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kirbykirby Group Title

Find k such that f(x) is a probability density function (pdf):

  • one year ago
  • one year ago

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  1. kirbykirby Group Title
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    \[f(x)=ke^{-\frac{x-\mu}{\theta}}\] for \[x>\mu\] Perhaps more legibly as: \[f(x)=k*\exp({-\frac{x-\mu}{\theta}})\]

    • one year ago
  2. kirbykirby Group Title
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    Usually there's a bound for x like 0<x<2, but here it's x>u, so do I integrate from u to infinity? (And set this integral equal to 1)

    • one year ago
  3. kirbykirby Group Title
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    I don't need to know how to solve the integral (at least I don't think so yet). I'm just wondering what the bounds on the integral should be.

    • one year ago
  4. hartnn Group Title
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    i think x> mu is just to tell that exponential term has negative exponent always, the bounds will be still -infinity to infinity.

    • one year ago
  5. kirbykirby Group Title
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    Oh ok I think you are right :) When I think about it now... I wouldn't have found a numerical answer had I used mu as one of the bounds ._.

    • one year ago
  6. hartnn Group Title
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    because if x<mu, the integral might diverge.

    • one year ago
  7. kirbykirby Group Title
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    Oh right I see

    • one year ago
  8. hartnn Group Title
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    yes, even thats true,

    • one year ago
  9. kirbykirby Group Title
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    Ok well thank you :)!

    • one year ago
  10. hartnn Group Title
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    welcome ^_^

    • one year ago
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