Density Function for X-Y on [0,1]

Click For Summary

Homework Help Overview

The discussion revolves around finding the density function for the random variable X - Y, where X and Y are independent and uniformly distributed on the interval [0,1]. Participants are exploring the implications of the distributions of X and Y on the resulting density function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to clarify the distributions of X and Y before proceeding. There is mention of using the Jacobi method for calculating the probability density function (pdf) of X - Y. One participant attempts to derive the pdf through convolution and presents a piecewise function based on their calculations.

Discussion Status

The discussion is active, with participants providing hints and suggestions for approaching the problem. There is a mix of attempts to derive the solution and requests for clarification on the distributions involved. No consensus has been reached yet, and multiple interpretations of the problem are being explored.

Contextual Notes

Participants are working under the assumption that X and Y are uniformly distributed on [0,1], but there is uncertainty regarding the specific distributions and the implications for the density function. Some participants emphasize the need for showing work before receiving further assistance.

flybyme
Messages
20
Reaction score
0
hi..

Homework Statement


what's the density function for X-Y if X and Y are independent and continously distributed on [0,1]?
 
Physics news on Phys.org
hi,

you must show some work before getting help; what have you tried?
 
The answer of course depends on how X and Y are distributed. Are you given two specific distributions or do you want to calculate the answer in all its generality? See the Jacobi method to get a way of calculating the pdf of X-Y.
 
ok... here's my try...

X has the uniform distribution f_X(x) = 1. to get the distribution for Y: F_{-Y}(y) = P(-Y \leq y) = P(Y \geq -y) = 1 - F_Y(-y) \Rightarrow f_Y(y) = f_Y(-y) = -1

the formula for convulsion in this case is f_Z(z) = \int_\infty^\infty f_X(z-y)f_Y(y)dy.

combining this with f_Y(y) leads to f_Z(z) = -\int_0^1 f_X(z-y)dy

the integrand is zero if the condition 0 <= z-y <= 1 (z-1 <= y <= z) isn't fulfilled.

we get three cases:

1. if 0 <= z <= 1: f_Z(z) = -\int^z_0 dy = -z
2. if 1 < z <= 2: f_Z(z) = -\int^1_{z-1} dy = z - 2
3. if z < 0 or z > 2: f_Z(z) = 0

it seems correct to me, but I'm not sure..
 
here is my hint:

1. first define X - Y as Z

2. then graph X - Y <= Z on a graph

3. find the limits of integration

4. solve it and this gives the cumulative distribution function

5. take the derivative of the CDF with respect to Z
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 27 ·
Replies
27
Views
2K
Replies
7
Views
1K
Replies
2
Views
2K
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K