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Bounded Probability Density Function

  1. Apr 2, 2008 #1
    Let the random variable X have the probability density function f(x). Suppose f(x) is
    continuous over its domain and Var[X] is bounded away from zero: 0 < c < Var[X].

    Claim: f(x) is bounded over its domain.

    Is this claim true?

    I don't think a counterexample like X ~ ChiSq_1 applies because, while f(0) is not bounded, the domain is x > 0. The question of continuity at 0 doesn't arise since 0 is not part of the domain, and therefore the lack of boundedness at 0 isn't relevant. But for any given c > 0, I can find an M such that f(c) < M. This example does show, however, why one can't claim "uniformly bounded."

    Similarly, a degenerate r.v. (all probability mass at the point X = c, say) is continuous (because it is discrete), but the variance is not bounded away from zero.

    So the question remains: is the claim true? How might one prove it?

    Thanks.
     
  2. jcsd
  3. Apr 3, 2008 #2
    That example isn't relevant, because a degenerate r.v. does not have a p.d.f.
     
  4. Jul 25, 2008 #3

    statdad

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    Homework Helper

    Let the random variable X have the probability density function f(x). Suppose f(x) is
    continuous over its domain and Var[X] is bounded away from zero: 0 < c < Var[X].

    Claim: f(x) is bounded over its domain.

    Is this claim true?
    No, it is not, and the [tex] \chi^2_1 [/tex] distribution you mention shows it. You do not need to worry about the pdf at [tex] x = 0 [/tex], since the domain is only [tex] (0, \infty) [/tex]. The variance of [tex] \chi^2_k [/tex] is [tex] \sigma^2 = 2k [/tex], so the variance of [tex] \chi^2_1 [/tex] is bounded away from zero. However, the density is not bounded on its domain.
    Statdad
     
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