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

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1Usually 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)

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1I 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.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1i think x> mu is just to tell that exponential term has negative exponent always, the bounds will be still infinity to infinity.

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1Oh 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 ._.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1because if x<mu, the integral might diverge.

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1Ok well thank you :)!
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