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PDF of a sine wave cycle

  1. Jan 16, 2009 #1
  2. jcsd
  3. Jan 18, 2009 #2


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    Pls clarify which is the rv.
  4. Jan 19, 2009 #3
    Hi ssd. Note sure what you mean by the rv.

    Actually I have now solved this problem. The pdf of a sine wave is given by:

    \textrm{P}(x) \textrm{ d}x= \frac{1}{\pi \sqrt{1-x^2}} \textrm{ d}x

  5. Jan 21, 2009 #4


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    R.V. is "random variable".
  6. Apr 20, 2010 #5
    Could you show the derivation?

    Hmmm... From http://en.wikipedia.org/wiki/Differ...ns#Differentiating_the_inverse_sine_function":

    [tex]\frac{d}{dx} \arcsin x & {}= \frac{1}{\sqrt{1-x^2}}\\[/tex]
    Last edited by a moderator: Apr 25, 2017
  7. May 5, 2010 #6
    Actually that makes sense. As the slope of the function increases, the likelihood of getting a point at that value increases, so it would seem that the PDF of a function is the derivative of the http://en.wikipedia.org/wiki/Inverse_function" [Broken] to figure it out, like using only a single cycle of the sine wave, which is what arcsin does.

    So for [tex]y = x^2[/tex], for instance, the inverse function is [tex]x = \pm\sqrt{y}[/tex], and the derivative of one side of this (since both positive and negative are identical) is [tex]1 \over {2 \sqrt{x}}[/tex]. Weight it so that the total area under the curve is 1, and it's the PDF.

    But what about functions that don't have inverses, and also aren't symmetrical or repetitious? They still have PDFs. Do you just break them up into piecewise functions at each http://en.wikipedia.org/wiki/Stationary_point" [Broken], and then sum them)
    Last edited by a moderator: May 4, 2017
  8. May 6, 2010 #7


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    Yes, I think you would have to do it that way.
    Last edited by a moderator: May 4, 2017
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