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Homework Help: Integral defined as zero implies function is zero

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int\limits_{\theta}^{\infty} f(x) g(x) dx = 0 [/tex]
    [tex]\theta > 0 [/tex]
    [tex]f(x) = ke^{-k(x-\theta)}[/tex]
    Show g(x) is identically 0.

    2. Relevant equations

    3. The attempt at a solution
    [tex]f[/tex] is always >= 0 since it behaves exponentially in the region of interest.
    From something like https://www.physicsforums.com/showthread.php?t=299145" I could say that IF [tex]f(x)g(x)\ge 0[/tex] in this region, then [tex]g(x)=0[/tex], but I don't know anything about g. I could say that assuming g is positive somewhere, then the integral wouldn't be zero, and that assuming g is negative somewhere, the integral wouldn't be zero, but what about the case where [tex]g(c_1) f(c_1) = k[/tex] and [tex]g(c_2) f(c_2) = -k[/tex]. Then g is positive somewhere and negative somewhere else such that the product cancels out. Wouldn't that allow the integral to be zero?
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Dec 4, 2009 #2


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    What are your assumptions? g continuous? True for all k and [itex]\theta[/itex]?
  4. Dec 5, 2009 #3


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    As it stands, this simply isn't true! Are you requiring that g be continuous?
  5. Dec 8, 2009 #4
    I guess I should explain a bit more. I'm trying to show that a family of pdfs is complete. I no longer have access to my paper (for the time being), so I'll reproduce what I can remember. The notation seems standard, so I'll just type out a definition:
    Let the random variable Z of either the continuous type or the discrete type have a pdf or pmf that is one member of the family [tex]\left\{h(z;\theta) : \theta \in \Omega\right\}[/tex]. If the condition [tex]E[u(Z)]=0[/tex], for every [tex]\theta \in\Omega[/tex] requires that u(z) be zero except on a set of points that has probability zero for each [tex]h(z;\theta),\;\theta\in\Omega[/tex], then the family [tex]\left\{h(z;\theta) : \theta \in \Omega\right\}[/tex] is called a complete family of probability density or mass functions.

    Here, my pdf is the f I typed in the first post, and I need to show it's complete. As far as I can tell, I know nothing more about g(x). So, being the naive little statistician that I am, I would try and set the integral of g*f equal to zero and show that g is identically zero.

    (ps the book sucks so there's a good chance the authors performed a little hand waving, although they stress this example so I would guess the error is on my part, not theirs)
    (pps the book is hogg mckean and craig)
    (ppps is there a way to use dollar signs $ instead of typing [ tex ] each time?)
  6. Dec 8, 2009 #5


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    Do you mean f(x)=k*exp(-k*|x-theta|)? Then as k->infinity f(x) behaves like a multiple of the delta function centered at theta. Is that what they are getting at?
  7. Dec 11, 2009 #6


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    The key to proving that is the fact that [itex]\theta[/itex] is arbitrary. The point is not that the integral is zero for some [itex]\theta[/itex], it's that it's zero for every [itex]\theta[/itex].

    You can prove it for you specific function by, for example, taking a derivative with respect to [itex]\theta[/itex]. Note, however, I think that it will require that g(x) be continuous over the integration region. I'm not sure if there is a more general proof for your function.
    Last edited: Dec 11, 2009
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