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Math Help - change in variable question

  1. #1
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    Angry change in variable question

    Y=sin(x)
    let X~uniform[0,pi/2]
    compute density of fy(y) for Y.

    my attempt
    ------------
    arcsin(Y) = X

    fx(arcsinY)
    -------------- = sqrt(1-y^2) * fx(arcsinY)
    (1/sqrt(1-y^2))

    then

    /\ arcsin(pi/2) sqrt(1-y^2) * fx(arcsinY) dy
    \
    .\
    \/ arcsin(0)

    /\ infinity sqrt(1-y^2) * fx(arcsinY) dy
    \
    .\
    \/ 0

    and i am stuck here as i cannot evaluate this, i probably did something wrong earlier...
    Last edited by Sneaky; October 16th 2010 at 03:53 PM.
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  2. #2
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    Quote Originally Posted by Sneaky View Post
    Y=sin(x)
    let X~uniform[0,pi/2]
    compute density of fy(y) for Y.

    my attempt
    ------------
    arcsin(Y) = X

    fx(arcsinY)
    -------------- = sqrt(1-y^2) * fx(arcsinY)
    (1/sqrt(1-y^2))

    then

    /\ arcsin(pi/2) sqrt(1-y^2) * fx(arcsinY) dy
    \
    .\
    \/ arcsin(0)

    /\ infinity sqrt(1-y^2) * fx(arcsinY) dy
    \
    .\
    \/ 0

    and i am stuck here as i cannot evaluate this, i probably did something wrong earlier...
    Just like I have previously suggested: The cdf is \displaystyle G(y) = \frac{2}{\pi} \int_0^{\sin^{-1} y} dy and the pdf is \displaystyle \frac{dG}{dy}. The support is 0 \leq y \leq 1.
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  3. #3
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    the answer is
    fY(y) = 2/(pi*sqrt(1-y^2)) for 0<=y<=1 and equals 0 otherwise.

    I don't understand how to get the answer like that.
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  4. #4
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    how did u get G(x) and where did u get 2/pi from?
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  5. #5
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    Quote Originally Posted by Sneaky View Post
    how did u get G(x) and where did u get 2/pi from?
    Have you understood any of my posts on this topic?

    My expression for G(y) in this thread comes from using exactly the same approach that I explained in more detail here: http://www.mathhelpforum.com/math-he...ge-159842.html. Have you read it? What don't you understand?

    As for where the 2/pi comes from, you said that X~uniform[0,pi/2]. Are you familiar with the pdf of a continuous uniform distribution? (I assumed that you were).

    Quote Originally Posted by Sneaky View Post
    the answer is
    fY(y) = 2/(pi*sqrt(1-y^2)) for 0<=y<=1 and equals 0 otherwise.

    I don't understand how to get the answer like that.
    Simply differentiate the cdf I gave in my previous reply.
    Last edited by mr fantastic; October 16th 2010 at 05:17 PM. Reason: Merged posts.
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  6. #6
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    Can you explain your steps in your second post as I am very lost in this topic.
    edit: i get where 2/pi came from now

    so now i don't understand how to evaluate the integral part

    you have integral from 0 to arcsin y of dy, how do you evaluate that?
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  7. #7
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    Quote Originally Posted by Sneaky View Post
    Can you explain your steps in your second post as I am very lost in this topic.
    edit: i get where 2/pi came from now
    I intended the answer I gave here:

    http://www.mathhelpforum.com/math-he...ge-159842.html

    to be an example you could follow when trying to answer other similar questions. Please try to apply each step of that answer to the current problem. Show all your work, say where you get stuck.

    Quote Originally Posted by Sneaky View Post
    [snip]
    so now i don't understand how to evaluate the integral part

    you have integral from 0 to arcsin y of dy, how do you evaluate that?
    You do exactly the same as if you were finding \displaystyle \int_a^b dy.
    Last edited by mr fantastic; October 16th 2010 at 05:32 PM. Reason: Merged posts.
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  8. #8
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    thanks i understand now
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