Mode of a Beta(3,5)

Hi there,
Could someone please break down the steps to finding the roots of the following equation?
Thanks a lot,
Tom

210x(1-x)^3(1-3x)

 

is it
\(210*x\left(1-x\right)^3 \left(1-3x\right) = ?\)
or
\( f(x) = 210*x\left(1-x\right)^3 \left(1-3x\right)\)

 

It's the top one. Just trying to find the roots when it's set to zero. Cheers

 

The equation will be equal to zero if either x=0 or (1-x) is equal to zero or (1-3x) is equal to zero.
Thus we get, x can be equal to zero or 1 or 1/3. ( By solving the 3 above components for zero)

 

Another way of solving for mode is to use constraint maximization of two variables i.e.

if \( X \sim Beta(3,5) \) then the mode of \(X\) is the solution of the problem
\begin{align*} \max_{x\in (0,1)}& B(3,5)x^2(1-x)^4 \\ & \Leftrightarrow \\ \max_{x,y\in (0,1)}& x^2y^4 \quad st. x+y=1\\ & \Leftrightarrow \\ \max_{x,y\in (0,1)}& 2\ln(x)+4\ln(y) \quad st. x+y=1\\ \end{align*}
Now by plotting the level curves of both the function and the constraint, we see the solution satisfies the following equations:

• \(\dfrac{y}{2x} = 1 \leftrightarrow y = 2x\)
• \( x+y = 1\)

This gives \( (x^*,y^*) = (1/3,2/3) \) therefore mode of \(X\) as \(1/3\)

 

Thanks very much everybody