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In my papers the probability density for a burr distribution is given as

$f(x) = \dfrac{\gamma \tau \alpha^{\gamma}x^{\tau - 1} }{(\alpha + x^{\tau})^{\gamma + 1}}$

however i have encountered this definition

$f(x \mid \alpha,c , k)= \dfrac{\frac{kc}{\alpha}(\frac{x}{\alpha})^{c-1} }{ \Big( 1+ (\frac{x}{\alpha})^c \Big)^{k+1}}$

Iam trying to write the former definition as the first one i gave above , but setting $\tau =c$ and $\gamma = k$ won't do it. anyone?

***update

what Iam trying to say that if it is possible to do this?

$\dfrac{\frac{kc}{\alpha}(\frac{x}{\alpha})^{c-1} }{ \Big( 1+ (\frac{x}{\alpha})^c \Big)^{k+1}} =\dfrac{\dfrac{\gamma \tau}{\alpha} (\dfrac{x}{\alpha})^{\tau-1}}{(1 + (\frac{x}{\alpha})^{\tau})^{\gamma+1}} = \dfrac{\gamma \tau x^{\tau-1}}{\alpha^{\tau} (1 + (\frac{x}{\alpha})^{\tau})^{\gamma+1}} = \dfrac{\gamma \tau x^{\tau-1}}{\alpha^{\tau} (1 + (\frac{x}{\alpha})^{\tau}) (1 + (\frac{x}{\alpha})^{\tau})^{\gamma}} = \cdots ?? \cdots = \dfrac{\gamma \tau \alpha^{\gamma}x^{\tau - 1} }{(\alpha + x^{\tau})^{\gamma + 1}}$

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  • $\begingroup$ In what sense will those substitutions "not do it"? (If you apply the principles in my answer at stats.stackexchange.com/a/136458 it should be almost immediately obvious that these two formulas must describe the same distributional families, because they have the same underlying functional forms.) $\endgroup$
    – whuber
    Feb 13, 2015 at 20:33
  • $\begingroup$ @whuber please check the edited text above $\endgroup$
    – Danny
    Feb 13, 2015 at 21:26

1 Answer 1

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The issue that is probably confusing you is that $\alpha$ doesn't represent the same thing in the two equations.

$$f(x) = \dfrac{\gamma \tau \alpha^{\gamma}x^{\tau - 1} }{(\alpha + x^{\tau})^{\gamma + 1}}$$

Let $\alpha=\beta^\tau$

$$= \dfrac{\gamma \tau \beta^{\tau\gamma}x^{\tau - 1} }{(\beta^\tau + x^{\tau})^{\gamma + 1}}$$

$$= \dfrac{\gamma \tau \beta^{\tau\gamma+\tau-1}(\frac{x}{\beta})^{\tau - 1} }{\beta^{\tau\gamma+\tau}(1 + (\frac{x}{\beta})^{\tau})^{\gamma + 1}}$$

$$= \dfrac{\frac{\gamma \tau }{\beta}(\frac{x}{\beta})^{\tau - 1} }{(1 + (\frac{x}{\beta})^{\tau})^{\gamma + 1}}$$

which is of the required form.

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