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Homework Help: Probability Density

  1. Apr 17, 2005 #1
    Here's the question:

    The needle on a broken car spedometer is free to swing, and bounces perfectly off the pins at either end, so that if you give it a flick it is equally likely to come to rest at any angle between 0 and [itex] \pi [/itex].

    Consider the x-coordinate of the needle point - that is, the "shadow," or "projection," of the needle on the horizontal line.

    What is the probability density [itex] \rho(x) [/itex] if r is the length of the needle?


    Conceptually, I'm trying to clarify the following:

    In a previous question, they asked to find [itex] \rho(\theta) [/itex], which I found as simply [itex] \frac{1}_{\pi} [/itex]. This implies the theta-derived answer moves the x coordinate from r to -r as [itex] \theta [/itex] goes from 0 to [itex] \pi [/itex].

    In calculating the probability for this question, I assumed x moved in the same fashion (ie, x goes from r at [itex] \theta = 0 [/itex] to -r at [itex] \theta = \pi [/itex]). This reverse movement gives a probability density that is always negative.

    My question is, is this possible? The total probability does equal 1 if I move from right to left, though I'm not sure if this is permissible.

    Likewise, could I simply change the sign from negative to positive for the probability density, only reasoning that if x moves left to right, the signs should be reversed?
  2. jcsd
  3. Apr 17, 2005 #2


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    The probability density is defined such that for a small interval dx, the probability of finding x in dx is p(x)dx, and for a large range you integrate this quantity. By defintion, dx is positive, because it is the size of the interval. Since a change dθ on this range corresponds to a negative change in x, you negate this to get the size of the change in x, which is what you're looking for.
    Last edited: Apr 17, 2005
  4. Apr 17, 2005 #3
    I guess the part I'm not understanding is this:

    the needle could technically go either direction - it could move right-to-left or left-to-right. One corresponds to positive changes in x, and the other negative.

    Likewise, one probability density is always positive, and the other is always negative. Given the symmetry of the system, should preference be given to the positive probability density? Or does it not matter?


    Are you saying that the positive result must always be chosen and that negative result has no meaning (ie, what does it mean that you have a -10% result for the probability density?)?
    Last edited: Apr 17, 2005
  5. Apr 17, 2005 #4
    x(\theta) = r cos \theta

    \frac{dx}{d \theta} = - r sin \theta

    d\theta = \frac{dx}_{-r sin \theta}

    This is where I get my negative from. Should I then just throw out the negative?
  6. Apr 18, 2005 #5


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    But we're not talking about a change in x. Nothing is changing. We're talking about the size of an interval that might be selected. This size is always positve. For a positive change in theta, there will be a negative change in x, but for a given interval of theta, the size of the corresponding interval of x is positive. So yes, take the absolute value.
  7. Apr 18, 2005 #6
    When I do the change of variables from theta to x, I do get a square root in the expression for rho(x) (if I did the algebra correctly), so you do need to take the positive root, obviously.
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