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I'm seeing a distribution where the pdf $F(x)$ for random variable $x$ is like this:

$$ F(x) = (bx)^a + o(x^a)$$

where $a>0$ and $b>0$.

EDIT: For context, this distribution appears near the beginning of this paper: https://www.sciencedirect.com/science/article/pii/0377221778900449 . It is accompanied by 3 assumptions:

  • $\inf \{F(x) > 0 \}=0$
  • $F(x) = (bx)^a + o(x^a),\,\, a>0, b>0$
  • $\int_0^{\infty} x*[1-F(x)]dx < \infty$

I understand that $o(x^a)$ is the highest upper bound of $x^a$. It doesn't seem to be one of the common distributions like geometric, uniform, normal, etc. I'm having difficulty identifying this distribution.

underdog987
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  • A beta distribution has two parameters but having two parameters does not make a beta distribution. – Henry Jun 21 '23 at 16:53
  • I see, thank you. – underdog987 Jun 21 '23 at 17:00
  • Is the support restricted? If the density is $(bx)^a$ on $[0,1]$ with no $o(x^a)$ term, then $b=(a+1)^{1/a}$ and the density is $(a+1)x^a$ which is a Beta distribution with parameters $\alpha =a+1$ and $\beta=1$, which some people call a "standard power function distribution". – Henry Jun 21 '23 at 17:00
  • Yes the support is restricted to $x>0$ and $x<\infty$ – underdog987 Jun 21 '23 at 17:23
  • If the support is on $(0,\infty)$ then $(bx)^a$ is not a probability density as it does not integrate to $1$: it is increasing in $x$ for $a>0,b>0$ – Henry Jun 21 '23 at 19:33
  • @Henry, My apologies for not providing enough detail initially. I've edited the question, which perhaps will clarify things. If you can elaborate a little more on what you mean, I'd like to understand better. – underdog987 Jun 21 '23 at 23:26
  • It is a paywalled article on the travelling salesman problem, so difficult to comment. It is also unclear whether your $F(x)$ is a density or a cumulative distribution function - the use of $1-F(x)$ in an assumption would suggest the latter. – Henry Jun 22 '23 at 10:22

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