1. The probability distribution table for X and Y is presented below. Y\X 0 1 2 0 0.05 0.1 0.03 1 0.21 0.11 0.19 2 0.08 0.15 0.08 1-a) compute COV(X 2 , Y)

To compute \( \text{COV}(X^2, Y) \), we follow these steps:

### Step 1: Recall the formula for covariance
The covariance between two random variables \( X^2 \) and \( Y \) is given by:
\[
\text{COV}(X^2, Y) = E[X^2 Y] - E[X^2] E[Y]
\]
We need to compute three things:
1. \( E[X^2 Y] \) — the expected value of the product of \( X^2 \) and \( Y \),
2. \( E[X^2] \) — the expected value of \( X^2 \),
3. \( E[Y] \) — the expected value of \( Y \).

### Step 2: Calculate \( E[X^2 Y] \)

Using the joint probability distribution table, we compute \( E[X^2 Y] \) as:
\[
E[X^2 Y] = \sum_x \sum_y (x^2 \cdot y \cdot P(X=x, Y=y))
\]
We apply the table values.

#### For \( X = 0 \):
- \( (0^2 \cdot 0 \cdot P(0,0)) = 0^2 \cdot 0 \cdot 0.05 = 0 \)
- \( (0^2 \cdot 1 \cdot P(0,1)) = 0^2 \cdot 1 \cdot 0.1 = 0 \)
- \( (0^2 \cdot 2 \cdot P(0,2)) = 0^2 \cdot 2 \cdot 0.03 = 0 \)

#### For \( X = 1 \):
- \( (1^2 \cdot 0 \cdot P(1,0)) = 1^2 \cdot 0 \cdot 0.21 = 0 \)
- \( (1^2 \cdot 1 \cdot P(1,1)) = 1^2 \cdot 1 \cdot 0.11 = 0.11 \)
- \( (1^2 \cdot 2 \cdot P(1,2)) = 1^2 \cdot 2 \cdot 0.19 = 0.38 \)

#### For \( X = 2 \):
- \( (2^2 \cdot 0 \cdot P(2,0)) = 2^2 \cdot 0 \cdot 0.08 = 0 \)
- \( (2^2 \cdot 1 \cdot P(2,1)) = 2^2 \cdot 1 \cdot 0.15 = 4 \cdot 0.15 = 0.6 \)
- \( (2^2 \cdot 2 \cdot P(2,2)) = 2^2 \cdot 2 \cdot 0.08 = 4 \cdot 2 \cdot 0.08 = 0.64 \)

Now, summing all the terms:
\[
E[X^2 Y] = 0 + 0 + 0 + 0 + 0.11 + 0.38 + 0 + 0.6 + 0.64 = 1.73
\]

### Step 3: Calculate \( E[X^2] \)

Now, calculate the expected value of \( X^2 \):
\[
E[X^2] = \sum_x (x^2 \cdot P(X=x))
\]

#### For \( X = 0 \):
- \( 0^2 \cdot P(X=0) = 0^2 \cdot (0.05 + 0.1 + 0.03) = 0 \)

#### For \( X = 1 \):
- \( 1^2 \cdot P(X=1) = 1^2 \cdot (0.21 + 0.11 + 0.19) = 1 \cdot 0.51 = 0.51 \)

#### For \( X = 2 \):
- \( 2^2 \cdot P(X=2) = 2^2 \cdot (0.08 + 0.15 + 0.08) = 4 \cdot 0.31 = 1.24 \)

Now, summing these values:
\[
E[X^2] = 0 + 0.51 + 1.24 = 1.75
\]

### Step 4: Calculate \( E[Y] \)

Now, calculate the expected value of \( Y \):
\[
E[Y] = \sum_y (y \cdot P(Y=y))
\]

#### For \( Y = 0 \):
- \( 0 \cdot P(Y=0) = 0 \cdot (0.05 + 0.21 + 0.08) = 0 \)

#### For \( Y = 1 \):
- \( 1 \cdot P(Y=1) = 1 \cdot (0.1 + 0.11 + 0.15) = 1 \cdot 0.36 = 0.36 \)

#### For \( Y = 2 \):
- \( 2 \cdot P(Y=2) = 2 \cdot (0.03 + 0.19 + 0.08) = 2 \cdot 0.30 = 0.60 \)

Now, summing these values:
\[
E[Y] = 0 + 0.36 + 0.60 = 0.96
\]

### Step 5: Calculate the covariance

Now, substitute everything into the covariance formula:
\[
\text{COV}(X^2, Y) = E[X^2 Y] - E[X^2] E[Y] = 1.73 - (1.75 \cdot 0.96)
\]
\[
\text{COV}(X^2, Y) = 1.73 - 1.68 = 0.05
\]

Thus, the covariance \( \text{COV}(X^2, Y) \) is \( \boxed{0.05} \).

Would you like further clarification or details on any step?

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Here are 5 related questions for further practice:
1. How would you calculate \( E[XY] \) using the same table?
2. What is the variance of \( X \)?
3. How would you find \( \text{COV}(X, Y) \) from the same table?
4. What is the joint probability distribution function?
5. How can you find the marginal distribution of \( X \)?

**Tip:** Covariance is a measure of how two random variables change together. If it's positive, they tend to increase together, and if it's negative, one increases while the other decreases.

The probability distribution table for X and Y is presented below. Y\X 0 1 2 0 0.05 0.1 0.03 1 0.21 0.11 0.19 2 0.08 0.15 0.08 1-a) compute COV(X^2, Y)

This problem involves calculating the covariance COV(X^2, Y) using a probability distribution table. It covers important concepts such as expected values, covariance, and probability distributions. Suitable for students in Grades 11-12.

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