1. The joint density function of the two continuous random variables X and Y is
𝑓𝑥,𝑦
(𝑥, 𝑦) = {
6𝑥, 0 < 𝑥 < 𝑦 < 1
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
1-a) Find 𝑓𝑥
(𝑥)
1-b) Find 𝐸(𝑦|𝑥)

We are given the joint probability density function (PDF) of two continuous random variables \( X \) and \( Y \):

\[
f_{X,Y}(x, y) = 
\begin{cases} 
6x & \text{if } 0 < x < y < 1, \\
0 & \text{otherwise}.
\end{cases}
\]

Let's solve the problems step by step:

### 1-a) Find the marginal density \( f_X(x) \)

To find the marginal density \( f_X(x) \), we integrate the joint density function \( f_{X,Y}(x, y) \) with respect to \( y \):

\[
f_X(x) = \int_{y=x}^{1} f_{X,Y}(x, y) \, dy
\]

Since the joint density is \( f_{X,Y}(x, y) = 6x \) for \( x < y < 1 \), we perform the integration:

\[
f_X(x) = \int_{x}^{1} 6x \, dy = 6x \int_{x}^{1} 1 \, dy = 6x(1 - x)
\]

Thus, the marginal density is:

\[
f_X(x) = 6x(1 - x), \quad \text{for } 0 < x < 1.
\]

### 1-b) Find the conditional expectation \( E(Y|X=x) \)

The conditional expectation \( E(Y|X=x) \) is the expected value of \( Y \), given \( X = x \). To find this, we use the conditional probability density function \( f_{Y|X}(y|x) \), which is defined as:

\[
f_{Y|X}(y|x) = \frac{f_{X,Y}(x, y)}{f_X(x)}.
\]

From part 1-a, we know that \( f_X(x) = 6x(1 - x) \), and the joint PDF is \( f_{X,Y}(x, y) = 6x \) for \( x < y < 1 \).

So, for \( x < y < 1 \):

\[
f_{Y|X}(y|x) = \frac{6x}{6x(1 - x)} = \frac{1}{1 - x}.
\]

Now, we compute the conditional expectation \( E(Y|X=x) \), which is the mean of the conditional distribution:

\[
E(Y|X=x) = \int_{y=x}^{1} y f_{Y|X}(y|x) \, dy.
\]

Substituting \( f_{Y|X}(y|x) = \frac{1}{1 - x} \):

\[
E(Y|X=x) = \int_{x}^{1} y \frac{1}{1 - x} \, dy = \frac{1}{1 - x} \int_{x}^{1} y \, dy.
\]

Now, calculate the integral of \( y \):

\[
\int_{x}^{1} y \, dy = \left[\frac{y^2}{2}\right]_x^1 = \frac{1}{2} - \frac{x^2}{2}.
\]

Thus, the conditional expectation becomes:

\[
E(Y|X=x) = \frac{1}{1 - x} \left( \frac{1}{2} - \frac{x^2}{2} \right) = \frac{1 - x^2}{2(1 - x)}.
\]

So, the conditional expectation \( E(Y|X=x) \) is:

\[
E(Y|X=x) = \frac{1 - x^2}{2(1 - x)}, \quad \text{for } 0 < x < 1.
\]

---

Would you like any further details or clarification? Here are some related questions you might explore:

1. How do we interpret the marginal distribution \( f_X(x) \)?
2. How would the joint density change if the region of integration were different?
3. What are the properties of conditional distributions in general?
4. How can we use the joint PDF to find \( E(X) \) and \( E(Y) \)?
5. What is the variance of \( Y \) given \( X = x \)?

**Tip:** When solving for conditional expectations, first ensure you have the correct conditional probability density function by dividing the joint PDF by the marginal PDF.

The joint density function of the two continuous random variables X and Y is f(x, y) = 6x, 0 < x < y < 1, and 0 otherwise. 1-a) Find f_X(x). 1-b) Find E(Y|X=x).

This problem explores the joint density function of two continuous random variables X and Y, focusing on calculating the marginal distribution f_X(x) and the conditional expectation E(Y|X=x). Detailed step-by-step solutions are provided, making it suitable for university-level statistics students.

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