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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).
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.
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
thanks
lanedance
Oct26-09, 08:51 AM
what are Z & W?
what are Z & W?
sorry i have missed it
Z=X-Y
W=X+Y
lanedance
Oct26-09, 09:37 AM
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
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
lanedance
Oct26-09, 09:45 AM
yes, think area...
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?
lanedance
Oct26-09, 09:59 AM
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
lanedance
Oct26-09, 10:03 AM
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 probabilty reduces to the independent probabilty,
P(A|B) = P(A)
however in this case looking at the areas, that doesn't appear so
can i use marginal density find f(Z,W)
f(z) = \int f(z,w) dw
lanedance
Oct26-09, 10:11 AM
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<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<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
thanks I'll give it a try
lanedance
Oct26-09, 10:36 AM
can i use marginal density find f(Z,W)
f(z) = \int f(z,w) dw
the marginal probabilty as I understand it is the probabilty 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)
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