Conditional PDF Calculation

Question

The joint p.d.f of $(X, Y)$ is

24 y (1-x), & 0 \leq x \leq 1, 0 \leq y \leq x, \\ 0, & \text{otherwise}. \end{cases} $$ and then $ f(y|x) = ((\_\_\_\_\_\_\_\_\_))). $ A. $$ f(y|x) = \begin{cases} \frac{2y}{x^2}, & 0 \leq x < 1, 0 \leq y \leq x, \\ 0, & \text{otherwise}. \end{cases} $$ B. $$ f(y|x) = \begin{cases} \frac{2y}{x^2}, & 0 < x < 1, 0 \leq y \leq x, \\ 0, & \text{otherwise}. \end{cases} $$ C. $$ f(y|x) = \begin{cases} \frac{2y}{x^2}, & 0 < x \leq 1, 0 \leq y \leq x, \\ 0, & \text{otherwise}. \end{cases} $$ D. $$ f(y|x) = \begin{cases} \frac{2y}{x^2}, & 0 \leq x \leq 1, 0 \leq y \leq x, \\ 0, & \text{otherwise}. \end{cases} $$

确定边缘概率密度函数

首先，我要找到边缘概率密度函数 $f_X(x)$ 。通过对联合 pdf 求积分：

$f_X(x) = \int_{0}^{x} 24 y (1 - x) \, dy$

积分结果是：

$f_X(x) = 12 x^2 (1 - x)$

确认条件

我已经找到 $f(y|x) = \frac{2y}{x^2}$ 这个表达式。然后，我检查了选项，发现在 $0 < x < 1$ 这个条件下，所有选项都是相同的，因此正确的选项是 D。

思考，持续几秒

To find the conditional probability density function $f(y|x)$ , we start with the given joint probability density function (p.d.f):

f(x, y) = \begin{cases} 24 y (1-x), & 0 \leq x \leq 1, \ 0 \leq y \leq x, \\ 0, & \text{otherwise}. \end{cases}

The marginal p.d.f of $X$ is obtained by integrating the joint p.d.f over $y$ :

f_X(x) = \int_{0}^{x} 24 y (1 - x) \, dy = 24 (1 - x) \int_{0}^{x} y \, dy = 24 (1 - x) \left[ \frac{y^2}{2} \right]_0^x = 12 x^2 (1 - x).

The conditional p.d.f $f(y|x)$ is given by:

f(y|x) = \frac{f(x, y)}{f_X(x)} = \frac{24 y (1 - x)}{12 x^2 (1 - x)} = \frac{2y}{x^2}, \quad \text{for } 0 < x < 1 \text{ and } 0 \leq y \leq x.

Reviewing the options provided:

Option B correctly specifies the conditions $0 < x < 1$ and $0 \leq y \leq x$ , matching our derived expression for $f(y|x)$ .

Answer:
B.

f(y|x) = \begin{cases} \frac{2y}{x^2}, & 0 < x < 1, \ 0 \leq y \leq x, \\ 0, & \text{otherwise}. \end{cases}