Math Problem Statement
Let X X and Y Y be continuous random variables with joint density
f X Y ( x , y ) f XY (x,y) { c x y for 0 < x < 2 , 1 < y < 3 0 otherwise { cxy 0
for 0<x<2,1<y<3 otherwise Calculate P ( ( X + Y ) < 3 ) P((X+Y)<3)
Solution
To calculate the probability , we need to integrate the joint probability density function (PDF) over the region where .
Step 1: Determine the Joint PDF
Given the joint density function:
\begin{cases} cxy & \text{for } 0 < x < 2 \text{ and } 1 < y < 3, \\ 0 & \text{otherwise}, \end{cases}$$ we first need to determine the constant $$c$$ by ensuring that the total probability over the entire valid range is 1. ### Step 2: Find the Constant $$c$$ We integrate the joint PDF over the entire valid region: $$\int_1^3 \int_0^2 cxy \, dx \, dy = 1.$$ First, integrate with respect to $$x$$: $$\int_0^2 cxy \, dx = cy \int_0^2 x \, dx = cy \left[\frac{x^2}{2}\right]_0^2 = cy \cdot 2 = 2cy.$$ Now, integrate with respect to $$y$$: $$\int_1^3 2cy \, dy = 2c \int_1^3 y \, dy = 2c \left[\frac{y^2}{2}\right]_1^3 = 2c \left(\frac{9}{2} - \frac{1}{2}\right) = 2c \cdot 4 = 8c.$$ Set this equal to 1: $$8c = 1 \quad \Rightarrow \quad c = \frac{1}{8}.$$ Thus, the joint PDF is: $$f_{XY}(x, y) = \begin{cases} \frac{1}{8}xy & \text{for } 0 < x < 2 \text{ and } 1 < y < 3, \\ 0 & \text{otherwise}. \end{cases}$$ ### Step 3: Calculate $$P((X + Y) < 3)$$ We now need to integrate $$f_{XY}(x, y)$$ over the region where $$X + Y < 3$$, with $$x$$ and $$y$$ within their respective ranges. This region is bounded by: - $$0 < x < 2$$ - $$1 < y < 3$$ - $$x + y < 3$$ These constraints define a triangular region. ### Step 4: Determine the Limits of Integration For $$y$$ in the region where $$X + Y < 3$$: - $$y > 1$$ - $$y < 3 - x$$ The integration limits are: $$\int_1^2 \int_1^{3-x} \frac{1}{8}xy \, dy \, dx.$$ ### Step 5: Evaluate the Integral 1. **Integrate with respect to $$y$$**: $$\int_1^{3-x} \frac{1}{8}xy \, dy = \frac{x}{8} \int_1^{3-x} y \, dy = \frac{x}{8} \left[\frac{y^2}{2}\right]_1^{3-x} = \frac{x}{8} \left(\frac{(3-x)^2}{2} - \frac{1}{2}\right).$$ Simplify: $$\frac{x}{8} \left(\frac{9 - 6x + x^2 - 1}{2}\right) = \frac{x}{8} \left(\frac{8 - 6x + x^2}{2}\right) = \frac{x(4 - 3x + \frac{x^2}{2})}{8}.$$ 2. **Integrate with respect to $$x$$**: $$\int_1^2 \frac{x(4 - 3x + \frac{x^2}{2})}{8} \, dx = \frac{1}{8} \int_1^2 x \left(4 - 3x + \frac{x^2}{2}\right) \, dx.$$ Distribute $$x$$ and integrate: $$\frac{1}{8} \left[\frac{4x^2}{2} - \frac{3x^3}{3} + \frac{x^4}{8}\right]_1^2 = \frac{1}{8} \left(8 - 8 + 2 - (2 - 1 + \frac{1}{8})\right).$$ ### Step 6: Final Probability Value Upon evaluating the integrals and simplifying, the probability $$P((X + Y) < 3)$$ will be: $$\frac{7}{48}.$$ Would you like further clarification on any of these steps or need additional details? **Relative Questions:** 1. What is the significance of the joint probability density function in probability theory? 2. How would the problem change if the limits for $$x$$ and $$y$$ were different? 3. Can you explain why the region where $$X + Y < 3$$ is triangular? 4. What happens to the joint PDF if $$c$$ was not calculated correctly? 5. How do you find the marginal distributions from a joint PDF? **Tip:** When integrating over a region defined by inequalities, carefully sketching the region can help visualize the limits of integration.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Theory
Joint Probability Density Function
Integration
Formulas
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Theorems
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Suitable Grade Level
Advanced Undergraduate Level
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