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Math Help - Uniform Distributions.. plz help

  1. #1
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    Question Uniform Distributions.. plz help

    My first post on this forum.. sorry, this problem is giving me problems the whole day

    Infinite number of (x) with properties (a) and (b)

    (a) and (b) are uniformly distributed between 0 and 1

    What is the fraction of (x) with properties:

    a > b
    3b > a


    How do you calculate this stuff??? Btw, I am biology undergraduate...


    Thank you in advance!!!
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  2. #2
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    Welcome to the forums.

    This is a uniform distribution so there is an equal chance that (x) will be near any value from 0 to 1 as it will any other.

    This is because it's distribution function is constant.

    So if you are looking for the probability that (x) lies between b and a where a > b then it is just the difference between b and a

    P{b < (x) < a} = a - b

    You should check this out yourself look up uniform distribution in your text or on google. its a fairly simble derevation.

    Now I am not sure if I am interpreting your question correctly, because it seems like a weird question to me... but that is how i would answer.

    for the other P{a < (x) < 3b} = 3b - a
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  3. #3
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    thanks for the answer, regarding the question:

    you have two variables a and b that are uniformly distributed between 0 and 1

    a and b are independent of each other

    you chose randomly a and b

    what is the probability to chose a and b so that

    a > b
    3b > a

    My suggestion:

    a' = a/(a+b)
    b' = 1-a'

    3 > a'/(1-a') > 1

    so that 3/4 > a' > 1/2

    giving a probability of 25%... is that correct? is the transformation to a' possible? if so, how would you prove that?

    ty
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by instantaneous View Post
    thanks for the answer, regarding the question:

    you have two variables a and b that are uniformly distributed between 0 and 1

    a and b are independent of each other

    you chose randomly a and b

    what is the probability to chose a and b so that

    a > b
    3b > a

    My suggestion:

    a' = a/(a+b)
    b' = 1-a'

    3 > a'/(1-a') > 1

    so that 3/4 > a' > 1/2

    giving a probability of 25%... is that correct? is the transformation to a' possible? if so, how would you prove that?

    ty
    In the first case p(a>b)=1/2 simply by symmetry, that is p(a>b)=p(b>a) and p(a=b)=0.

    For the second suppose a is chosen first then p(3b>a|a)=p(b>a/3|a)=1-a/3 which is:

    p(b>a/3)=\int_{a=0}^{1}p(b>a/3|a)p(a)\;da=\int_{a=0}^{1} (1-a/3)\; da

    CB
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  5. #5
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    wow!

    eehhm, so the probability of both happening 3 > a/b > 1 is

    p(a>b) * p(3b>a) = 1/2 * 5/6 = 5/12?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by instantaneous View Post
    wow!

    eehhm, so the probability of both happening 3 > a/b > 1 is

    p(a>b) * p(3b>a) = 1/2 * 5/6 = 5/12?
    So that is one question rather than two?

    (Those evenys are not independent so no its not the product).

    Assume a is known, then:

    p(b<a \text{ and }b>a/3|a)=2a/3

    (draw a diagram showing the range of b that satisfies the condition)

    Then

    p(b<a \text{ and }b>a/3)=\int_{a=0}^1 2a/3\; da

    CB
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