Cumulative distribution function

[ulissess]

i have two random variables x e y independent and they're uniform on the interval [0, 1] find cumulative distribution function of Z= (x+y)/(x-y)

i just try to solve...

[PLAIN]http://img202.imageshack.us/img202/5647/97250438.jpg

is it right?

 

[D H]

For a function F(z) to qualify as a CDF,

• The function must be monotonic: F(z) ≥ F(a) for all z>a
• The function must be zero at the low end of the range: F(z_min) = 0
• The function must be one at the high end of the rangeL F(z_max) = 1.

Given that, does your result look like a CDF?

A couple of hints:
1. Your limits of integration aren't right.
2. What values can z take on? Can it be a large negative number? A large positive number? Zero?

 

[ulissess]

i don't understand how find the limits of integration :(

if z=0 => F_Z(z)=1/2
if z--> oo ==> F_Z(z)=-1/2

so this isn't a cdf.. can i try change of variables? how can i undestand what to use for solve these type of exercises?

 

[D H]

How can z ever be zero? Think about it.

 

[D H]

A couple of other hints:

1. Note that z is negative when y>x, positive when y<x.

2. z can never be zero. In fact, there is some interval that contains z=0 that cannot be reached by any x,y in [0,1]. What is this interval?

 

[ulissess]

thank you for the help..

(x+y)/(x-y)< z ==> i have 2 disequations x-y > 0 and x+y - z(x-y) < 0 with x>y my disequation is valid. So, i believe, i must use that 2 disequations for my integral.
i don't understand your 2. hint, i don't believe exist: [0,1] is an interval all positive.

 

[D H]

Regarding hint #2: With x and y both constrained to [0,1], name any (x,y) pair that yields z=0 or z=1/2 (just to pick two impossible z values). There is an interval of z values that cannot be attained. Your CDF must reflect this interval.

 

[ulissess]

with X=-Y i'll have Z=0 but negative values i cannot have it because my interval is [0,1]

 

[D H]

Exactly. Now what about z=1/2? Can you find an (x,y) pair with both x and y restricted to [0,1] that yields z=1/2? What about z=-1/2 or z=3/4?

 

[ulissess]

y=1, x=-3 ==> z=1/2 but x is negative and it's < 1
y=1, x=-7 ==> z=3/4 .. it's impossible .. so? any value is impossible.. my interval is (-oo,+oo)?

 

[D H]

y=1, x=-3 ==> z=1/2 but x is negative and it's < 1
y=1, x=-7 ==> z=3/4 .. it's impossible .. so? any value is impossible.. my interval is (-oo,+oo)?

You are missing my point. Forget about the fact that the probability of obtaining a specific value is "impossible". There is a big difference between drawing 1/2 from U(0,1) than drawing a value of 4. The difference is that the probability of drawing a value in a small but finite neighborhood of 1/2 is non-zero while the probability of drawing a value in a small but finite neighborhood of 4 is zero. Back to the problem at hand, the probability of obtaining a z value in a small but finite neighborhood of zero is zero.

 

[ulissess]

now i have understood hehehe (i don't speak very well english language), so z can to have values in (0,1)... so my 2° integral in dy is wrong.. how can i find this interval?

 

[D H]

How about z = -1/2 ?

One thing you need to do is to get your mapping from (x,z) to y correct. It isn't.

It might be easier to find the PDF f(z) and integrate it to form F(z) rather than finding F(z) directly. To find f(z), try to determine the probability that Z is between z and z+dz, where dz is small (infinitesimally small).

 

[ulissess]

[PLAIN]http://img338.imageshack.us/img338/4998/34928228.jpg

[PLAIN]http://img89.imageshack.us/img89/2320/93533692.jpg

z= -1/2 it's negative, the probability is 0, it's right?

 

[D H]

As is, your F_z(z) is not a CDF. It is, however, part of the answer.

To arrive at the answer you will need to find the range of z and incorporate this into your result.

 

[ulissess]

maybe, i believe ( 0< z < 1 --> -1/(z-1) ), ( z<=0 --> 0 ), ( z=>1 --> 1 )

but if i want solve directly F(z) what interval i use?

 

[D H]

One more time: Can z ever be -1/2? How about -3/4?

Don't just guess! Think about the problem.

 

[ulissess]

i believe z=-1/2 and z=-3/4 cannot be, because the Domain is (0,1) so only positive quantity between 0 to 1 are accepted

 

[D H]

maybe, i believe ( 0< z < 1 --> -1/(z-1) ), ( z<=0 --> 0 ), ( z=>1 --> 1 )

I don't know what you are saying here.

You have yet to answer my question: What (x,y) pairs (if any) with x and y restricted to [0,1] yields a z value of -1/2?

You need to know what values of z are attainable here. Knowing this is crucial to answering the problem.

 

[ulissess]

all z$$\neq$$1 are attainable here.. because with z=1 become impossibile this fraction -1/(z-1)

 

[D H]

NO!

What about z=0? 1/2? -1/2?

 

[ulissess]

y=-7x i yield z=-4/3
y=- 3x i yield z=-1/2
y=-1/3x i yield z=1/2
y=-x i yield z=0
y=x=0 i yield z=1

 

[vela]

Try plotting z vs. x for some fixed value of y, say y=0.5. What's the range of values for z?

 

[ulissess]

[PLAIN]http://img251.imageshack.us/img251/552/97845663.jpg

is it right?

 

[D H]

Have you tried to the suggestion in post #23?

You might also want to plot z as a function of y for different fixed values of x.

 

[ulissess]

Z=(x+y)/(x-y)

D: {R^2 \ (x-y$$\neq$$0)} so if y=0.5 ==> x$$\neq$$0.5

why is it important ? Is the range all the values x-y$$\neq$$0 ?

 

[vela]

You're talking about the domain, not the range. You first have to understand what the allowed values of Z are before you have a hope to figure out its cdf. That's why we are suggesting you plot the function to see the range. You don't need to plot the function to figure out its domain.

 

[ulissess]

yes i have understood... but i don't know how to plot a function in 3d.. it's difficult to image it.

 

[vela]

That's why I suggested fixing y at some value, so you only have to do a 2D plot. Once you understand the shape of that basic graph, you should be able to understand how it changes as you vary y.

 

[ulissess]

z=(x+0.5)/(x-0.5)

[PLAIN]http://img34.imageshack.us/img34/6961/37903189.jpg

the function is positive in (-oo; -0.5] U (0.5; +oo) and negative in (-0.5 ; 0.5) so my range is my positive interval?
So my range of my function z=(x+y)/(x-y) is (-oo; -y] U (y; +oo) is it right??

 

[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)?