To solve this problem, we first need to analyze the given joint PDF and its implications regarding the stick-breaking problem.
Part (a)
Given the joint probability density function (PDF):
\[ f_{X, Y}(x, y) = \frac{1}{x} \quad \text{for } 0 \leq y \leq x \leq 1 \]
we want to find the conditional PDF of \(X\) given \(Y = \frac{2}{3}\).
Step 1: Determine the support for \(Y = \frac{2}{3}\)
From the condition \(0 \leq Y \leq X\), we can substitute \(Y = \frac{2}{3}\):
\[ 0 \leq \frac{2}{3} \leq x \leq 1 \]
Thus, the constraints on \(X\) when conditioned on \(Y = \frac{2}{3}\) are:
\[ \frac{2}{3} \leq x \leq 1 \]
Hence, we can identify:
\[ a = \frac{2}{3}, \quad b = 1 \]
Part (b)
To find the conditional PDF \(f_{X|Y}(x | \frac{2}{3})\):
Step 1: Compute the marginal PDF \(f_Y(y)\)
To find \(f_{Y}(y)\), we need to integrate the joint PDF over \(x\):
\[ f_Y(y) = \int_{y}^{1} f_{X,Y}(x,y) , dx = \int_{y}^{1} \frac{1}{x} , dx \]
Calculating the integral:
\[ f_Y(y) = \left[ \ln(x) \right]_{y}^{1} = \ln(1) - \ln(y) = -\ln(y) \]
Thus, the marginal PDF \(f_Y(y)\) is:
\[ f_Y(y) = -\ln(y) \quad \text{for } 0 < y \leq 1 \]
Step 2: Compute the conditional PDF \(f_{X|Y}(x | \frac{2}{3})\)
Using the definition of conditional PDFs:
\[ f_{X|Y}(x | y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} \]
Substituting \(y = \frac{2}{3}\):
\[ f_{X | Y}(x | \frac{2}{3}) = \frac{f_{X,Y}(x, \frac{2}{3})}{f_Y(\frac{2}{3})} = \frac{\frac{1}{x}}{-\ln(\frac{2}{3})} \]
Therefore, for \(\frac{2}{3} \leq x \leq 1\):
\[ f_{X|Y}(x | \frac{2}{3}) = \frac{1}{-\ln(\frac{2}{3})} \cdot \frac{1}{x} \]
Step 3: Identify the form of \(f_{X|Y}(x | \frac{2}{3})\)
Now we want to express this as \(c \cdot x^d\). Comparing:
\[ f_{X|Y}(x | \frac{2}{3}) = \frac{1}{-\ln(\frac{2}{3})} \cdot x^{-1} \]
Here we consider \(c = \frac{1}{-\ln(\frac{2}{3})}\) and we see:
\[ d = -1 \]
Final Answers
Thus, the final answers for the problem are:
a = \(\frac{2}{3}\)
b = \(1\)
d = \(-1\)