Math Problem Statement
Solve
Solution
The problem involves finding the probability density functions (pdfs) , , and over specified ranges. Let's address each question and clarify the correct approach:
1. Find :
We need to define over three intervals:
- or
From the provided image:
- For , .
- For , the answer was incorrect. We should identify the correct function over this interval, likely based on continuity or normalization conditions.
- For or , the answer provided was also incorrect, meaning it might involve setting the pdf to zero or specifying an alternative value for the non-overlapping regions.
2. Find :
This represents the conditional density function of given :
- For , , which is correct.
- For or , the conditional pdf might be zero outside the defined range, but 1 was marked as incorrect, indicating a different relationship or condition that needs to be considered.
3. Find :
This is the conditional pdf of given :
- For , .
- For , .
- For or , the provided answer of 0.75 was incorrect, suggesting a zero value outside this range or another specific density function.
To finalize these functions, it would be helpful to have the context of the problem or additional information like normalization conditions or the nature of the joint distribution between and .
Would you like to go through these in detail, or focus on a specific part? Here are some related questions that might help deepen understanding:
- How do we normalize a probability density function over a given interval?
- What are the conditions for defining a piecewise pdf?
- How is the conditional pdf derived from joint distributions?
- What happens to the pdf outside its defined interval?
- How can we verify if a pdf is valid (non-negative and integrates to 1)?
Tip: When solving piecewise pdf problems, always check for continuity at the interval boundaries and ensure the entire function integrates to 1 over its defined range.
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Math Problem Analysis
Mathematical Concepts
Probability Density Functions
Conditional Probability
Piecewise Functions
Formulas
f_X(x) = x/2 for 0 < x <= 1
f_{Y|X}(y | 0.5) = 2 for 0 < y < 1/2
f_{X|Y}(x | 0.5) = 0.5 for 1/2 < x < 1
Theorems
Basic PDF Properties
Conditional Probability Formula
Suitable Grade Level
Grades 11-12 or Early University Level
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