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.