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kirbykirbyBest ResponseYou've already chosen the best response.1
\[f(x)=ke^{\frac{x\mu}{\theta}}\] for \[x>\mu\] Perhaps more legibly as: \[f(x)=k*\exp({\frac{x\mu}{\theta}})\]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
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

kirbykirbyBest ResponseYou've already chosen the best response.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.
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
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

kirbykirbyBest ResponseYou've already chosen the best response.1
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

hartnnBest ResponseYou've already chosen the best response.1
because if x<mu, the integral might diverge.
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
Ok well thank you :)!
 one year ago
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