Joint Probability of Z & W: Homework Statement

  • Thread starter Thread starter aruna1
  • Start date Start date
  • Tags Tags
    Joint Probability
Click For Summary

Homework Help Overview

The problem involves finding the joint probability density function (pdf) of two random variables, Z and W, derived from independent and identically distributed random variables X and Y, which are uniformly distributed over the interval (0,2). The variables are defined as Z = X - Y and W = X + Y. The discussion also explores the relationship between Z and W, particularly focusing on their correlation and independence.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definitions of Z and W and their relationships to X and Y. There are attempts to express the pdf of W as a convolution and to derive the pdf of Z similarly. Some participants question the implications of independence and correlation between Z and W. Others explore geometric interpretations related to areas in the probability space.

Discussion Status

The discussion is ongoing, with participants providing various insights and approaches to the problem. Some have suggested methods for deriving the joint pdf and exploring conditional distributions, while others are still clarifying their understanding of the concepts involved. There is no explicit consensus yet, but several productive lines of reasoning are being explored.

Contextual Notes

Participants note the uniform distribution of X and Y over the interval (0,2) as a relevant constraint. There are also discussions about the implications of independence and the geometric interpretation of the problem, particularly regarding the areas in the probability space.

aruna1
Messages
110
Reaction score
0

Homework Statement



If and where X,Y are independent and identically distributed (i.i.d) random variables uniformly distributed over the interval (0,2).

Z=X-Y
W=X+Y

(a) Find the joint pdf of Z and W .
(b) Show that even though the random variables Z,W are uncorrelated they are not independent.

Homework Equations


The Attempt at a Solution



I have no idea how to this.
i know E[XY]=E[X].E[Y] but don't know if it is gong to help here.

can some one point me how to solve this
thanks
 
Last edited:
Physics news on Phys.org
what are Z & W?
 
lanedance said:
what are Z & W?
sorry i have missed it

Z=X-Y
W=X+Y
 
not too sure how to show the joint pdf, but as a start

but for W = X+Y, you can write the pdf of W as a convolution
f_W(w) = \int dx f_X(x) f_Y(w-x)

simlarly for Z = X-Y
f_Z(z) = \int dx f_X(x) f_Y(z+x)

the mean values will be
E[Z] = <Z> = <X-Y> = <X> - <Y>
E[W] = <W> = <X+Y> = <X> + <Y>

the convariance will be given by
E[(Z-<Z>)(W-<W>)]
substituting in for Z & Y should get the desired result
 
lanedance said:
not too sure how to show the joint pdf, but as a start

but for W = X+Y, you can write the pdf of W as a convolution
f_W(w) = \int dx f_X(x) f_Y(w-x)

simlarly for Z = X-Y
f_Z(z) = \int dx f_X(x) f_Y(z+x)

the mean values will be
E[Z] = <Z> = <X-Y> = <X> - <Y>
E[W] = <W> = <X+Y> = <X> + <Y>

the convariance will be given by
E[(Z-<Z>)(W-<W>)]
substituting in for Z & Y should get the desired result

well is there any thing we can do with given info like unifomely distributed in range of 0,2
 
yes, think area...
 
lanedance said:
yes, think area...

well i get f(x)=1/2
and f(y)=1/2

but what is the use of them in here
 
if X,Y independent the does Z and W independent too?
 
now i haven't totally put it together yet, but what i meant in terms of area, consider the poistive quandrant in the plane bounded by X,Y from 0 to 2. Each section of area is directly proportional to the probability of finding (X,Y) in that area

The line X=Y, represents Z=X-Y=0, and there will be cumulative probability F(z<0) = 1/2, the probability density of z=Z is a maximum here, and decreases linearly to zero at Z=-2, Z=2

Similary consider the line from (0,2) to (2,0), that represents W = 2, and W can take values from 0 to 4.

also from a quick google http://en.wikipedia.org/wiki/Conditional_distribution. if you can find the conditional distribution, you're pretty much there
 
  • #10
aruna1 said:
if X,Y independent the does Z and W independent too?

i don't think so in this case

if two variables are independent then the conditional probability reduces to the independent probability,
P(A|B) = P(A)

however in this case looking at the areas, that doesn't appear so
 
  • #11
can i use marginal density find f(Z,W)

f(z) = \int f(z,w) dw
 
  • #12
now consider first w<2, the cumulative distribution will be the area bounded by the triangle of side w, divided by2x2=4 to normalise
for w<2
P_W(W&lt;w) = \frac{w^2}{2.4}

then differentiating gives
f_W(W=w)dw = \frac{w}{4}dw

by the symmetry for w>2
f_W(W=w)dw = \frac{1-w}{4}dw

now can you find the conditional distribution
P_{Z|W}(Z&lt;z|W=w) = ?
in the geomoetric picture, i think you might be considering lengths of lines...

as soon as you can show the probability density is not independent of W, you have shown they are not independent variables

now that I think of it though, you can probably do it by substituting into the original integrals, though its not as intuitive... but actually it looks tricky
 
Last edited:
  • #13
thanks I'll give it a try
 
  • #14
aruna1 said:
can i use marginal density find f(Z,W)

f(z) = \int f(z,w) dw

the marginal probability as I understand it is the probability of W, regardless of Z, it is the same as the previously given f_W(W=w)

for further geometric insight compare f_W(W=w) with the length of the line, given by W=X+Y = w

now what is the conditional probability of Z=z, given W=w, written as f_{Z|W}(Z=z|W=w)

once again think of the length of lines

The joint pdf will then be given by
f_{W,Z}(W=w, Z=z) = f_{Z|W}(Z=z|W=w) f_W(W=w)
(see previous reference)
 
  • #15
thanks :)
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 43 ·
2
Replies
43
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K