Math Problem Statement

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.

Solution

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

Problem 1: Joint Probability Density Function

Given: f(x,y)=15(x+2y)for 0<x<1 and 0<y<2f(x, y) = \frac{1}{5} (x + 2y) \quad \text{for } 0 < x < 1 \text{ and } 0 < y < 2 f(x,y)=0elsewheref(x, y) = 0 \quad \text{elsewhere}

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

To find E(xy)E(xy), we use: E(xy)=0102xyf(x,y)dydxE(xy) = \int_{0}^{1} \int_{0}^{2} x y f(x, y) \, dy \, dx Substitute f(x,y)=15(x+2y)f(x, y) = \frac{1}{5}(x + 2y): E(xy)=0102xy15(x+2y)dydxE(xy) = \int_{0}^{1} \int_{0}^{2} x y \cdot \frac{1}{5} (x + 2y) \, dy \, dx

Steps:

  1. Expand the integrand xy15(x+2y)x y \cdot \frac{1}{5}(x + 2y).
  2. Integrate with respect to yy first, from 00 to 22.
  3. Integrate the result with respect to xx, from 00 to 11.

Problem 2: Joint Probability Mass Function

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

Part (a): Find P(x+y<2)P(x + y < 2)

This probability is the sum of p(x,y)p(x, y) values where x+y<2x + y < 2. From the table:

  • For (x=0,y=0):x+y=0(x = 0, y = 0): x + y = 0
  • For (x=0,y=1):x+y=1(x = 0, y = 1): x + y = 1
  • For (x=1,y=0):x+y=1(x = 1, y = 0): x + y = 1

So, P(x+y<2)=p(0,0)+p(0,1)+p(1,0)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=0.42P(x + y < 2) = 0.1 + 0.25 + 0.07 = 0.42

Part (b): Find the Marginal Probability Density of xx

The marginal probability of xx, p(x)p(x), is found by summing the probabilities across all values of yy for each xx:

  • For x=0x = 0: p(x=0)=p(0,0)+p(0,1)+p(0,2)+p(0,3)=0.1+0.25+0.2+0.1=0.65p(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=1x = 1: p(x=1)=p(1,0)+p(1,1)+p(1,2)+p(1,3)=0.07+0.0+0.15+0.13=0.35p(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 xx is: p(x=0)=0.65,p(x=1)=0.35p(x=0) = 0.65, \quad p(x=1) = 0.35

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


Related Questions

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

Tip

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

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Math Problem Analysis

Mathematical Concepts

Joint Probability Density Function
Expected Value
Marginal Probability

Formulas

Expected Value E(xy) = ∫∫ xy * f(x, y) dy dx
Marginal Probability p(x) = Σ p(x, y) over y

Theorems

Properties of Joint Probability Distributions
Marginal Probability Calculation

Suitable Grade Level

College Level - Probability and Statistics