I want to compute the CDF of a Pareto distribution
In particular, I have a Pareto distribution with p.d.f.
$$f(x) = \lambda x^{-(\lambda + 1)},$$
and I want to find the CDF of $f$. When I perform straightforward integration, I get
$$F(x) = \int_{-\infty}^{w} \lambda w^{-\lambda + 1} \mathop{dw} = (-w^{-\lambda})\Big|_{-\infty}^{x} = -x^{-\lambda}$$
Is this right? Wolfram Alpha says my integration is correct, but it differs from the CDF provided on the Pareto Distribution's Wikipedia page. The wiki page suggests it should be $1 - x^{-\lambda}$. Any help is appreciated.
$\endgroup$2 Answers
$\begingroup$In general if $X$ has Pareto distribution with scale parameter $x_m>0$ and shape parameter $\lambda>0$ then its density is$$ f_X(x) = \lambda x_m^\lambda x^{-(\lambda+1)}\cdot\mathsf 1_{(x_m,\infty)}(x). $$For any $x>x_m$, it follows by definition the density of an absolutely continuous random variable that the distribution function is given by\begin{align} F_X(x) :&= \mathbb P(X\leqslant x) \\ &= \int_{-\infty}^y f_X(y)\ \mathsf dy\\ &= \int_{x_m}^x \lambda x_m^\lambda y^{-(\lambda+1)}\ \mathsf dy\\ &= x_m^\lambda (x_m^{-\lambda}-x^{-\lambda})\\ &= 1 - \left(\frac{x_m}{x}\right)^\lambda. \end{align}In the case where $x_m=1$, this reduces to $$F_X(x) = \left(1-x^{-\lambda}\right)\cdot\mathsf 1_{(1,\infty)}(x).$$
$\endgroup$ $\begingroup$For pdf you have$$f(x)=\begin{cases} 0, & x\leqslant 1\\ \lambda x^{-(\lambda + 1)}, & x> 1 \end{cases}$$so$$\int\limits_{-\infty}^{x} f(t)\,dt = \int\limits_{1}^{x} \lambda t^{-(\lambda + 1)}\,dt = 1 - x^{-\lambda} $$and we obtain$$F(x)=\begin{cases} 0, & x\leqslant1\\ 1-x^{-\lambda }, & x>1 \end{cases}$$
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