Joint Probability of Z & W: Homework Statement

[aruna1]

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

 

[lanedance]

what are Z & W?

 

[aruna1]

what are Z & W?

sorry i have missed it

Z=X-Y
W=X+Y

 

[lanedance]

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

 

[aruna1]

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]

yes, think area...

 

[aruna1]

yes, think area...

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

but what is the use of them in here

 

[aruna1]

if X,Y independent the does Z and W independent too?

 

[lanedance]

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

 

[lanedance]

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

 

[aruna1]

can i use marginal density find f(Z,W)

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

 

[lanedance]

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

 

[aruna1]

thanks I'll give it a try

 

[lanedance]

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)

 

[aruna1]

thanks :)