tuloskan kembali dg bahasa indonesia =Step 1: Identify the support of the joint density

The joint density is non-zero for the following pairs:

[ {(-2, 0), (-1, 0), (1, 0), (2, 0), (0, -2), (0, -1), (0, 1), (0, 2)} ]

Step 2: Calculate the expected values ( E[X] ) and ( E[Y] )

Given the uniform probability of ( \frac{1}{8} ) for each of the points, we can calculate the expected values as follows.

For ( E[X] ):

[ E[X] = \sum_{(x,y)} x \cdot f(x,y) = (-2) \cdot \frac{1}{8} + (-1) \cdot \frac{1}{8} + 1 \cdot \frac{1}{8} + 2 \cdot \frac{1}{8} + 0 \cdot \frac{1}{8} + 0 \cdot \frac{1}{8} + 0 \cdot \frac{1}{8} + 0 \cdot \frac{1}{8} ]

Calculating this gives:

[ E[X] = \frac{-2 - 1 + 1 + 2 + 0 + 0 + 0 + 0}{8} = \frac{0}{8} = 0 ]

For ( E[Y] ):

[ E[Y] = \sum_{(x,y)} y \cdot f(x,y) = 0 \cdot \frac{1}{8} + 0 \cdot \frac{1}{8} + (-1) \cdot \frac{1}{8} + 1 \cdot \frac{1}{8} + (-2) \cdot \frac{1}{8} + 2 \cdot \frac{1}{8} ]

Calculating this gives:

[ E[Y] = \frac{0 - 1 + 1 - 2 + 2}{8} = \frac{0}{8} = 0 ]

Step 3: Calculate ( E[XY] )

Next, we calculate:

[ E[XY] = \sum_{(x,y)} x \cdot y \cdot f(x,y) = (-2)(0) \cdot \frac{1}{8} + (-1)(0) \cdot \frac{1}{8} + (1)(0) \cdot \frac{1}{8} + (2)(0) \cdot \frac{1}{8} + (0)(-2) \cdot \frac{1}{8} + (0)(-1) \cdot \frac{1}{8} + (0)(1) \cdot \frac{1}{8} + (0)(2) \cdot \frac{1}{8} ]

Since all terms involve multiplication by zero, we have:

[ E[XY] = 0 ]

Step 4: Calculate the covariance ( \text{Cov}(X, Y) )

Using the formula for covariance:

[ \text{Cov}(X, Y) = E[XY] - E[X]E[Y] ]

Substituting the values we found:

[ \text{Cov}(X, Y) = 0 - 0 \cdot 0 = 0 ]

Step 5: Check for independence

Two random variables ( X ) and ( Y ) are independent if the joint probability density function factors into the product of the marginal densities:

[ f(x, y) = f_X(x) \cdot f_Y(y) ]

Calculating the marginal densities:

Marginal of ( X ):

[ fX(x) = \sum{y} f(x, y) = f(x, 0) + f(x, 2) + f(x, -2) + f(x, 1) + f(x, -1) ]

Since ( f(x, 0) ) is non-zero for ( x = -2, -1, 1, 2 ), we have:

For ( x = -2, -1, 1, 2 ):

[ f_X(-2) = f(-2, 0) = \frac{1}{8} ] [ f_X(-1) = f(-1, 0) = \frac{1}{8} ] [ f_X(1) = f(1, 0) = \frac{1}{8} ] [ f_X(2) = f(2, 0) = \frac{1}{8} ]

So, the marginal density ( f_X(x) ) is uniform over these values.

Marginal of ( Y ):

[ fY(y) = \sum{x} f(x, y) = f(0, y) + f(2, y) + f(-2, y) + f(1, y) + f(-1, y) ]

Again, since ( f(0, y) ) is non-zero for ( y = -2, -1, 1, 2 ), we can conclude similarly that the marginal densities will not be independent because the outcomes of ( X ) and ( Y ) are dependent on their defined values.

Conclusion

The covariance of ( X ) and ( Y ) is ( \text{Cov}(X, Y) = 0 ).

( X ) and ( Y ) are not independent.