Math Problem Statement
joint DF is given by FXY (x,y) = {(1-e^-ax) (1-e^-by) x,y>=0 and a,b>=0 and if 0 else
Determine if X and Y are dependent or independent. Also, what are P(X<=x, Y<Y)
- P(X>x, Y>y)
Solution
Given the joint distribution function as:
\begin{cases} (1 - e^{-ax})(1 - e^{-by}) & \text{if } x, y \geq 0 \text{ and } a, b \geq 0, \\ 0 & \text{otherwise}. \end{cases}$$ ### 1. Determine if $$X$$ and $$Y$$ are independent or dependent: Two random variables $$X$$ and $$Y$$ are independent if the joint cumulative distribution function (CDF) can be written as a product of the marginal CDFs: $$F_{XY}(x, y) = F_X(x) \cdot F_Y(y).$$ From the given CDF: $$F_{XY}(x, y) = (1 - e^{-ax})(1 - e^{-by}).$$ Now, let’s check the marginal CDFs: - The marginal CDF of $$X$$, $$F_X(x)$$, is obtained by setting $$y \to \infty$$ (since $$F_{XY}(x, \infty) = F_X(x)$$): $$F_X(x) = 1 - e^{-ax}.$$ - Similarly, the marginal CDF of $$Y$$, $$F_Y(y)$$, is obtained by setting $$x \to \infty$$: $$F_Y(y) = 1 - e^{-by}.$$ Thus, we can see that the joint CDF $$F_{XY}(x, y)$$ is indeed the product of the marginal CDFs: $$F_{XY}(x, y) = F_X(x) \cdot F_Y(y).$$ Therefore, $$X$$ and $$Y$$ are **independent**. --- ### 2. $$P(X \leq x, Y \leq y)$$: By definition, this is the joint CDF: $$P(X \leq x, Y \leq y) = F_{XY}(x, y) = (1 - e^{-ax})(1 - e^{-by}).$$ --- ### 3. $$P(X > x, Y > y)$$: This probability can be computed as: $$P(X > x, Y > y) = 1 - P(X \leq x \text{ or } Y \leq y).$$ Since $$X$$ and $$Y$$ are independent, we can express this as: $$P(X > x, Y > y) = P(X > x) \cdot P(Y > y).$$ From the marginal CDFs, we know: - $$P(X > x) = 1 - F_X(x) = e^{-ax}$$, - $$P(Y > y) = 1 - F_Y(y) = e^{-by}$$. Thus, $$P(X > x, Y > y) = e^{-ax} \cdot e^{-by} = e^{-ax - by}.$$ --- ### Summary: 1. $$X$$ and $$Y$$ are **independent** because their joint CDF is the product of their marginal CDFs. 2. $$P(X \leq x, Y \leq y) = (1 - e^{-ax})(1 - e^{-by})$$. 3. $$P(X > x, Y > y) = e^{-ax - by}$$. --- Would you like further details or any clarification? Here are 5 questions related to this: 1. How can you derive the marginal probability density functions (PDFs) for $$X$$ and $$Y$$? 2. What is the joint probability density function (PDF) of $$X$$ and $$Y$$? 3. How would the solution change if $$X$$ and $$Y$$ were dependent? 4. Can we compute $$P(X > x, Y \leq y)$$? 5. How does the value of parameters $$a$$ and $$b$$ influence the probabilities? **Tip**: For independent random variables, you can always express joint probabilities as the product of marginal probabilities.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Probability Theory
Joint Distribution Function
Independence of Random Variables
Formulas
F_{XY}(x, y) = (1 - e^{-ax})(1 - e^{-by})
P(X \leq x, Y \leq y) = F_{XY}(x, y) = (1 - e^{-ax})(1 - e^{-by})
P(X > x, Y > y) = e^{-ax - by}
Theorems
Independence of Random Variables (joint CDF as product of marginals)
Suitable Grade Level
Undergraduate (Probability and Statistics)
Related Recommendation
Solving Joint Probability Density Function Problems with Exponential Distributions
Finding P(X ≥ Y ≥ 2) Using Joint Exponential Distribution
Joint Probability Distribution and Independence: Solve for P(X < Y)
Joint Density Function and Conditional Expectation of X and Y
Joint PMF Analysis for Discrete Random Variables X and Y with Independence Check