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Thinking of a lower bound for a function

  1. Apr 30, 2014 #1
    A function ##f:\mathbb{R}^3_+\to[0,1]## defined as ##f(\lambda,\beta,x)=1-e^{-\frac{\lambda}{\beta}\left(1-e^{-\beta x}\right)}## serves a lot of pain under integration.
    As this function is used to describe a lower bound, could anyone suggest another non-zero function that would be smaller than ##f##?
     
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  3. May 1, 2014 #2

    mfb

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    Making f smaller <-> making the outer exponential larger <-> making the argument larger <-> making the negative argument smaller <-> making the inner exponential larger
    Hmm, bad direction.

    You can use approximations like ##e^{-x}<max\left(\frac{1}{e},1-\frac{x}{2}\right)## for example (possible for both exponentials), but I'm not sure if that makes calculations easier.
     
  4. May 1, 2014 #3
    Having one exponential is perfectly acceptable, it's the ##\exp(...\exp(...))## bit that's causing the integration problems.
    It seems that ##g(x)=(1-e^{-\frac{\lambda}{\beta}})\cdot\min(\frac{1}{e},1-e^{-\beta x})## is a good enough lower bound. Thank you.

    EDIT: Just found an even better bound: ##f(\lambda,\beta,x)>(1-e^{-\frac{\lambda}{\beta}})(1-e^{-\beta x})##.
     
    Last edited: May 1, 2014
  5. May 2, 2014 #4

    mfb

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    This is equivalent to $$1-q^r > (1-q)r$$ for 0<r,q<1. I don't find a counterexample, but I wonder why this is true (and how to show it).

    (##q=\exp(-\frac \lambda \beta)## and ##r=1-\exp(-\beta x)##)
     
  6. May 2, 2014 #5

    micromass

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  7. May 3, 2014 #6

    mfb

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    Ah nice. It needs some re-writing, but then it is really just Bernoulli's inequality. Thanks.
     
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