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kirbykirby

  • 2 years ago

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

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

  2. kirbykirby
    • 2 years ago
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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)

  3. kirbykirby
    • 2 years ago
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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.

  4. hartnn
    • 2 years ago
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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.

  5. kirbykirby
    • 2 years ago
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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 ._.

  6. hartnn
    • 2 years ago
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    because if x<mu, the integral might diverge.

  7. kirbykirby
    • 2 years ago
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    Oh right I see

  8. hartnn
    • 2 years ago
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    yes, even thats true,

  9. kirbykirby
    • 2 years ago
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    Ok well thank you :)!

  10. hartnn
    • 2 years ago
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    welcome ^_^

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