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The mode of beta distribution

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  • The mode of beta distribution

    hi all,

    I have a question,

    A random variable has a beta distribution with parameters a>1 and b>1.
    Determine the mode.

    the answer is (a-1)/(a+b-2)

    Why? how to find it??

    Thanks for all your help

    T
    Tears stream down your face
    I promise you I will learn from my mistakes

  • #2
    Originally posted by Double T
    hi all,

    I have a question,

    A random variable has a beta distribution with parameters a>1 and b>1.
    Determine the mode.

    the answer is (a-1)/(a+b-2)

    Why? how to find it??

    Thanks for all your help

    T
    This looked interesting so I had to try it. The mode is where the pdf reaches its maximum, so differentiate the pdf and set it equal to zero. The pdf is given by:
    (Gamma(a+b)/[Gamma(a)*Gamma(b)]) * x^(a-1) * (1-x)^(b-1)
    Differentiating yields via product rule (and setting equal to zero):
    (Gamma(a+b)/[Gamma(a)*Gamma(b)]) * [(a-1)*x^(a-2)*(1-x)^(b-1) + x^(a-1)*(-1)*(b-1)*(1-x)^(b-2)] = 0
    Simplify:
    (a-1)*x^(a-2)*(1-x)^(b-1) + x^(a-1)*(-1)*(b-1)*(1-x)^(b-2) = 0
    (a-1)*(1-x) - x*(b-1) = 0
    a - ax - 1 + x - xb + x = 0
    x(2 - a - b) = 1 - a
    x = (1-a)/(2-a-b) = (a-1)/(a+b-2)

    Hopefully that all looks good to you!

    Comment


    • #3
      Originally posted by Greg1983
      This looked interesting so I had to try it. The mode is where the pdf reaches its maximum, so differentiate the pdf and set it equal to zero. The pdf is given by:
      (Gamma(a+b)/[Gamma(a)*Gamma(b)]) * x^(a-1) * (1-x)^(b-1)
      Differentiating yields via product rule (and setting equal to zero):
      (Gamma(a+b)/[Gamma(a)*Gamma(b)]) * [(a-1)*x^(a-2)*(1-x)^(b-1) + x^(a-1)*(-1)*(b-1)*(1-x)^(b-2)] = 0
      Simplify:
      (a-1)*x^(a-2)*(1-x)^(b-1) + x^(a-1)*(-1)*(b-1)*(1-x)^(b-2) = 0
      (a-1)*(1-x) - x*(b-1) = 0
      a - ax - 1 + x - xb + x = 0
      x(2 - a - b) = 1 - a
      x = (1-a)/(2-a-b) = (a-1)/(a+b-2)

      Hopefully that all looks good to you!
      Thank you Greg,

      But i have one more question, how about (Gamma(a+b)/[Gamma(a)*Gamma(b)])??
      We don't need to differentiate this part??

      Thanks,
      T
      Tears stream down your face
      I promise you I will learn from my mistakes

      Comment


      • #4
        Originally posted by Double T
        Thank you Greg,

        But i have one more question, how about (Gamma(a+b)/[Gamma(a)*Gamma(b)])??
        We don't need to differentiate this part??

        Thanks,
        T
        No because you're differentiating with respect to x. It stays out the front as a constant then you can just get rid of it when you set the whole thing equal to zero.

        Comment


        • #5
          Originally posted by Greg1983
          No because you're differentiating with respect to x. It stays out the front as a constant then you can just get rid of it when you set the whole thing equal to zero.
          ''', how come i didn't think about it??

          Great, thank you very much, Greg..........
          Tears stream down your face
          I promise you I will learn from my mistakes

          Comment

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