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Normalization: discrete vs. continuous

  1. Apr 29, 2010 #1


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    So, I'm taking an EE class and my teacher is terribly handwavy. She couldn't really explain this to me (not homework, lecture). I detect a fundamental problem in the math, coming from a science background, but it could just be my ignorance:

    Here's her lecture:

    physical setup: a continuous roulette wheel returns a random variable: o < x < 2pi


    int{Pdx} = 1, the x range is 2*pi, so for the total area to equal one, the probability is constantly 1/(2*pi) for every value of x.

    here is where my red flag goes up. If x is truly continuous, wouldn't the probability of hitting any particular value of x be 0 since there are infinite values of x between 0 and x?

    This implies to me, that x isn't continuous and that there is actually some delta-x instead of dx.

    What is my issue here?
  2. jcsd
  3. Apr 29, 2010 #2
    A roulette wheel represents a set of discrete outcomes. If you know calculus, you know about continuous functions and evaluating the integral


    Clearly if F(b)=F(a), the integral equals zero. So for a continuous probability distribution, you can't have a non zero probability of a point. What you can have is a probability density between two distinct points on a probability density function (PDF). Note one point can be at infinity depending on the PDF. If both points are at infinity, then the probability density is 1 if the PDF is defined over that range.

    However, for a continuous uniform distribution such as your roulette wheel with an infinite number of points, there can't be points at infinity since every equal interval would have to have the same probability density and that would be zero if there were points at infinity. (Points at infinity is just a term for a limit at infinity.)

    Therefore, you can define your intervals as small as you wish, but to define a non zero probability density the intervals must be non zero and the number of intervals must be finite if each interval has the same (non zero) probability density.

    So for example your wheel of [0,[tex]2\pi[/tex]] radians would be recalibrated to [0.1] broken down to n equal intervals, each with a probability density of 1/n. You can see why n cannot be infinite.
    Last edited: Apr 30, 2010
  4. Apr 30, 2010 #3


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    Ok, that's what I thought. But since the normalization leads to 1/(2pi) doesn't that determine our intervals? Which is kind of difficult since it's not an inverse integer.

    Also, I'm still confused why we call it continuous when it's not.
  5. Apr 30, 2010 #4
    In terms of radian measure the width of your intervals will be 2pi/(n) but your probabilities are based on the the entire space having an integral of one, so you need to transform from radian measure to probability measure.

    The interior measure of the intervals is still continuous, but your probability density measure applies to the width of the interval. It's continuous in the sense that you define the intervals [a,b] any way you want, whereas with a discrete countable set you're sort of "stuck" with what you have.

    EDIT: With a set of n discrete elements, you cannot divide the elements. If you change the number of elements, you change the cardinality of the set. All continuous sets have the same cardinality, so you are free to define intervals without changing the mathematical properties of the set
    Last edited: May 1, 2010
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