Functions of two or more random variables

Click For Summary

Homework Help Overview

The problem involves two random variables, X1 and X2, each uniformly distributed on the interval [0, 1]. The objective is to find the probability density function (p.d.f.) of the sum Y = X1 + X2.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the necessity of knowing the dependence between X1 and X2, with a consensus leaning towards assuming independence. There are inquiries about the concept of isolines and how to visualize the regions defined by Y = X1 + X2 within the unit square.

Discussion Status

Some participants have offered guidance on visualizing the problem and understanding the regions of integration. There is ongoing exploration of the boundaries for different values of Y and the implications of these boundaries on the p.d.f. Participants are actively questioning and clarifying their understanding of the concepts involved.

Contextual Notes

There is mention of the importance of defining cumulative distribution functions (cdf) correctly and the potential confusion arising from different conventions. Participants express a need for further review of calculus concepts related to the problem.

TheMathNoob
Messages
189
Reaction score
4

Homework Statement


Supposethat X1and X2 are .random variables and that each of them has the uniform distribution on the interval [0, 1]. Find the p.d.f. of Y =X1+X2.

Homework Equations


Find cdf of Y and then the pdf

The Attempt at a Solution


the joint pdf would be f(x1,x2)= 1 0<x1<1 0<x2<1
0 otherwise
so I have to compute the prob that pr(Y<y)= pr(x1+x2<y). If we graph y=x1+x2 over the x2 , x1 axes, then we can say that x2=y-x1 and y would be like the shifting value of the line. The defined region is a square and the region of integration would the area defined by this line and the square. I don't understand how to set the boundaries and the integrals. They are confusing me a lot. I would accept an explanation on a scratch paper to see the big picture of it. I would appreciate if you explain how to approach those kind of exercises that involve two random variables using the definition and the logic of pr(x1+x2<y). The manual solution gives two different pdfs.
 
Physics news on Phys.org
Firstly, the problem cannot be done unless we know the dependence between X1 and X2. Almost certainly, the problem wishes you to assume they are independent, but it's very important to get into the habit of noting such things because before long you'll start to get problems where that is not the case, and everything changes. For now, let's assume that they are independent.

Think about the 'isolines' of Y: the subsets of the unit square on which Y is constant. Draw them on a diagram. Now draw the regions of that square where Y<a for any given a in the possible range of Y. With luck, that should give you enough intuition to work out the pdf.
 
andrewkirk said:
Firstly, the problem cannot be done unless we know the dependence between X1 and X2. Almost certainly, the problem wishes you to assume they are independent, but it's very important to get into the habit of noting such things because before long you'll start to get problems where that is not the case, and everything changes. For now, let's assume that they are independent.

Think about the 'isolines' of Y: the subsets of the unit square on which Y is constant. Draw them on a diagram. Now draw the regions of that square where Y<a for any given a in the possible range of Y. With luck, that should give you enough intuition to work out the pdf.
I should have stated that the the random variables are independent. Sorry I did not realize that you answered that. I will think about it
 
andrewkirk said:
Firstly, the problem cannot be done unless we know the dependence between X1 and X2. Almost certainly, the problem wishes you to assume they are independent, but it's very important to get into the habit of noting such things because before long you'll start to get problems where that is not the case, and everything changes. For now, let's assume that they are independent.

Think about the 'isolines' of Y: the subsets of the unit square on which Y is constant. Draw them on a diagram. Now draw the regions of that square where Y<a for any given a in the possible range of Y. With luck, that should give you enough intuition to work out the pdf.
Can you go over isolines a little bit more?
 
TheMathNoob said:
Can you go over isolines a little bit more?
Draw the unit square on a number plane with X1 being the horizontal axis and X2 the vertical. Now draw the line that has all the points where Y=1 (ie X1+X2=1). Then draw another one for Y=1.5 and another for Y=0.5. Those lines are called isolines because for any such line, the value of Y is the same everywhere on the line.
 
TheMathNoob said:

Homework Statement


Supposethat X1and X2 are .random variables and that each of them has the uniform distribution on the interval [0, 1]. Find the p.d.f. of Y =X1+X2.

Homework Equations


Find cdf of Y and then the pdf

The Attempt at a Solution


the joint pdf would be f(x1,x2)= 1 0<x1<1 0<x2<1
0 otherwise
so I have to compute the prob that pr(Y<y)= pr(x1+x2<y). If we graph y=x1+x2 over the x2 , x1 axes, then we can say that x2=y-x1 and y would be like the shifting value of the line. The defined region is a square and the region of integration would the area defined by this line and the square. I don't understand how to set the boundaries and the integrals. They are confusing me a lot. I would accept an explanation on a scratch paper to see the big picture of it. I would appreciate if you explain how to approach those kind of exercises that involve two random variables using the definition and the logic of pr(x1+x2<y). The manual solution gives two different pdfs.

Just so you know: cdf's are almost always defined (by modern convention) as ##P(Y \leq y)##, not ##P(Y < y)##. Every book I own adopts this convention, and it is important to pay attention to it (not to just sloppily write "##< y##" when you really mean "##\leq y##"). The reason for being careful is so that more general cases, such as mixed discrete-continuous random variables can be treated properly. Anyway, for ##y \leq 1## the region of relevance is that part of ##\{ x_1 + x_2 \leq y \}## lying in the unit square ##S = \{ (x_1,x_2)\, : \, 0 \leq x_1 \leq 1, 0 \leq x_2 \leq 1 \}##, and the probability (the cdf) will just be the area of that region.

Draw it out for yourself, but here are some hints.
(1) For ##y \leq 1##, what does the region look like? What is its shape? What are its boundaries? What is its area? You know how to draw the line ##x_1 + x_2 = y##; you know you must be on or below that line; and you know you must be inside the unit square. I see no reason why you cannot draw that.
(2) For ##1 < y \leq 2##, what does the region look like, what are its boundaries, and what is its area? Why, for ##y## between 1 and 2 is it easier to look at the complementary region ##\{ x_1 + x_2 > y \}##?
(3) For ##y < 0## what does the region look like? What about for ##y > 1##?
 
Ray Vickson said:
Just so you know: cdf's are almost always defined (by modern convention) as ##P(Y \leq y)##, not ##P(Y < y)##. Every book I own adopts this convention, and it is important to pay attention to it (not to just sloppily write "##< y##" when you really mean "##\leq y##"). The reason for being careful is so that more general cases, such as mixed discrete-continuous random variables can be treated properly. Anyway, for ##y \leq 1## the region of relevance is that part of ##\{ x_1 + x_2 \leq y \}## lying in the unit square ##S = \{ (x_1,x_2)\, : \, 0 \leq x_1 \leq 1, 0 \leq x_2 \leq 1 \}##, and the probability (the cdf) will just be the area of that region.

Draw it out for yourself, but here are some hints.
(1) For ##y \leq 1##, what does the region look like? What is its shape? What are its boundaries? What is its area? You know how to draw the line ##x_1 + x_2 = y##; you know you must be on or below that line; and you know you must be inside the unit square. I see no reason why you cannot draw that.
(2) For ##1 < y \leq 2##, what does the region look like, what are its boundaries, and what is its area? Why, for ##y## between 1 and 2 is it easier to look at the complementary region ##\{ x_1 + x_2 > y \}##?
(3) For ##y < 0## what does the region look like? What about for ##y > 1##?
Hi ray, how did you figure out the boundaries of y<=1 and 1<y<=2. I think I have to review calc 3 some more. If you have the stewart book, can you tell me where can I practice this?
 
TheMathNoob said:
Hi ray, how did you figure out the boundaries of y<=1 and 1<y<=2. I think I have to review calc 3 some more. If you have the stewart book, can you tell me where can I practice this?

Draw a picture and you will see why!

I don't have the Stewart book. You can find out lots of information by doing the 21st Century version of going to the library---namely, searching on-line. The internet is full of free sites that review all this material, ranging from tutorial pages to entire elementary calculus textbooks.
 
Ray Vickson said:
Draw a picture and you will see why!

I don't have the Stewart book. You can find out lots of information by doing the 21st Century version of going to the library---namely, searching on-line. The internet is full of free sites that review all this material, ranging from tutorial pages to entire elementary calculus textbooks.
Hi Ray, now I understand what is going on. If we assume the boundaries to be x1>0 and x2>0 then we will just have a single integral going from 0 to y. In the square region that we have, we just can do this until y=1 otherwise we would get bad areas or areas that don't belong to the region. x2=y-x1 keeps moving until y=2, so we have to find a way to integrate in the interval 1<=y<=2. I came up with the integral 1 - double integral(y-1 to 1 and y-x1 to 1) of 1 dx2 dx1. Is that right?. That's for the cdf
 
Last edited:

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K