Cumulative distribution function

[D H]

You did not restrict x to [0,1] in that plot. What happens to z when x is outside of [0,1] is irrelevant.

 

[ulissess]

my range is (0.5; 1], if x $$\in$$ [0,1]

 

[D H]

No, it isn't. Look at the plot. Look at the equation.

 

[ulissess]

if x=1 ==> z=3 so exist a point of the function in x $$\in$$ [0,1]... or z $$\in$$ [0,1] too? If z and x $$\in$$ [0,1] doesn't exist the function in this interval

 

[vela]

When x=1, you have z=3. As x decreases toward 0.5, the graph goes up, which means z increases. There's a vertical asymptote at x=0.5, so as you get closer and closer to x=0.5, z grows without bound, right? So you know z takes on some value in [3,∞) when x is in (0.5,1]. What values of z do you get when x is in [0,0.5)?

 

[ulissess]

oh yes, sorry ... (-∞,-1] when x is in [0,0.5).. so my z has values in (-∞, -1] U [z,∞) ... z=1 if y=0 .
So i can integrate my PDF in those intervals, in (-1,1) instead my PDF is zero because i haven't my function in that interval. All right?
Slowly I'm coming near the solution, thank you !

 

[D H]

Finally! Very good.

 

[ulissess]

thank you! i have tried to do the integrals...

[PLAIN]http://img51.imageshack.us/img51/4570/97869363.jpg

in [1,+oo] my PDF change? why? have i used good the integrals?

 

[D H]

No, you haven't.

You need to show your work, and please explain your variables and your steps when you do so. If all you show are your final results it is a bit tough to figure out where you went wrong.

 

[ulissess]

yes sorry.. i was thinking this...( 2w/(z-1)^2 is my jacobian)

[PLAIN]http://img600.imageshack.us/img600/79/26405949.jpg

so this is my PDF but now i don't know what interval use for the CDF

 

[D H]

You still have something very wrong there. Look back to your post #14, where you essentially got it right for z<-1. It should be obvious what the pdf looks like from z=-1 to z=1.

The only remaining issue is what the pdf looks like for z>1. Hint: Is there any symmetry that you can use here?

 

[ulissess]

[PLAIN]http://img121.imageshack.us/img121/5523/84830083.jpg
so maybe this is correct, but i don't understand any things. I've integrated f(z) in dw=dy (z is constant): y is in (0,1) so i have integrated in this interval, the second f(z) I've integrated in dw but i have limited x in (0,1) and i have seen how vary the y. right? How do i understand why the second f(z) is used in z>1 and -1<z<1 and the first in z<-1? I believed the pdf in -1<z<1 was zero, why has it a pdf if it doesn't exist in x (0,1)?

 

[D H]

You need to explain why you are making the steps you are making.

One problem here is that you are assuming that one simple expression describes the pdf from -infinity to infinity. That this is not the case should be obvious given that the pdf has to be identically zero for z between -1 and 1.

 

[ulissess]

Well, i don't know.. i just try to find a solution. I'm thinking from 3 hours. (z-1)/(z+1) i believe is the curve between 1 and +inf so F(z) use that density but for z<-1 use another pdf between (0,1) so .. I'm lost. Suggestion?

 

[vela]

It might help you to figure out what the curves z=constant look like in the xy plane.

 

[ulissess]

[PLAIN]http://img689.imageshack.us/img689/6431/55725049.jpg

[PLAIN]http://img262.imageshack.us/img262/1655/46800016.jpg

i've solved so..

 

[vela]

What is that diagram supposed to represent?

 

[ulissess]

$$y=(z+1)/(z-1)*x$$

 

[ulissess]

sorry I'm very shallow...

with z in (-oo,-1)

[PLAIN]http://img256.imageshack.us/img256/1354/unoir.jpg

$$F_z(z)=1*(z+1)/(z-1)$$*0.5 (area of triangle)

with z in (1,+oo)

[PLAIN]http://img26.imageshack.us/img26/4328/duep.jpg

$$F_z(z)=1*(z-1)/(z+1)$$*0.5 (area of triangle)

 

[vela]

Ah, that's better, but you got the fractions upside down. From earlier in the thread, you correctly had

$$y=\frac{1-z}{1+z}x$$

which doesn't match with what you have on your diagrams.

If you want to go straight to F_Z(z) by computing areas, you need to make sure you're looking at the right areas, so think about which lines z=±1 and z=±∞ correspond to. (Note you can also use this to figure out what the correct limits on your earlier integrals should be.)

 

[ulissess]

oh sure .. $$y=\frac{z-1}{z+1}x$$



with z in (1,+oo)
[PLAIN]http://img525.imageshack.us/img525/2809/unoj.jpg

if (z->+oo) ==> $$F_z(z)=\int_{0}^{1}xdx=0.5$$

if z=1 ==>$$F_z(z)=\int_{0}^{0}1dx=0$$

$$F_z(z)=\int_{0}^{1}\frac{z-1}{z+1}dx=\frac{z-1}{2(z+1)}$$

with z in (-oo,-1)
[PLAIN]http://img689.imageshack.us/img689/4300/duex0.jpg

$$F_z(z)=\int_{0}^{1}\frac{z+1}{z-1}dy=\frac{z+1}{2(z-1)}$$

if (z->-oo) ==>$$F_z(z)=\int_{0}^{1}ydy=0.5$$

if z=-1 ==>$$F_z(z)=1$$ because y->+oo

 

[vela]

F_Z(z) = Pr(Z<z), right? So are you really claiming the following?

F_Z(∞) = Pr(Z<∞) = 0.5
F_Z(1) = Pr(Z<1) = 0
F_Z(-1) = Pr(Z<-1) = 1
F_Z(-∞) = Pr(Z<-∞) = 0.5

 

[D H]

If you want to go straight to F_Z(z) by computing areas, you need to make sure you're looking at the right areas, so think about which lines z=±1 and z=±∞ correspond to. (Note you can also use this to figure out what the correct limits on your earlier integrals should be.)

Since you are having so much trouble with this, ulissess, I help out a bit more than we normally do here.

That graphical view is a good way to go, but you do need to get things right. (Nothing special there; you need to get things right when you are using an integral approach as well.)

Using that unit square with the x and y axes in their standard positions (e.g. exactly the way you have drawn it),

1. Draw a diagonal line from the origin to (1,1). Note this line separates negative z values from positive ones. The upper triangle (left triangle) represents (x,y) pairs that map to negative z while the lower triangle (right triangle) represents (x,y) pairs that map to positive z values.
2. Except for the origin, all points on the x-axis will map to the same z value. What is this z value for points on the x axis?
3. Except for the origin, all points on the y-axis will also map to a single z value. What is this z value for points on the y axis?
4. In fact, all points on any line emanating from the origin except for the origin itself will map to a single z value. Prove this.
5. With this in mind, draw that diagonal line segment from the origin to (1,1) and another line segment from the origin to a point (x,1) on the upper edge of the unit square.
   • Suppose you want this line to represent a specific z value, obviously a negative z value given the discussion in item 1 above. What is the value of x that makes all points on this second line segment represent the (x,y) pairs that map to this value of z?
   • Note that these two line segments split the unit square into three triangles. One of these triangles represents all z values from -∞ to your selected z value. Which of the three triangles is this, and what is its area?
6. To deal with positive z values, do the same thing but replace the line to (x,1) with a line to (1,y). This line segment plus the diagonal once again divides the unit square into three triangles, each representing some set of z values. What does this tell you about the CDF for positive z values?

 

[ulissess]

Except for the origin, all points on the x-axis will map to the same z value. What is this z value for points on the x axis?

1

Except for the origin, all points on the y-axis will also map to a single z value. What is this z value for points on the y axis?

-1

Suppose you want this line to represent a specific z value, obviously a negative z value given the discussion in item 1 above. What is the value of x that makes all points on this second line segment represent the (x,y) pairs that map to this value of z?

$$\frac{z+1}{z-1}$$

Note that these two line segments split the unit square into three triangles. One of these triangles represents all z values from -∞ to your selected z value. Which of the three triangles is this, and what is its area?

[PLAIN]http://img683.imageshack.us/img683/7565/duegu.jpg (the black area)
$$1=F_z(z)+\frac{1}{2}+\frac{z+1}{2(z-1)}=>F_z(z)=\frac{1}{2}-\frac{z+1}{2(z-1)}=\frac{1}{1-z}$$

To deal with positive z values, do the same thing but replace the line to (x,1) with a line to (1,y). This line segment plus the diagonal once again divides the unit square into three triangles, each representing some set of z values. What does this tell you about the CDF for positive z values?

[PLAIN]http://img694.imageshack.us/img694/5990/unokc.jpg
my function is $$y=\frac{z-1}{z+1}x$$ so ..
when z=1 => y=0
when z->+oo ==> y=x
so, my $$\frac{z-1}{z+1}$$ go near y=0 for z->1 values and go near y=x for z->+oo values

 

[D H]

Finally! You have yet to obtain the CDF for z>1. What is it?

Now put it all together: What is the CDF for all z, from -∞ to +∞?

 

[ulissess]

[PLAIN]http://img687.imageshack.us/img687/742/unos.jpg

$$1=F_z(z)+\frac{1}{2}+\frac{z-1}{2(z+1)}=>F_z(z)=\frac{1}{2}-\frac{z-1}{2(z+1)}=\frac{1}{1+z}$$

for z in (-1,1)

[PLAIN]http://img217.imageshack.us/img217/1504/trehc.jpg
$$F_z(z)=\frac{1}{2}$$

 

[vela]

If the black areas are meant to represent F_Z(z), your pictures aren't correct. Again, keep in mind that F_Z(z) is the probability that Z<z.

Also, your expression for F_Z(z) for z>1 isn't correct either. The CDF is an increasing function of z and its limit as z→∞ is 1. Neither of those is true for your F_Z(z).

 

[ulissess]

$$F_Z(z)= P(Z<z)$$

i don't understand find this relation.. maybe for z>1

$$F_z(Z)= P(Z<z)= P( \frac{x+y}{x-y}<=z>1 )$$
right? but i don't know how find the right area.. z>1 is all triangle (0,1)*(1,1)*0.5 but i don't know how to continue..

 

[D H]

In your first diagram in post #56, what values of z does your shaded triangle represent? Hint: One edge is the diagonal. What kind of z value do you obtain for a point just to the right (or just below) that diagonal line?

 

[ulissess]

$$y>\frac{z-1}{z+1}x==>z>\frac{y+x}{x-y}$$