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Looking for an example of a random variable that does not have a prob density fn

  1. Jan 19, 2010 #1
    "If a random variable has a probability density function, then the characteristic function is its Fourier transform" - http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)#Definition".

    I have never come across a random variable that did not have a probability density function. Can someone give an example of a random variable that does not have a probability density function?
     
    Last edited by a moderator: Apr 24, 2017
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  3. Jan 19, 2010 #2

    statdad

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    A mixed distribution corresponds to a random variable that is discrete over part of its domain and continuous over another part.

    More technical: If a random variable is continuous, its distribution function is an absolutely continuous function, and doesn't have any jumps from the left:

    [tex]
    \sum_{x \in \mathcal{R}} [F(x) - F(x-)] = 0
    [/tex]

    Since F is absolutely continuous it has a derivative, which is the density.

    On the other hand, if a random variable is discrete, every x with non-zero probability is an "atom", and

    [tex]
    \sum_{x \in \mathcal{R}} [F(x) - F(x-)] = 1
    [/tex]

    Here F does not have a density - the mass function is not, technically, the derivative of the cdf.

    For a mixed distribution it is true that

    [tex]
    0 < \sum_{x \in \mathcal{R}} [F(x) - F(x-)] < 1
    [/tex]

    The cdf is a "mixture" of a continuous and discrete function; again, there is no density.
     
    Last edited: Jan 19, 2010
  4. Jan 19, 2010 #3

    jambaugh

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    Note however that even in this mixed case one can use generalized functions (distributions) in particular Dirac delta functions to write a pdf analogue. It will then also have a fourier transform. Formally the Dirac delta function delta(x-a) is the derivative of a unit upward jump at a.

    Here is, I think, a counter example.
    Let the random variable X take on only rational values in the interval [0,1].

    Recall the rational numbers are countable, i.e. you can index them with the natural numbers. Pick some random indexing of the rational values in [0,1], say x_1, x_2, ...
    and let:
    [tex]Pr(X=x_k) = 1/2^k[/tex]

    [tex] \sum p_k = \sum_{k=1}^\infty 1/2^k = 1[/tex]

    The distribution won't have a density function because the probabilities, though discrete are defined on a dense set in the unit interval. The cumulative probability will be Highly Discontinuous.
     
  5. Jan 19, 2010 #4

    statdad

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    this is a discrete distribution - it is not mixed since there is no continuous portion.
     
  6. Jan 20, 2010 #5

    jambaugh

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    Yes.... and your point is?
     
  7. Jan 20, 2010 #6

    statdad

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    I took your post (given the topic of the OP) as a presentation of a mixed distribution. clearly I misunderstood the thrust of your message.
     
  8. Jan 20, 2010 #7

    jambaugh

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    Ahhh, I see. To my mind even a discrete distribution (within a continuous space such as the reals) has a pdf in the more general sense of a distribution rather than a true function, by utilizing delta functions one may be able to handle the discrete components. So a purely discrete or a mixed may have a pdf. However as with my example you can cook up a random variable with well defined probabilities but due to topological issues no well defined pdf even in the weaker sense I'm using. However upon thinking about you can probably still work with delta functions (at least formally) in my example and even define a Fourier transform of it. My counter example may not be a counter example in this weaker sense.

    I'll have to think about it.
     
  9. Jan 20, 2010 #8

    statdad

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    I've never seen delta functions used as to represent analogs of densities - the measure theory based stat and probability courses I took used the typical approach (Lebesgue-Stieltjes integrals, etc) exclusively (I assume you're well aware of this, so please don't take this as a "lecture" or "talking down".)

    As a final comment, if you haven't seen it, there is a very interesting, widely used, construction of a singular continuous distribution on the Cantor set. (Pages 12-13 of Chung, second edition, is one source).

    No more from me, however, as it is likely I have already gone too far astray from the question in the OP's first post.
     
  10. Jan 20, 2010 #9
    EDIT: just saw now the final comment of Statdat's post; I believe this is what is referring to.

    One example would be a random variable with the devil's staircase (or Cantor's function) as a distribution function. For a description see:

    http://en.wikipedia.org/wiki/Cantor_function" [Broken]

    It's a continuous function, differentiable almost everywhere, with a null derivative (almost everywhere). Therefore, it doesn't have a pdf, not even in a generalized sense.
     
    Last edited by a moderator: May 4, 2017
  11. Jan 20, 2010 #10

    jambaugh

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    My understanding of the delta function is that, as it is not a true function, it is only meaningful as a distribution.

    That would be the "Devil's staircase"? I recall it from Real Analysis some time back as an exotic example of a continuous function useful for many counter examples.

    Edit: Here's its pic on Wolfram Math:
    DevilsStaircase_1000.gif
     
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