1. Beta distribution mode

$\displaystyle X$ has a beta distribution with p.d.f:

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

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

But how do I find the mode? I can see from looking at the graph that it is $\displaystyle x=0.4$ , but how do I find it mathematically?

2. Hello,
Originally Posted by billym
$\displaystyle X$ has a beta distribution with p.d.f:

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

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

But how do I find the mode? I can see from looking at the graph that it is $\displaystyle x=0.4$ , but how do I find it mathematically?
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

3. Yeah I thought of that before, but when I set it to zero:

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

I could only see the roots 0 and 1.

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

$\displaystyle -60x(-1+x)^2(-2+5x)$

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:

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

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.

4. Originally Posted by billym
Yeah I thought of that before, but when I set it to zero:

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

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

so you can factor (1-x)² :

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

can you finish it ?

5. Originally Posted by Moo
Note that $\displaystyle (1-x)^3=(1-x)^2(1-x)$

so you can factor (1-x)² :

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

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