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Let's break down and solve each part of this problem.

### Problem 1: Joint Probability Density Function

Given:
\[
f(x, y) = \frac{1}{5} (x + 2y) \quad \text{for } 0 < x < 1 \text{ and } 0 < y < 2
\]
\[
f(x, y) = 0 \quad \text{elsewhere}
\]

We are asked to find the expected value \( E(xy) \).

To find \( E(xy) \), we use:
\[
E(xy) = \int_{0}^{1} \int_{0}^{2} x y f(x, y) \, dy \, dx
\]
Substitute \( f(x, y) = \frac{1}{5}(x + 2y) \):
\[
E(xy) = \int_{0}^{1} \int_{0}^{2} x y \cdot \frac{1}{5} (x + 2y) \, dy \, dx
\]

#### Steps:
1. Expand the integrand \( x y \cdot \frac{1}{5}(x + 2y) \).
2. Integrate with respect to \( y \) first, from \( 0 \) to \( 2 \).
3. Integrate the result with respect to \( x \), from \( 0 \) to \( 1 \).

### Problem 2: Joint Probability Mass Function

The table of the joint probability mass function \( p(x, y) \) is given for \( x \in \{0, 1\} \) and \( y \in \{0, 1, 2, 3\} \).

#### Part (a): Find \( P(x + y < 2) \)
This probability is the sum of \( p(x, y) \) values where \( x + y < 2 \). From the table:
- For \( (x = 0, y = 0): x + y = 0 \)
- For \( (x = 0, y = 1): x + y = 1 \)
- For \( (x = 1, y = 0): x + y = 1 \)

So,
\[
P(x + y < 2) = p(0, 0) + p(0, 1) + p(1, 0)
\]
Using the table values:
\[
P(x + y < 2) = 0.1 + 0.25 + 0.07 = 0.42
\]

#### Part (b): Find the Marginal Probability Density of \( x \)

The marginal probability of \( x \), \( p(x) \), is found by summing the probabilities across all values of \( y \) for each \( x \):
- For \( x = 0 \):
  \[
  p(x=0) = p(0, 0) + p(0, 1) + p(0, 2) + p(0, 3) = 0.1 + 0.25 + 0.2 + 0.1 = 0.65
  \]
- For \( x = 1 \):
  \[
  p(x=1) = p(1, 0) + p(1, 1) + p(1, 2) + p(1, 3) = 0.07 + 0.0 + 0.15 + 0.13 = 0.35
  \]

So, the marginal probability density of \( x \) is:
\[
p(x=0) = 0.65, \quad p(x=1) = 0.35
\]

Would you like further details on the integration steps for \( E(xy) \) or any specific part?

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### Related Questions
1. What is the marginal probability density of \( y \) in Problem 2?
2. How do we interpret the expected value \( E(xy) \) in a real-world context?
3. What conditions make a joint probability density function valid?
4. How would \( E(x) \) and \( E(y) \) be computed for the distribution in Problem 1?
5. In Problem 2, how would we calculate \( P(x + y = 3) \)?

### Tip
For joint distributions, remember that marginal probabilities are calculated by summing or integrating over the unwanted variable.

1. The joint probability density of two random variables is given by f(x, y) = (1/5)(x + 2y) for 0 < x < 1 and 0 < y < 2, and 0 elsewhere. Find the expected value of xy, which is E(xy).

2. The joint probability mass function is given by p(x, y) in a table for x = 0, 1 and y = 0, 1, 2, 3. Find (a) P(x + y < 2), (b) marginal probability density of x.

Solve problems involving joint probability density and mass functions. Calculate the expected value E(xy) for continuous variables and determine the marginal probability of x in a discrete distribution.

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