Math Problem Statement

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.

Solution

Langkah 1: Identifikasi Dukungan dari Kepadatan Gabungan

Kepadatan gabungan bernilai bukan nol untuk pasangan berikut:

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

Langkah 2: Hitung Nilai Harapan (E[X]) dan (E[Y])

Diberikan probabilitas uniform sebesar 18\frac{1}{8} untuk setiap titik, kita dapat menghitung nilai harapan sebagai berikut.

Untuk (E[X]):

E[X]=(x,y)xf(x,y)=(2)18+(1)18+118+218+018+018+018+018E[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}

Menghitung ini menghasilkan:

E[X]=21+1+2+0+0+0+08=08=0E[X] = \frac{-2 - 1 + 1 + 2 + 0 + 0 + 0 + 0}{8} = \frac{0}{8} = 0

Untuk (E[Y]):

E[Y]=(x,y)yf(x,y)=018+018+(1)18+118+(2)18+218E[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}

Menghitung ini menghasilkan:

E[Y]=01+12+28=08=0E[Y] = \frac{0 - 1 + 1 - 2 + 2}{8} = \frac{0}{8} = 0

Langkah 3: Hitung (E[XY])

Selanjutnya, kita menghitung:

E[XY]=(x,y)xyf(x,y)=(2)(0)18+(1)(0)18+(1)(0)18+(2)(0)18+(0)(2)18+(0)(1)18+(0)(1)18+(0)(2)18E[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}

Karena semua hasil perkalian melibatkan nol, maka:

E[XY]=0E[XY] = 0

Langkah 4: Hitung Kovarians (Cov(X,Y)\text{Cov}(X, Y))

Menggunakan rumus kovarians:

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

Memasukkan nilai yang telah kita temukan:

Cov(X,Y)=000=0\text{Cov}(X, Y) = 0 - 0 \cdot 0 = 0

Langkah 5: Periksa Kemandirian

Dua variabel acak XX dan YY independen jika fungsi kepadatan probabilitas gabungan dapat difaktorkan menjadi hasil kali dari kepadatan marginal:

f(x,y)=fX(x)fY(y)f(x, y) = f_X(x) \cdot f_Y(y)

Menghitung kepadatan marginal:

Marginal dari (X):

fX(x)=yf(x,y)=f(x,0)+f(x,2)+f(x,2)+f(x,1)+f(x,1)f_X(x) = \sum_{y} f(x, y) = f(x, 0) + f(x, 2) + f(x, -2) + f(x, 1) + f(x, -1)

Karena f(x,0)f(x, 0) bukan nol untuk x=2,1,1,2x = -2, -1, 1, 2, kita memiliki:

Untuk x=2,1,1,2x = -2, -1, 1, 2:

fX(2)=f(2,0)=18,fX(1)=f(1,0)=18,fX(1)=f(1,0)=18,fX(2)=f(2,0)=18f_X(-2) = f(-2, 0) = \frac{1}{8}, \quad f_X(-1) = f(-1, 0) = \frac{1}{8}, \quad f_X(1) = f(1, 0) = \frac{1}{8}, \quad f_X(2) = f(2, 0) = \frac{1}{8}

Jadi, kepadatan marginal fX(x)f_X(x) adalah uniform pada nilai-nilai ini.

Marginal dari (Y):

fY(y)=xf(x,y)=f(0,y)+f(2,y)+f(2,y)+f(1,y)+f(1,y)f_Y(y) = \sum_{x} f(x, y) = f(0, y) + f(2, y) + f(-2, y) + f(1, y) + f(-1, y)

Sekali lagi, karena f(0,y)f(0, y) bukan nol untuk y=2,1,1,2y = -2, -1, 1, 2, kita dapat menyimpulkan bahwa hasil marginal tidak akan independen karena hasil dari XX dan YY bergantung pada nilai yang telah ditentukan.

Kesimpulan

Kovarians dari XX dan YY adalah Cov(X,Y)=0\text{Cov}(X, Y) = 0.

XX dan YY tidak independen.

Apakah Anda ingin rincian tambahan atau ada pertanyaan?

Pertanyaan terkait:

  1. Bagaimana cara menentukan dukungan untuk kepadatan gabungan lainnya?
  2. Mengapa nilai harapan bisa nol dalam kasus tertentu?
  3. Apa artinya jika kovarians bernilai nol?
  4. Bagaimana kita menentukan kemandirian dalam variabel acak lainnya?
  5. Bagaimana cara menghitung kepadatan marginal untuk variabel lain?

Tip: Ketika kovarians antara dua variabel acak nol, mereka tidak selalu independen, terutama dalam distribusi diskrit atau non-normal.

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Math Problem Analysis

Mathematical Concepts

Joint Density
Expected Value
Covariance
Independence of Random Variables

Formulas

E[X] = sum_{(x,y)} x * f(x,y)
E[Y] = sum_{(x,y)} y * f(x,y)
Cov(X, Y) = E[XY] - E[X]E[Y]

Theorems

Covariance indicates dependence if non-zero, though zero covariance does not imply independence.

Suitable Grade Level

Undergraduate Probability and Statistics