1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Joint probability

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data

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


    (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.

    2. Relevant equations

    3. The attempt at a solution

    I have no idea how to this.
    i know E[XY]=E[X].E[Y] but dont know if it is gong to help here.

    can some one point me how to solve this
    Last edited: Oct 26, 2009
  2. jcsd
  3. Oct 26, 2009 #2


    User Avatar
    Homework Helper

    what are Z & W?
  4. Oct 26, 2009 #3
    sorry i have missed it

  5. Oct 26, 2009 #4


    User Avatar
    Homework Helper

    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
    [tex] f_W(w) = \int dx f_X(x) f_Y(w-x) [/tex]

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

    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
    substituting in for Z & Y should get the desired result
  6. Oct 26, 2009 #5
    well is there any thing we can do with given info like unifomely distributed in range of 0,2
  7. Oct 26, 2009 #6


    User Avatar
    Homework Helper

    yes, think area...
  8. Oct 26, 2009 #7
    well i get f(x)=1/2
    and f(y)=1/2

    but what is the use of them in here
  9. Oct 26, 2009 #8
    if X,Y independent the does Z and W independent too?
  10. Oct 26, 2009 #9


    User Avatar
    Homework Helper

    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 probabilty of finding (X,Y) in that area

    The line X=Y, represents Z=X-Y=0, and there will be cumulative probabilty F(z<0) = 1/2, the probabilty 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
  11. Oct 26, 2009 #10


    User Avatar
    Homework Helper

    i don't think so in this case

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

    however in this case looking at the areas, that doesn't appear so
  12. Oct 26, 2009 #11
    can i use marginal density find f(Z,W)

    f(z) = [tex]\int f(z,w) dw[/tex]
  13. Oct 26, 2009 #12


    User Avatar
    Homework Helper

    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
    [tex] P_W(W<w) = \frac{w^2}{2.4} [/tex]

    then differentiating gives
    [tex] f_W(W=w)dw = \frac{w}{4}dw [/tex]

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

    now can you find the conditional distribution
    [tex] P_{Z|W}(Z<z|W=w) = ? [/tex]
    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: Oct 26, 2009
  14. Oct 26, 2009 #13
    thanks I'll give it a try
  15. Oct 26, 2009 #14


    User Avatar
    Homework Helper

    the marginal probabilty as I understand it is the probabilty of W, regardless of Z, it is the same as the previously given [itex] f_W(W=w) [/itex]

    for further geometric insight compare [itex] f_W(W=w) [/itex] 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 [itex] f_{Z|W}(Z=z|W=w) [/itex]

    once again think of the length of lines

    The joint pdf will then be given by
    [tex] f_{W,Z}(W=w, Z=z) = f_{Z|W}(Z=z|W=w) f_W(W=w) [/tex]
    (see previous reference)
  16. Oct 26, 2009 #15
    thanks :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook