Cumulative distribution function

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  • #1
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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 [Broken]

is it right?
 
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  • #2
D H
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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(zmin) = 0
  • The function must be one at the high end of the rangeL F(zmax) = 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?
 
  • #3
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i dont 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?
 
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  • #4
D H
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How can z ever be zero? Think about it.
 
  • #5
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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?
 
  • #6
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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.
 
  • #7
D H
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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.
 
  • #8
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with X=-Y i'll have Z=0 but negative values i cannot have it because my interval is [0,1]
 
  • #9
D H
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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?
 
  • #10
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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)?
 
  • #11
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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.
 
  • #12
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now i have understood hehehe (i dont 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?
 
  • #13
D H
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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).
 
  • #14
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[PLAIN]http://img338.imageshack.us/img338/4998/34928228.jpg [Broken]

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

z= -1/2 it's negative, the probability is 0, it's right?
 
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  • #15
D H
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As is, your Fz(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.
 
  • #16
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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?
 
  • #17
D H
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One more time: Can z ever be -1/2? How about -3/4?

Don't just guess! Think about the problem.
 
  • #18
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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
 
  • #19
D H
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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.
 
  • #20
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all z[tex]\neq[/tex]1 are attainable here.. because with z=1 become impossibile this fraction -1/(z-1)
 
  • #21
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NO!

What about z=0? 1/2? -1/2?
 
  • #22
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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
 
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  • #23
vela
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Try plotting z vs. x for some fixed value of y, say y=0.5. What's the range of values for z?
 
  • #24
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[PLAIN]http://img251.imageshack.us/img251/552/97845663.jpg [Broken]

is it right?
 
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  • #25
D H
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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.
 

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