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Let $f(x)$ be the probability density function (pdf) of the standard beta distribution on $(0,1)$. And let $f_d(x)$ be the pdf of the generalized beta distribution on $(0,d)$. I know that,

$$f_d(x) = d \cdot f(\frac{x}{d})$$

The cumulative distribution function (cdf) of the generalized beta distribution is,

$$F_d(x) = \int^x_{u=0} f_d(u) \, du.$$

I am tempted to replace $f_d(u)$ by $\left( d \cdot f(\frac{x}{d}) \right)$ in the integral to get,

$$F_d(x) = \int^x_{u=0} df(\frac{u}{d}) \, du = d \cdot \int^x_{u=0} f(\frac{u}{d}) \, du = d \cdot F(\frac{x}{d}).$$

Where $F$ is the cdf of the standard beta distribution. This is wrong because taking $x = d$ gives me $F_d(d) = d$ since $F(1) = 1$.

Question: Where have I gone wrong? Why I can't I replace $f_d(u)$ by $\left( d \cdot f(\frac{x}{d}) \right)$ in the integral?

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1 Answer 1

up vote 2 down vote accepted

OP wrote:


Let $f(x)$ be the probability density function (pdf) of the standard beta distribution on $(0,1)$. And let $f_d(x)$ be the pdf of the generalized beta distribution on $(0,d)$. I know that,

$$f_d(x) = d \cdot f(\frac{x}{d})$$


Scaled Beta

If $X$~$Beta(a,b)$ (a standard Beta on (0,1)), then the pdf of the scaled Beta ... $Y = u(X) = c*X$ will be:

$$f_c(y) = \frac{1}{c} \cdot f(\frac{y}{c}) ..... \text{not .... } c \cdot f(\frac{y}{c})$$

This is because, by the method of transformations, the pdf of $Y=u(X)$ is:

$$ |J| f( u^{-1}(y)) $$

where $x=u^{-1}(y)$ is the inverse of the transformation equation $Y = c X$, and $J=\frac{d u^{-1}(y)}{dy}$ denotes the Jacobian of the transformation . You have calculated the Jacobian of the transformation incorrectly.

P.S. What is described here is more correctly termed a scaled Beta, not a generalised Beta (which has generalised upper and lower bounds).

Addendum

In response to the comments below:

Let $X$ ~ standard Beta with pdf $f(x)$:

        f = (x^(a-1) (1-x)^(b-1))/Beta[a, b];
 domain[f] = {x, 0, 1} && {a > 0, b > 0}; 

... then, using Mathematica here, you wish to confirm:

Integrate[(1/c) f /. x -> y/c, {y, 0, c}, Assumptions -> {a > 0, b > 0, c > 0}] 

1

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Thanks! But $\int_0^c f_c(x) \, dx = 1$ and $\int_0^c \frac{1}{c} \cdot f(\frac{x}{c}) \, dx = \frac{1}{c}$? –  Legendre Jun 5 '13 at 21:03
    
@Legendre I get unity for the latter integral. –  wolfies Jun 5 '13 at 21:05
    
$\int^{x=c}_{x=0} \frac{1}{c} f(\frac{x}{c}) \, dx = \frac{1}{c} \int^{x=c}_{x=0} f(\frac{x}{c}) \, dx = \frac{1}{c} \cdot 1$ because $\int^{x=c}_{x=0} f(\frac{x}{c}) \, dx = 1$ since its the standard beta pdf integrated from $0$ to $1$? –  Legendre Jun 5 '13 at 21:09
1  
Because you need to adjust by the Jacobian ... –  wolfies Jun 5 '13 at 21:24
1  
Just the chain rule :) –  wolfies Jun 5 '13 at 21:33

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