## kirbykirby 2 years ago Find k such that f(x) is a probability density function (pdf):

1. kirbykirby

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

2. kirbykirby

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

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

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

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

because if x<mu, the integral might diverge.

7. kirbykirby

Oh right I see

8. hartnn

yes, even thats true,

9. kirbykirby

Ok well thank you :)!

10. hartnn

welcome ^_^