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

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\[f(x)=ke^{-\frac{x-\mu}{\theta}}\] for \[x>\mu\] Perhaps more legibly as: \[f(x)=k*\exp({-\frac{x-\mu}{\theta}})\]
Usually there's a bound for x like 0u, so do I integrate from u to infinity? (And set this integral equal to 1)
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.

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Other answers:

i think x> mu is just to tell that exponential term has negative exponent always, the bounds will be still -infinity to infinity.
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 ._.
because if x
Oh right I see
yes, even thats true,
Ok well thank you :)!
welcome ^_^

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