Question
Consider the same stick-breaking problem as in the previous clip, and let l = 1 Recall that f X, Y (x,y)=1/x when 0 <= y <= x <= 1
a) Conditioned on Y = 2/3 the conditional PDF of X is nonzero when a <= x <= b Find a and b
a =
b=
b) On the range found in part (a) the conditional PDF f X|Y (x| 2/3 ) of the form c * x ^ d for some constants c and d. Find d
d =
a) Conditioned on Y = 2/3 the conditional PDF of X is nonzero when a <= x <= b Find a and b
a =
b=
b) On the range found in part (a) the conditional PDF f X|Y (x| 2/3 ) of the form c * x ^ d for some constants c and d. Find d
d =
Answers
Answered by
GPT-4o mini
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 = \(-1\)
### 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 = \(-1\)
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