# Joint probability

1. Oct 26, 2009

### aruna1

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

Last edited: Oct 26, 2009
2. Oct 26, 2009

### lanedance

what are Z & W?

3. Oct 26, 2009

### aruna1

sorry i have missed it

Z=X-Y
W=X+Y

4. Oct 26, 2009

### 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

5. Oct 26, 2009

### aruna1

well is there any thing we can do with given info like unifomely distributed in range of 0,2

6. Oct 26, 2009

### lanedance

yes, think area...

7. Oct 26, 2009

### aruna1

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

but what is the use of them in here

8. Oct 26, 2009

### aruna1

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

9. Oct 26, 2009

### 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 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

10. Oct 26, 2009

### lanedance

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

11. Oct 26, 2009

### aruna1

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

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

12. Oct 26, 2009

### 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

Last edited: Oct 26, 2009
13. Oct 26, 2009

### aruna1

thanks I'll give it a try

14. Oct 26, 2009

### lanedance

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)

15. Oct 26, 2009

thanks :)