A How to change the support of a probability density function?

[Ad VanderVen]

TL;DR Summary

Given the support [a, b] of a probability density function. How can I change the formula for the probability density function with a support [u, v]?

Given the support [a, b] of a probability density function. How can I change the formula for the probability density function with a support [u, v]? Example: Given the beta distribution with support [a=0,b=1]:
$$\frac{x^{p-1} (1-x)^{q-1}}{Beta(p,q)}$$
Then the beta distribution with support [u,v] is given by
$$\frac{(x-u)^{p-1} (v-x)^{q-1}}{Beta(p,q) (b-a)^{p+q-1}}$$
My question is how should you make the transformation for an arbitrary probability density function?

 

[mathman]

What is the nature of the transformation - simply shifting and stretching the interval? If so, the answer is yes, if done right.

 

[Ad VanderVen]

Yes, the nature of the transformation is simply shifting and stretching the interval, but my question was how should you make the transformation for an arbitrary probability density function?

 

[mjc123]

Define the variable y = (x-u)/(v-u), so 0≤y≤1
For slices of equal area, f(y)dy = f(x)dx
Hence f(x) = f(y).dy/dx = f(y)/(v-u)
Replace x by y in your first equation, and you get f(x) as your second equation. (I assume you meant (v-u) in the denominator, not (a-b).)

 

[Stephen Tashi]

Example: Given the beta distribution with support [a=0,b=1]:
$\frac{x^{p-1} (1-x)^{q-1}}{Beta(p,q)}$
Then the beta distribution with support [u,v] is given by
$\frac{(x-u)^{p-1} (v-x)^{q-1}}{Beta(p,q) (b-a)^{p+q-1}}$

did you mean $\frac{(x-u)^{p-1} (v-x)^{q-1}}{Beta(p,q)(v-u)^{p+q-1}}$ ?

but my question was how should you make the transformation for an arbitrary probability density function?

The phrase "make the transformation" is ambiguous even if you specify the transformation of coordinates has the form $x' = Ax + B$ and that $[a,b]$ is transformed to $[u,v]$.

Given a probability density $f(x)$ and a transformation of coordinates $T(x) = Ax + B$ such that $T(a) = u$ and $T(b) = v$, I think what you want is to define a probability density $g(x)$ satisfying the requirement that $\int_{u_1}^{v_1} g(x) dx = \int_{T^{-1}(u_1)}^{T^{-1}(v_1)} f(x) dx$ for all intervals $[u_1,v_1]$

The $g(x)$ satisfying that requirement is $g(x) = (1/A) f(T^{-1}(x)) = (1/A) f(\frac{x-B}{A})$

In the integral $\int_{T^{-1}(u_1)}^{T^{-1}(v_1)}(1/A) f(\frac{x-B}{A}) dx$ make the change of variables $y = T^{-1}(x) = \frac{x-B}{A}$ , $dy = \frac{dx}{A}$, $dx = A\ dy$.

The integral is transformed to $\int_a^b (1/A) f(y)\ A dy = \int_a^b f(y)dy = \int_a^b f(x) dx$

(If $T$ was a more complicated transformation the choice $g(x) = (1/A) f(T^{-1}(x)$ might not work since a substitution might not produce the simple relation $dx = A\ dy$)

The conditions:
$T(a) = u = Aa + B$
$T(b) = v = Ab + B$
imply
$A = \frac{v - u}{b-a}$
$B = u - Aa = u - \frac{v-u}{b-a}(a)$