Consider the following joint PDF of 𝑋 and 𝑌 :
𝑓(𝑥, 𝑦) = 𝑐(𝑥 + 3𝑦) 0 < 𝑥 < 1, 0 < 𝑦 < 1
and zero otherwise. Find
(a) the value of 𝑐

To find the value of \( c \), we use the fact that the total probability of the joint probability density function (PDF) over the entire range of \( X \) and \( Y \) must equal 1. This means that we need to integrate the given joint PDF \( f(x, y) = c(x + 3y) \) over the region where it is defined, which is \( 0 < x < 1 \) and \( 0 < y < 1 \).

### Step 1: Set up the integral
The total probability is given by:

\[
\int_0^1 \int_0^1 f(x, y) \, dx \, dy = 1
\]

Substitute \( f(x, y) = c(x + 3y) \):

\[
\int_0^1 \int_0^1 c(x + 3y) \, dx \, dy = 1
\]

### Step 2: Solve the integral
First, we integrate with respect to \( x \):

\[
\int_0^1 c(x + 3y) \, dx = c \int_0^1 (x + 3y) \, dx
\]

Now, calculate the integral with respect to \( x \):

\[
\int_0^1 (x + 3y) \, dx = \left[ \frac{x^2}{2} + 3yx \right]_0^1 = \frac{1^2}{2} + 3y(1) - \left( \frac{0^2}{2} + 3y(0) \right)
\]

This simplifies to:

\[
\frac{1}{2} + 3y
\]

### Step 3: Integrate with respect to \( y \)
Now, integrate the result with respect to \( y \):

\[
\int_0^1 \left( \frac{1}{2} + 3y \right) \, dy = \left[ \frac{y}{2} + \frac{3y^2}{2} \right]_0^1
\]

Evaluating this:

\[
\frac{1}{2}(1) + \frac{3}{2}(1^2) - \left( \frac{1}{2}(0) + \frac{3}{2}(0^2) \right) = \frac{1}{2} + \frac{3}{2} = 2
\]

### Step 4: Solve for \( c \)
Now, substitute the result back into the equation:

\[
c \times 2 = 1
\]

Thus,

\[
c = \frac{1}{2}
\]

### Final Answer:
The value of \( c \) is \( \frac{1}{2} \).

---

Would you like to explore any specific details of the calculation or related concepts? Here are a few related questions:

1. What are the marginal distributions of \( X \) and \( Y \)?
2. How do you compute the expected value of \( X \) and \( Y \) given the joint PDF?
3. What is the covariance of \( X \) and \( Y \)?
4. How can we find the conditional PDF of \( X \) given \( Y = y \)?
5. How do you check if \( X \) and \( Y \) are independent?

**Tip:** When solving joint PDFs, remember that the integral of the entire PDF over the defined range must always equal 1.

Consider the following joint PDF of X and Y: f(x, y) = c(x + 3y) 0 < x < 1, 0 < y < 1 and zero otherwise. Find (a) the value of c.

Learn how to find the value of the normalizing constant c for the joint probability density function f(x, y) = c(x + 3y), where 0 < x < 1 and 0 < y < 1. This problem involves key concepts such as joint PDFs, integration, and normalization. Suitable for undergraduate students studying probability and statistics.

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