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Math Help - random variable X and random variable X^2

  1. #1
    Member Danshader's Avatar
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    random variable X and random variable X^2

    Well this is the question in the tutorial:

    A random variable X has the density function f(x) = c/(x^2 +1), x is a real number.
    a) Find the value of the constant c (already solved) [ans:1/pi]
    b) Find the probability that X^2 lies between 1/3 and 1 [ans: 1/6]

    well i understand what they want for part (a) which i had already solved but i don't understand part (b) especially on the X^2. Don't really know how it affects the function given for random variable X. Anyone mind explaining?
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    Hint:

    If 1/3 \leq X^2 \leq 1, what are the possible values of X?
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    er... sqrt(1/3) < x < sqrt(1)?
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    Moo
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    Hello,

    Quote Originally Posted by Danshader View Post
    er... sqrt(1/3) < x < sqrt(1)?
    Only if x is a positive variable
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    Member Danshader's Avatar
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    well the question said that x is a real number so it can be both negative and positive.
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    Moo
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    Ok, so :

    x^2 \geq 1/3 \text{ iff } \{ [x \geq \sqrt{1/3} \text{ and } x \geq -\sqrt{1/3}] \Longleftrightarrow x \geq \sqrt{1/3} \} OR \{ [x \leq \sqrt{1/3} \text{ and } x \leq -\sqrt{1/3}] \Longleftrightarrow x \leq -\sqrt{1/3} \}

    The same goes for x^2 \leq 1...

    If you want to continue this way, you will have 4 probabilities (maybe 3) to calculate...

    I dunno if there is an easier way.
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    Member Danshader's Avatar
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    i just did this:

    1/3 < x^2 <1
    sqrt(1/3) < x < sqrt(1) and -sqrt(1/3) > x > -sqrt(1)

    so this kinda give me two probabilities to solve o.o"

    well i solved for the first part and i got 1/12 so since the intervals are of the same range i multiplied the value obtained by 2 to get 1/6 o.o" is
    it correct?
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    Moo
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    Wait, I just have a problem :

    x^2 > 1/3

    x^2-1/3>0

    (x-\sqrt{1/3})(x+\sqrt{1/3})>0

    So x-\sqrt{1/3}>0 \ and \ x+\sqrt{1/3}>0 or x-\sqrt{1/3}<0 \ and \ x+\sqrt{1/3}<0

    This seems strange to me that the bounds are like this, but I can't find the mistake, and this makes me disagree with your probabilities :/



    Edit : nevermind... I was just confused >_<

    I didn't check the calculus, but it's correct, since the second inequality is the same as the first one (with x instead of -x) and the density function is even
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    ok i just confused myself >.<.... shouldn't the range be:

    sqrt(1/3) < x < 1
    -sqrt(1/3) < x < -1

    ???
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    Moo
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    Look at my edit

    To resume :

    P(1/3<X^2<1)=P(\sqrt{1/3}<X<1)+P(-\sqrt{1/3}>X>-1) =P(\sqrt{1/3}<X<1)+P(\sqrt{1/3}<-X<1)=2P(\sqrt{1/3}<X<1)
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    ok i understand the second line since it is an even function so we can add them up but is the second range of x correct? O.o?

    P(-sqrt(1/3) > x > -1)

    i think i am gonna read my high school maths back... >.>
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  12. #12
    Moo
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    Quote Originally Posted by Danshader View Post
    ok i understand the second line since it is an even function so we can add them up but is the second range of x correct? O.o?

    P(-sqrt(1/3) > x > -1)

    i think i am gonna read my high school maths back... >.>
    Oh it seems correct...
    Just multiply all of it by -1 to get the right thing.
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    ok >.> i will take your word for it... or else i am screwed in exam xD
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  14. #14
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    Erm... This is too much responsibility for me >.<

    But I'm pretty sure it's ok though ^^
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