Math Problem Statement

Solve

Find the probability density functions (pdfs) f_X(x), f_{Y|X}(y | 0.5), and f_{X|Y}(x | 0.5) over specified intervals, with errors indicated for incorrect solutions.

Solution

The problem involves finding the probability density functions (pdfs) fX(x)f_X(x), fYX(y0.5)f_{Y|X}(y | 0.5), and fXY(x0.5)f_{X|Y}(x | 0.5) over specified ranges. Let's address each question and clarify the correct approach:

1. Find fX(x)f_X(x):

We need to define fX(x)f_X(x) over three intervals:

  • 0<x10 < x \leq 1
  • 1<x<21 < x < 2
  • x<0x < 0 or x2x \geq 2

From the provided image:

  • For 0<x10 < x \leq 1, fX(x)=x2f_X(x) = \frac{x}{2}.
  • For 1<x<21 < x < 2, the answer was incorrect. We should identify the correct function over this interval, likely based on continuity or normalization conditions.
  • For x<0x < 0 or x2x \geq 2, 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 fYX(y0.5)f_{Y|X}(y | 0.5):

This represents the conditional density function of YY given X=0.5X = 0.5:

  • For 0<y<1/20 < y < 1/2, fYX(y0.5)=2f_{Y|X}(y | 0.5) = 2, which is correct.
  • For y<0y < 0 or y>1/2y > 1/2, 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 fXY(x0.5)f_{X|Y}(x | 0.5):

This is the conditional pdf of XX given Y=0.5Y = 0.5:

  • For 1/2<x<11/2 < x < 1, fXY(x0.5)=0.5f_{X|Y}(x | 0.5) = 0.5.
  • For 1<x<3/21 < x < 3/2, fXY(x0.5)=1.5f_{X|Y}(x | 0.5) = 1.5.
  • For x<1/2x < 1/2 or x>3/2x > 3/2, 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 XX and YY.

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:

  1. How do we normalize a probability density function over a given interval?
  2. What are the conditions for defining a piecewise pdf?
  3. How is the conditional pdf derived from joint distributions?
  4. What happens to the pdf outside its defined interval?
  5. 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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

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