Good morning. I need to create a pdf and estimate its parameters by the MLE method. The distribution is a power law ($f(x)=C(k)\;x^{-k}$) and I need to estimate the $k$ parameter and $C(k)$ is determinated by the normalization condition $\int_a^b\;C(k)\;x^{-k}=1$. I have tried to define the power distribution with ProbabilityDistribution function using, for example, C=5, but when I want to plot it, Mathematica doesn't display the plot. What I've been writing is
distr=ProbabilityDistribution[5/x^1.05,{x,0,Infinity}]
PDF[distr,x]
Plot[%,{x,0,1}]
NOTE: when I try to define the distribution function I'm setting $k=1.05$, just for knowing if I'm doing it well.
Could you help me, please?
Thanks for your answers.
Thanks for your help. I just want to understand something.
You told me that I can aproximate my power law function as a Pareto Distribution. When I search for Pareto Distribution I found that Distribution very different from my power law, but I found too that Mathematica uses the form $$f(x)=\left(\frac{x}{k}\right)^{-\alpha}$$ when $k$ is the constant. If I rewrite this Pareto Distribution form in order to compare with the power law that I'm using $f(x)=k(\alpha)x^{-\alpha}$, I supose that it would be $$f(x)=k^{\alpha}x^{-\alpha}$$ Am I right? Now, Bob Hanlon defined that Distribution with $a$ instead of $k$ and $k-1$ instead of $\alpha$. I mean, $$f(x)=\left(\frac{x}{a}\right)^{-(k-1)}=a^{k-1}x^{-(k-1)}$$ Am I right? What I mean is, the Pareto Distribution is my power law function if I take $k$ (or $a$, according to Bob) as my $k(\alpha)$ (or $a(k-1)$) and $\alpha$ (or $k-1$) as my exponent? If I do that, is the Pareto Distribution calculated by Mathematica tha same power law function tha I need?
Thank for make me this most clear.
5/x^1.05
to integrate to 1, do you not need for the lower limit to be10^40
? In other words,Integrate[5/x^(105/100), {x, 10^40, \[Infinity]}] = 1
. $\endgroup$