin{cases}
\frac{2}{3} & \text{for } 0 < x < 1 \text{ and } 0 < y < 3x \\
0 & \text{otherwise}
\end{cases}
\]

Would you like more details or have any questions? Here are some related questions to consider:

1. How do we find the marginal densities \( f_X(x) \) and \( f_Y(y) \)?
2. What is the expectation \( E[X] \) for this distribution?
3. How can we find the covariance \( \text{Cov}(X, Y) \) of this distribution?
4. What is the conditional density \( f_{Y|X}(y|x) \)?
5. Can we determine if \( X \) and \( Y \) are independent?
6. How do we compute the cumulative distribution function (CDF) \( F_{XY}(x, y) \)?
7. How would the density change if the region \( D \) was different, e.g., \( 2X > Y \) instead of \( 3X > Y \)?
8. What are some real-world examples of distributions similar to this one?

**Tip**: When dealing with joint distributions, always check the bounds of the variables and the region they cover to correctly identify the support of the density function.

Learn how to find the joint density function of X and Y given a uniform distribution over a specified region defined by 3X > Y, 0 < X < 1, and Y > 0. This problem involves understanding the concept of joint probability density functions and uniform distributions.

Math Problem Statement

Let ( X , Y ) ∼ Uniform ( D ) (X,Y)∼Uniform(D), where D :

f X Y ( x , y )

f X Y ( x , y )

f X Y ( x , y )

f X Y ( x , y )

Solution

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