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Math Help - Beta distribution mode

  1. #1
    Member billym's Avatar
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    Beta distribution mode

    X has a beta distribution with p.d.f:

    f(x)=\left\{<br />
\begin{array}{lr}<br />
kx^2(1-x)^3&0\le x\le 1\\<br />
0&otherwise<br />
\end{array}<br />
\right.

    First I had to find k so I integrated for the interval 0\le x\le 1 and set it equal to 1 which gave k=60 . Correct?

    But how do I find the mode? I can see from looking at the graph that it is x=0.4 , but how do I find it mathematically?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by billym View Post
    X has a beta distribution with p.d.f:

    f(x)=\left\{<br />
\begin{array}{lr}<br />
kx^2(1-x)^3&0\le x\le 1\\<br />
0&otherwise<br />
\end{array}<br />
\right.

    First I had to find k so I integrated for the interval 0\le x\le 1 and set it equal to 1 which gave k=60 . Correct?
    Yes

    But how do I find the mode? I can see from looking at the graph that it is x=0.4 , but how do I find it mathematically?
    Quote Originally Posted by Wikipedia
    The mode of a continuous probability distribution is the value x at which its probability density function attains its maximum value, so, informally speaking, the mode is at the peak.
    So differentiate the pdf and find where it attains its maximum
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  3. #3
    Member billym's Avatar
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    Yeah I thought of that before, but when I set it to zero:

    120x(1-x)^3-180x^2(1-x)^2=0

    I could only see the roots 0 and 1.

    I can see on wolfrom the (wonderful) alternate form:

    -60x(-1+x)^2(-2+5x)

    where I can see the root I'm looking for, 0.4.

    I've crudely expanded it but the closest I can get to the other form is:

    -60x(5x^3-12x^2+9x-2)

    I can see that I'm almost there... can somebody show me the final step?

    Surely there is a more civilized way of doing this? I remember being able to do this stuff 3 years ago, but that was 3 three years ago.
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  4. #4
    Moo
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    Quote Originally Posted by billym View Post
    Yeah I thought of that before, but when I set it to zero:

    120x(1-x)^3-180x^2(1-x)^2=0

    I could only see the roots 0 and 1.
    Note that (1-x)^3=(1-x)^2(1-x)

    so you can factor (1-x) :

    =(1-x)^2[120x(1-x)-180x^2]=60x(1-x)^2[2(1-x)-3x]

    can you finish it ?
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  5. #5
    Member billym's Avatar
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    Quote Originally Posted by Moo View Post
    Note that (1-x)^3=(1-x)^2(1-x)

    so you can factor (1-x) :

    =(1-x)^2[120x(1-x)-180x^2]=60x(1-x)^2[2(1-x)-3x]

    can you finish it ?
    [2(1-x)-3x]=(2-5x)

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