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Pdf of exponetial distribution

  1. Jun 1, 2009 #1
    I am confused as hell. If you look at exponential distribution at wikipedia or elsewhere, you can see that the pdf can attain values larger than 1.

    How is it possible? This basically implies that probability density of some value is larger than 1? :mad:
     
  2. jcsd
  3. Jun 1, 2009 #2

    statdad

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    Why is that a problem? All that is needed is for the density to be continuous and

    [tex]
    \int_0^\infty p(x) \, dx = 1
    [/tex]

    (0 to infinity here since the exponential r.v. only takes on positive values)
     
  4. Jun 1, 2009 #3
    I think I get it.


    PDF refers to probability DENSITY function. So one cannot say the PDF of X, if X is discrete random variable, right?
     
    Last edited: Jun 1, 2009
  5. Jun 1, 2009 #4

    statdad

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    No, you won't get a probability of more than one. If you look at the form of the distribution function (the c.d.f), you will see it is

    [tex]
    1 - \text{ non-zero expression}
    [/tex]

    so the probability is never above one.
     
  6. Jun 1, 2009 #5
    sorry for the ninja edit above
     
  7. Jun 1, 2009 #6
    So a discrete random variable does not have a probability density function, right? How is it called then? Probability mass function?
     
  8. Jun 1, 2009 #7

    statdad

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    I'm not sure we're on the same thought here.
    a) An exponential distribution is continuous, and it has a density and cumulative distribution function. it is possible for the density to have a maximum that is larger than one, as long as
    it still integrates to one. the density function values are not probabilities - those come from the distribution function. If [tex] p(x) [/tex] is the density, then the distribution function is

    [tex]
    P(x) = P(X \le x) = \int_{-\infty}^x p(t) \, dt
    [/tex]
    b) A discrete distribution has a mass function (the discrete analog of a density). the individual values of the mass function are probabilities; each one of them must be smaller than one, and the sum of all values of the mass function has to equal one. If [tex] m(x) [/tex] is the mass function, then for each value of the random variable

    [tex]
    m(a) = P(X = a)
    [/tex]

    A discrete distribution also has distribution function. If [tex] t [/tex] is a real number,

    [tex]
    M(t) = P(X \le t) = \sum_{x_i \le t} m(x_i)
    [/tex]

    where I mean the sum extends over all values that are [tex] \le t [/tex].

    Primary differences:

    1) the density for a continuous random variable doesn't give probability directly - the distribution function does that
    2) the density for a continuous random variable can be larger than one, as long as it integrates to one
    3) the mass function for a discrete random variable does give probability for individual values, and must always be [tex] \le 1[/tex]
    4) a discrete random variable also has a distribution function

    does this help?
     
  9. Jun 1, 2009 #8
    Yes, totally clear now. Thanks a lot.
     
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