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)

$y&space;=&space;\frac{((x-a)^{(c-1)}&space;(b-x)^{(d-1)})}{(b-a)^{(c+d-1)}}$

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

${y}'=y\frac{(\alpha&space;+\beta&space;-2&space;)-(\alpha&space;-1)}{(x-1)x}$

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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.

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When $\alpha_1$ and  $\alpha_2$ are not equal the curve doesn’t approach 0 asymptotically at that end but has a peak closer to that end.

This means if $\alpha_1$ is smaller then the peak will be close to a.

Beta curve with $\alpha_1$=1 is $\alpha_2$=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:

$\frac{ \alpha - 1 }{ \alpha + \beta - 2 } \le \text{median} \le \frac{ \alpha }{ \alpha + \beta } ,$
$\text{median} \approx \frac{\alpha - \tfrac{1}{3}}{\alpha + \beta - \tfrac{2}{3}} \text{ for } \alpha, \beta \ge 1.$

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In general with $\alpha_1$=2 is $\alpha_2$=3

y=((x-a)(b-x)^2)

is a solution to

with $\alpha_1$=2 is $\alpha_2$=3 a=0 b=5

y=(x(5-x)^2)

is a solution to:

In general with $\alpha_1$=2 is $\alpha_2$=3 and a=0

y=((x)(b-x)^2)

is a solution to:

In general with $\alpha_1$=3 is $\alpha_2$=2 and a=0

y=((x)^2(b-x))

is a solution to:

In general with $\alpha_1$=3 is $\alpha_2$=3

y=((x-a)^2(b-x)^2)

is a solution to

This will have a peak at (a+b)/2

In general with $\alpha_1$=5 is $\alpha_2$=5

y=((x-a)^4(b-x)^4)

is a solution to

This will have a peak at (a+b)/2

the mean is $Mean=a+\frac{\alpha_1}{\alpha_1+\alpha_2}(b-a)$

$mode=\frac{\alpha_1-1}{\alpha_1+\alpha_2-2}(b-a))$

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 $\alpha_1$=$\alpha_2$=3 is bell shaped and $\alpha_1$=$\alpha_2$=5 is very similar to normal it always has 0 skewness and kurtosis will approach 0 when $\alpha_1$=$\alpha_2$=infinity 🙂

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Below is the evolution of Beta as shape parameters change

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