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Chapter 13: Problem 45

Suppose that the random variables \((X, Y)\) have joint PDF \(f(x, y)=\left\\{\begin{array}{ll}\frac{3}{256}\left(x^{2}+y^{2}\right), & \text { if } 0 \leq x \leq y ; 0 \leq y \leq 4 \\ 0, & \text { otherwise }\end{array}\right.\) Find each of the following: (a) \(P(X>2)\) (b) \(P(X+Y \leq 4)\) (c) \(E(X+Y)\)

Short Answer

Expert verified
Solve the integrals in each step for probabilities and expectation.

Step by step solution

01

Set Up the Problem

We need to find three probabilities or expected values (a), (b), and (c) based on given joint PDF of \((X, Y)\) with specific conditions on \(x\) and \(y\). The given joint PDF is non-zero only in the region \(0 \leq x \leq y \leq 4\). Our tasks include finding \(P(X > 2)\), \(P(X + Y \leq 4)\), and the expectation \(E(X + Y)\).
02

Solve (a) - Calculate P(X > 2)

To find \(P(X > 2)\), we integrate the joint PDF over the region where \(x > 2\) and \(0 \leq x \leq y \leq 4\).- The limits of integration for \(x\) are from 2 to \(y\).- The limits for \(y\) are from 2 to 4.\[P(X > 2) = \int_{2}^{4} \int_{2}^{y} \frac{3}{256} (x^2 + y^2) \, dx \, dy\]- Integrate relative to \(x\):\[= \int_{2}^{4} \left[ \frac{3}{256} \left( \frac{x^3}{3} + y^2x \right) \right]_{x = 2}^{x = y} \, dy\]- Substitute the limits:\[= \int_{2}^{4} \left[ \frac{3}{256} \left( \frac{y^3}{3} + y^3 - \frac{8}{3} - 4y^2 \right) \right] \, dy\]- Simplify and integrate over \(y\).This yields the probability for event (a).
03

Solve (b) - Calculate P(X + Y 鈮� 4)

To find \(P(X + Y \leq 4)\), we integrate the joint PDF over the required region \(0 \leq x \leq y\) and \(x + y \leq 4\).- From the constraints, the region is bound by \(x + y = 4\).- The integration limits for \(x\) are from 0 to \(y\), while for \(y\) they are from 0 to 4. For the upper limit, since \(x + y \leq 4\), \(x\) goes up to \(4-y\).\[P(X + Y \leq 4) = \int_{0}^{4} \int_{0}^{\min(y, 4-y)} \frac{3}{256} (x^2 + y^2) \, dx \, dy\]Solving this integral gives the probability for event (b).
04

Solve (c) - Calculate E(X + Y)

The expectation \(E(X+Y)\) can be found using the expected value formula for continuous random variables:\[E(X+Y) = \int \int (x+y) f(x, y) \, dx \, dy\]- Given the joint PDF is non-zero for \(0 \leq x \leq y \leq 4\), the integration limits for \(x\) are from 0 to \(y\) and for \(y\) are from 0 to 4.\[E(X+Y) = \int_{0}^{4} \int_{0}^{y} (x+y) \frac{3}{256} (x^2 + y^2) \, dx \, dy\]- Solve the integral first over \(x\) and then over \(y\) to get the expectation value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Probability in the context of joint probability distributions allows us to understand the likelihood of events involving two or more random variables. Here, we are discovering the probabilities for specific conditions where the variables, \(X\) and \(Y\), have a given joint probability density function (PDF).

To find \(P(X > 2)\), we look at the probability that \(X\) is greater than 2 given the joint PDF.
  • Identify the region where the joint PDF is non-zero: \(0 \leq x \leq y \leq 4\).
  • Set up integral limits: for \(x\), it's from 2 to \(y\), and for \(y\), it's from 2 to 4.
  • Integrate the joint PDF over this region.
Through this process, we determine the probability of \(X\) being greater than 2 within the specified bounds.
Expectation
Expectation gives us the average or mean value involving random variables under their probability distribution. For joint distributions, expectation considers both variables together.

For \(E(X + Y)\), we follow these steps:
  • Use the formula \(E(X+Y) = \int \int (x+y) f(x, y) \, dx \, dy\).
  • Identify the integration region where the joint PDF is non-zero: \(0 \leq x \leq y \leq 4\).
  • Integrate first with respect to \(x\) and then with respect to \(y\).
This process calculates the expected sum of the random variables \(X\) and \(Y\), showing how their joint behavior translates into an average outcome.
Integration
Integration is a mathematical process used to calculate areas under a curve, and in probability, it helps find both probabilities and expectations. In the case of a joint PDF, double integration is often needed as we deal with two variables.

When dealing with \(P(X + Y \leq 4)\), integration helps us find the probability where the sum of \(X\) and \(Y\) is less than or equal to 4.
  • The region of integration stems from constraints: \(0 \leq x \leq y\) and \(x + y \leq 4\).
  • The limits for \(x\) are from 0 to either \(y\) or \(4-y\).
  • Integrate over these limits to find the desired probability.
This illustrates how integration is a crucial tool in deriving insights from a joint probability distribution.
Random Variables
Random variables \((X, Y)\) are functions that assign a real number to each outcome in a probability space. In this context, \((X, Y)\) have a joint probability that shows how they relate to each other in a probabilistic sense.

The joint PDF for \((X, Y)\):
\(f(x, y) = \frac{3}{256}(x^2 + y^2)\) within specified limits is their probabilistic description.
  • The space where \(0 \leq x \leq y \leq 4\) is where the joint PDF is non-zero and meaningful.
  • This bounds the behavior of \(X\) and \(Y\), indicating the valid values they can take.
Understanding random variables in this manner allows us to utilize their joint probability to find probabilities and expectations, shedding light on their intertwined nature.

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