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During my studies we define the Pareto distribution as

$F(x) = 1-(1+\frac{x}{\beta})^{-\alpha}$

with density function

$ f(x)= 1-(1+\frac{x}{\beta})^{-\alpha} *1_{x>o} $.

I don´t understand how we get this density function. If i calculate the derivative of $F(x)$ i get:

$F^\prime (x) =f(x) = - -\alpha(1+\frac{x}{\beta})^{-\alpha-1} * \frac{1}{\beta} = \frac{\alpha}{\beta}(1+\frac{x}{\beta})^{-(\alpha+1)}$

i have tryed to reformulate the formula and get so far:

$f(x)=\frac{\alpha}{\beta}(1+\frac{x}{\beta})^{-(\alpha+1)} = \frac{\alpha}{\beta+x} \frac{\beta+x}{\beta} (1+\frac{x}{\beta})^{-(\alpha+1)} = \frac{\alpha}{\beta+x} (1+\frac{x}{\beta}) (1+\frac{x}{\beta})^{-(\alpha+1)} = \frac{\alpha}{\beta+x} (1+\frac{x}{\beta})^{-\alpha}$

can someone tell me how to get the formula? I'm not getting anywhere. Thanks in advance!

Mufasa
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    Your formula for $f(x)$ is correct but the one given in the definition is not correct. Since $1-(1+\frac x {\beta})^{-\alpha)1_{x>0}} \to 1$ as $ x \to \infty$ this is, not even a denisity function. – Kavi Rama Murthy Jan 21 '20 at 08:19
  • thank you very much! I was also wondering. – Mufasa Jan 21 '20 at 08:27

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