Banner Ad 1

Collapse

Announcement

Collapse
No announcement yet.

Mgf

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Mgf

    Why can't I get the MGF for exponential function? What am I doing wrong?
    Assume hazard rate = 1

    =∫ exp(t*x)*exp(-x) dx

    limit x-> infinity { [exp(t*x-x)]/(t-1) }


    which does not = 1/(1-t)

    Please help

  • #2
    Originally posted by jacky View Post
    Why can't I get the MGF for exponential function? What am I doing wrong?
    Assume hazard rate = 1

    =∫ exp(t*x)*exp(-x) dx

    limit x-> infinity { [exp(t*x-x)]/(t-1) }


    which does not = 1/(1-t)

    Please help
    You need to integrate between 0 and infinity so

    \int_{0}^{\infty} e^{tx} e^{-x} dx = \int_{0}^{\infty} e^{x(t-1)} dx = e^{x(1-t)}/(t-1) from 0 to infinity.

    Now assume 1 - t < 0 i.e. t > 1 we know that the upper bound converges to 0 and the lower bound converges to 1/(t-1) so you get 0 - 1/(t-1) = 1/(1-t)

    Comment


    • #3
      thank you very much for your help

      Comment


      • #4
        Hi I just have an irrelevant question.
        How did you make that integral sign?
        What did you type?

        Comment


        • #5
          Originally posted by ronaldy27 View Post
          Hi I just have an irrelevant question.
          How did you make that integral sign?
          What did you type?
          It was copied and pasted!!

          Comment


          • #6
            Originally posted by ronaldy27 View Post
            Hi I just have an irrelevant question.
            How did you make that integral sign?
            What did you type?
            If you have a Mac you can also hold alt and then letter b simultaneously to write an integral sign ∫.

            Comment

            Working...
            X