Math Problem Statement
The joint density function of X and Y is given by f(x, y) = {xe^(-x(y+1)), x > 0, y > 0; 0, otherwise}. Find the conditional probability density of X given Y = y, and that of Y given X = x. Find the probability density function of Z = XY.
Solution
Let's address the given question step by step.
Given Joint Density Function:
\begin{cases} xe^{-x(y+1)}, & x > 0, y > 0, \\ 0, & \text{otherwise}. \end{cases}$$ --- ### **(a) Conditional Probability Densities** #### 1. **Conditional PDF of $$X$$ given $$Y = y$$:** The conditional probability density function of $$X$$ given $$Y = y$$ is: $$f_{X|Y}(x|y) = \frac{f(x, y)}{f_Y(y)},$$ where $$f_Y(y)$$ is the marginal density of $$Y$$, found by integrating the joint density $$f(x, y)$$ over $$x$$: $$f_Y(y) = \int_0^\infty f(x, y) \, dx.$$ For $$x > 0$$ and $$y > 0$$, substitute $$f(x, y) = xe^{-x(y+1)}$$: $$f_Y(y) = \int_0^\infty xe^{-x(y+1)} \, dx.$$ Using the substitution $$z = x(y+1)$$, $$dz = (y+1)dx$$, we have: $$f_Y(y) = \int_0^\infty \frac{z}{y+1} e^{-z} \frac{1}{y+1} \, dz = \frac{1}{(y+1)^2} \int_0^\infty z e^{-z} \, dz.$$ The integral $$\int_0^\infty z e^{-z} \, dz = 1$$ (gamma function property), so: $$f_Y(y) = \frac{1}{(y+1)^2}, \quad y > 0.$$ Now, substitute back into the conditional density formula: $$f_{X|Y}(x|y) = \frac{xe^{-x(y+1)}}{1/(y+1)^2} = x(y+1)^2 e^{-x(y+1)}, \quad x > 0, y > 0.$$ --- #### 2. **Conditional PDF of $$Y$$ given $$X = x$$:** Similarly, the conditional probability density of $$Y$$ given $$X = x$$ is: $$f_{Y|X}(y|x) = \frac{f(x, y)}{f_X(x)},$$ where $$f_X(x)$$ is the marginal density of $$X$$, found by integrating the joint density $$f(x, y)$$ over $$y$$: $$f_X(x) = \int_0^\infty f(x, y) \, dy.$$ Substitute $$f(x, y) = xe^{-x(y+1)}$$: $$f_X(x) = x \int_0^\infty e^{-x(y+1)} \, dy.$$ Using the substitution $$z = x(y+1)$$, $$dz = x \, dy$$, we have: $$f_X(x) = x \int_x^\infty \frac{1}{x} e^{-z} \, dz = \int_0^\infty e^{-z} \, dz = 1.$$ Thus, $$f_X(x) = 1$$, and: $$f_{Y|X}(y|x) = \frac{xe^{-x(y+1)}}{1} = xe^{-x(y+1)}, \quad x > 0, y > 0.$$ --- ### **(b) PDF of $$Z = XY$$** To find the PDF of $$Z = XY$$, we use the transformation technique. 1. Define $$Z = XY$$ and $$W = X$$ (as a helper variable). Then, the transformation is: $$x = w, \quad y = \frac{z}{w}, \quad J = \left| \frac{\partial(x, y)}{\partial(z, w)} \right| = \frac{1}{w}.$$ 2. The joint density of $$Z$$ and $$W$$ is: $$f_{Z,W}(z, w) = f_{X,Y}\left(w, \frac{z}{w}\right) \cdot \left|\frac{1}{w}\right|.$$ Substitute $$f_{X,Y}(x, y) = xe^{-x(y+1)}$$: $$f_{Z,W}(z, w) = w e^{-w\left(\frac{z}{w} + 1\right)} \cdot \frac{1}{w} = e^{-z - w}, \quad z > 0, w > 0.$$ 3. The marginal density of $$Z$$ is found by integrating out $$W$$: $$f_Z(z) = \int_0^\infty f_{Z,W}(z, w) \, dw = \int_0^\infty e^{-z - w} \, dw.$$ $$f_Z(z) = e^{-z} \int_0^\infty e^{-w} \, dw = e^{-z}, \quad z > 0.$$ --- ### Final Answers: (a) - $$f_{X|Y}(x|y) = x(y+1)^2 e^{-x(y+1)}, \quad x > 0, y > 0.$$ - $$f_{Y|X}(y|x) = xe^{-x(y+1)}, \quad x > 0, y > 0.$$ (b) $$f_Z(z) = e^{-z}, \quad z > 0.$$ --- Would you like me to go into further detail on any step or concept? ### Related Questions: 1. How do marginal densities relate to conditional probabilities in multivariate distributions? 2. Can the PDF of $$Z = XY$$ be derived using a different transformation method? 3. What are the practical implications of finding $$f_{X|Y}(x|y)$$ in real-world problems? 4. How do the properties of exponential functions simplify joint density integration? 5. What are other methods for solving for the PDF of a transformed random variable? ### Tip: When working with joint densities, always verify that the integrals over the entire domain equal 1 to ensure proper normalization.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Theory
Joint Probability Densities
Conditional Probability Densities
Random Variable Transformations
Formulas
f_{X|Y}(x|y) = f(x, y) / f_Y(y)
f_{Y|X}(y|x) = f(x, y) / f_X(x)
f_Z(z) using transformation technique
Theorems
Marginalization for joint probability density functions
Gamma function for integral simplification
Suitable Grade Level
Undergraduate (Mathematics or Statistics)
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