http://en.wikipedia.org/wiki/Beta_distribution
The general from of the function is:
Between a AND b with coefficient c and d
y = ((x-a)^(c-1) (b-x)^(d-1))/(b-a)^(c+d-1) y'=(x-a)^(c-2) (b-x)^(d-2) (b-a)^(-c-d+1) (a (d-1)+b (c-1)-x (c+d-2))
y'=(x-a)^(c-1) (b-x)^(d-1) (b-a)^(-c-d+1) (a (d-1)+b (c-1)-x (c+d-2))/((x-a)(b-x))
therefore the differential equation is: y’=y (a (d-1)+b (c-1)-x (c+d-2))/((x-a)(b-x))
if a=0 and b=1 it is a solution to
is a solution to this differential equation
================================================================================= To make sure that the area below y is one and convert it to a probability density function we find the
integral_a^b (x-a)^(c-1) (b-x)^(d-1) (b-a)^(-c-d+1) dx
which we cal B(c,d) then
f(x)=(1/((gamma(c)*gamma(d))/Gamma(c+d)))*((x-a)^(c-1) (b-x)^(d-1))/(b-a)^(c+d-1)
is a probability density function with an area of 1. ========================================================================
When and are not equal the curve doesn’t approach 0 asymptotically at that end but has a peak closer to that end.
This means if is smaller then the peak will be close to a.
Beta curve with =1 is =3 b=1 f=1/B*(x-a)^0(b-x)^2
f=1/B*(x-a)^0(5-x)^2
f=(5-x)^2
is a derivative of:
f'(x)=2(x-5)
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The following is similar to y=e^-x
Exponential curve which will be 0 at infinity and is a solution to:
When power of positive x is less than 1 and less than the power negative x the curve is monotonous decreasing.
When power of positive x is less than the power negative x the curve is skewed to right
If 1 < α < β then mode ≤ median ≤ mean. Expressing the mode (only for α, β > 1), and the mean in terms of α and β:
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In general with =2 is =3
y=((x-a)(b-x)^2)
is a solution to
with =2 is =3 a=0 b=5
y=(x(5-x)^2)
is a solution to:
In general with =2 is =3 and a=0
y=((x)(b-x)^2)
is a solution to:
In general with =3 is =2 and a=0
y=((x)^2(b-x))
is a solution to:
In general with =3 is =3
y=((x-a)^2(b-x)^2)
is a solution to
This will have a peak at (a+b)/2
In general with =5 is =5
y=((x-a)^4(b-x)^4)
is a solution to
This will have a peak at (a+b)/2
the mean is
This dynamic will be dictated by the distance from (a+b)/2, the curve will have a peak at (a+b)/2 and y will asymptotically reach 0 at x=a and x=b.
Beta distribution with ==3 is bell shaped and ==5 is very similar to normal it always has 0 skewness and kurtosis will approach 0 when ==infinity 🙂
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Below is the evolution of Beta as shape parameters change