kirbykirby
  • kirbykirby
Find k such that f(x) is a probability density function (pdf):
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
kirbykirby
  • kirbykirby
\[f(x)=ke^{-\frac{x-\mu}{\theta}}\] for \[x>\mu\] Perhaps more legibly as: \[f(x)=k*\exp({-\frac{x-\mu}{\theta}})\]
kirbykirby
  • kirbykirby
Usually there's a bound for x like 0u, so do I integrate from u to infinity? (And set this integral equal to 1)
kirbykirby
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.