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

Question

kirbykirby · Answer

$$f(x)=ke^{-\frac{x-\mu}{\theta}}$$ for $$x>\mu$$

Perhaps more legibly as: $$f(x)=k*\exp({-\frac{x-\mu}{\theta}})$$

kirbykirby · Answer

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)

kirbykirby · Answer

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.

hartnn · Answer

i think x> mu is just to tell that exponential term has negative exponent always, the bounds will be still -infinity to infinity.

kirbykirby · Answer

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 ._.

hartnn · Answer

because if x<mu, the integral might diverge.

kirbykirby · Answer

Oh right I see

hartnn · Answer

yes, even thats true,

kirbykirby · Answer

Ok well thank you :)!

hartnn · Answer

welcome ^_^