Mapping the uniform distribution to an inverse power distribution

Question

I have looked up a lot about how to do inverse transform sampling and still cannot figure this out. I have an inverse power distribution $f(x)=\frac{1}{(x+1)^n}$ where $n$ is an integer greater than 1. I've done the integral from 0 to infinity and gotten $\frac{-1}{n-1}$, and know the CDF from negative infinity to x is $\frac{-1}{(n-1)(x+1)^{n-1}}$. My understanding is that all I need to do is get the inverse of the CDF and replace the variable (random variable I guess) with U(0,1). In particular I want it for $n=4$. I've gotten both $\frac{-ln(U)}{ln(3)}-2$ and $(\frac{-1}{(n-1)U})^{-3}-1$. These may not be exactly right - I've done it many times and gotten slightly different answers, but the point is neither of these seem remotely close to reproducing the original distribution.

Answer 1

So yes $F(x) = \frac{-1}{(n-1)(x+1)^{n-1}}$ just solve for $x^{-1}(F=U[0,1])$ and

$$x=\left( (1-n)U[0,1]\right)^{\frac{1}{1-n}} -1$$

Answer 2

Let $$f_X(x \mid n) \propto (1+x)^{-n}, \quad x > 0, \; n > 1.$$ Then we require $$1 = \int_{x=0}^\infty f_X(x \mid n) \, dx \propto \left[\frac{1}{(1-n)(1+x)^{n-1}}\right]_{x=0}^\infty = 0 - \frac{1}{1-n} = \frac{1}{n-1}.$$ Therefore, the correct constant of proportionality is $n-1$, namely $$f_X(x \mid n) = (n-1)(1+x)^{-n}, \quad x > 0, \; n > 1,$$ and the CDF is simply $$F_X(x \mid n) = \int_{t=0}^x f_X(t \mid n) \, dt = \left[- \frac{1}{(1+t)^{n-1}} \right]_{t=0}^x = -\frac{1}{(1+x)^{n-1}} + 1 = 1 - (1+x)^{1-n}.$$ I do not understand how you could have obtained your result, as it is nonsensical: the CDF requires $F_X(x \mid n) \to 1$ as $x \to \infty$, and both the PDF and CDF need to be nonnegative over the support of $X$.

All that remains is for you to calculate $$X = F^{-1}_X(U \mid n)$$ for some $U \sim \operatorname{Uniform}(0,1)$; this random variable will have the distribution of $X$.