Cumulative distribution function

[ulissess]

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



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

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 [Broken]

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

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 [Broken] (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 [Broken]
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 [Broken]

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

for z in (-1,1)

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

 

[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]

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

 

[D H]

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

In other words, your shaded triangle represents Pr(z>Z). That obviously is not the CDF, but it should also be obvious that it is closely related to the CDF. How?

 

[ulissess]

i don't understand very well..