To find the conditional PDF \( f_{X|Y}(x | 0.5) \), we first need to determine the normalizing constant \( c \) for the joint PDF \( f_{X,Y}(x,y) = cx y \) in the specified region \( 0 \leq x \leq y \leq 1 \).
Step 1: Find the normalizing constant \( c \)
The joint PDF must integrate to 1 over the specified region. We need to set up the double integral: \[ \int_0^1 \int_0^y cxy , dx , dy \]
First integrate with respect to \( x \): \[ \int_0^y cxy , dx = c y \int_0^y x , dx = c y \left[ \frac{x^2}{2} \right]_0^y = c y \left( \frac{y^2}{2} \right) = \frac{c y^3}{2} \]
Now, integrate with respect to \( y \): \[ \int_0^1 \frac{c y^3}{2} , dy = \frac{c}{2} \left[ \frac{y^4}{4} \right]_0^1 = \frac{c}{2} \cdot \frac{1}{4} = \frac{c}{8} \]
Setting the integral equal to 1 for normalization: \[ \frac{c}{8} = 1 \implies c = 8 \]
Step 2: Determine the conditional PDF \( f_{X|Y}(x | 0.5) \)
We already have the joint PDF: \[ f_{X,Y}(x,y) = 8xy, \quad \text{for } 0 \leq x \leq y \leq 1. \]
Now we can find the marginal PDF \( f_Y(y) \): \[ f_Y(y) = \int_0^y f_{X,Y}(x,y) , dx = \int_0^y 8xy , dx = 8y \int_0^y x , dx = 8y \left[ \frac{x^2}{2} \right]_0^y = 8y \cdot \frac{y^2}{2} = 4y^3. \]
The marginal PDF \( f_Y(y) \) is valid for \( 0 \leq y \leq 1 \).
Now we find \( f_{X|Y}(x|y) \): \[ f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}. \]
For \( y = 0.5 \): \[ f_Y(0.5) = 4(0.5)^3 = 4 \cdot \frac{1}{8} = \frac{1}{2}. \]
Now compute \( f_{X,Y}(x, 0.5) \): \[ f_{X,Y}(x, 0.5) = 8x(0.5) = 4x, \quad \text{for } 0 \leq x \leq 0.5. \]
Thus, the conditional PDF \( f_{X|Y}(x | 0.5) \) becomes: \[ f_{X|Y}(x|0.5) = \frac{4x}{\frac{1}{2}} = 8x, \quad \text{for } 0 \leq x \leq 0.5. \]
Step 3: Form \( ax^b \)
Comparing with the form \( ax^b \): \[ f_{X|Y}(x|0.5) = 8x^1 \] gives \( a = 8 \) and \( b = 1 \).
Thus, the answers are: \[ \boxed{8} \quad \text{and} \quad \boxed{1}. \]