To solve part (f) of question 7, we need to show that the conditional variance $\text{Var}(Y \mid X = x)$ is given by

$\text{Var}(Y \mid X = x) = \frac{x^2 - 2x + 1}{12}, \quad \text{for } 0 < x < 1.$

Step-by-Step Solution:

1. Recall the formula for conditional variance:

[ \text{Var}(Y \mid X = x) = \mathbb{E}[Y^2 \mid X = x] - (\mathbb{E}[Y \mid X = x])^2. ]

2. Determine the conditional PDF $f_{Y|X}(y \mid x)$:

The joint PDF is: $f_{X,Y}(x, y) = c, \quad \text{for } 0 < x < y < 1.$ The marginal PDF of $X$ is required to normalize $f_{X,Y}$. From part (c), the marginal PDF of $X$ is: $f_X(x) = \int_x^1 c \, dy = c(1 - x).$ Thus, the conditional PDF is: $f_{Y|X}(y \mid x) = \frac{f_{X,Y}(x, y)}{f_X(x)} = \frac{c}{c(1 - x)} = \frac{1}{1 - x}, \quad \text{for } x < y < 1.$

3. Compute $\mathbb{E}[Y \mid X = x]$:

The expectation is: $\mathbb{E}[Y \mid X = x] = \int_x^1 y f_{Y|X}(y \mid x) \, dy = \int_x^1 y \frac{1}{1 - x} \, dy.$ Factoring out $1/(1-x)$: $\mathbb{E}[Y \mid X = x] = \frac{1}{1 - x} \int_x^1 y \, dy.$ The integral of $y$ is: $\int_x^1 y \, dy = \left[\frac{y^2}{2}\right]_x^1 = \frac{1^2}{2} - \frac{x^2}{2} = \frac{1 - x^2}{2}.$ Thus: $\mathbb{E}[Y \mid X = x] = \frac{1}{1 - x} \cdot \frac{1 - x^2}{2} = \frac{1 + x}{2}.$

4. Compute $\mathbb{E}[Y^2 \mid X = x]$:

The expectation is: $\mathbb{E}[Y^2 \mid X = x] = \int_x^1 y^2 f_{Y|X}(y \mid x) \, dy = \int_x^1 y^2 \frac{1}{1 - x} \, dy.$ Factoring out $1/(1-x)$: $\mathbb{E}[Y^2 \mid X = x] = \frac{1}{1 - x} \int_x^1 y^2 \, dy.$ The integral of $y^2$ is: $\int_x^1 y^2 \, dy = \left[\frac{y^3}{3}\right]_x^1 = \frac{1^3}{3} - \frac{x^3}{3} = \frac{1 - x^3}{3}.$ Thus: $\mathbb{E}[Y^2 \mid X = x] = \frac{1}{1 - x} \cdot \frac{1 - x^3}{3} = \frac{1 + x + x^2}{3}.$

5. Substitute into the variance formula:

[ \text{Var}(Y \mid X = x) = \mathbb{E}[Y^2 \mid X = x] - (\mathbb{E}[Y \mid X = x])^2. ] Substitute $\mathbb{E}[Y \mid X = x] = \frac{1 + x}{2}$ and $\mathbb{E}[Y^2 \mid X = x] = \frac{1 + x + x^2}{3}$: $\text{Var}(Y \mid X = x) = \frac{1 + x + x^2}{3} - \left(\frac{1 + x}{2}\right)^2.$ Simplify $\left(\frac{1 + x}{2}\right)^2$: $\left(\frac{1 + x}{2}\right)^2 = \frac{(1 + x)^2}{4} = \frac{1 + 2x + x^2}{4}.$ Thus: $\text{Var}(Y \mid X = x) = \frac{1 + x + x^2}{3} - \frac{1 + 2x + x^2}{4}.$ Find a common denominator (12): $\text{Var}(Y \mid X = x) = \frac{4(1 + x + x^2)}{12} - \frac{3(1 + 2x + x^2)}{12}.$ Simplify: $\text{Var}(Y \mid X = x) = \frac{4 + 4x + 4x^2 - 3 - 6x - 3x^2}{12} = \frac{x^2 - 2x + 1}{12}.$

Final Answer:

$\text{Var}(Y \mid X = x) = \frac{x^2 - 2x + 1}{12}.$

Would you like detailed explanations for any specific part of this solution?

Related Questions:

1. How is conditional variance used in statistics, and why is it significant in this problem?
2. What is the geometric interpretation of the conditional variance?
3. Can you explain the derivation of the marginal distribution $f_X(x)$ in more detail?
4. How does independence between $X$ and $Y$ relate to the variance in this problem?
5. What are practical examples where conditional variance calculations are important?

Tip:

When working with conditional distributions, always ensure that the marginal distributions are properly normalized to maintain consistency.