I'm writing an application where I have a circular annulus with outer radius $R$ and inner radius $r$. I need to pick a new random radius $r_{new}$ such that $r < r_{new}\leq R$ in a way that radii closer to the outer one are much more likely to be selected. I have a random number generator and can simulate any distribution such that inverse transform sampling can be applied.
Initially I thought I could use an exponential distribution to generate a number in $x\in\left[0,1 \right]$ and then map the outcome to $(r, R]$ the following way: $x\mapsto x*(R-r) + r$. But the problem with this approach is that exponential random variables' support is $[0,+\infty)$ and I'm unsure how to transform the generated number to the desired interval... maybe I should pick a $\lambda$ such that the generated value is less or equal than $1$ with probability $p\geq \alpha$ for large enough $\alpha$ (e.g. $\alpha =0.95$) and then go on generating a new point until my condition is met?
Is there any better approach to my problem? One of my biggest concerns is that I'll have to use the probability distribution I'm going to use to produce a theoretical analysis about results, so for each $r_{new}\in (r,R]$ I would like to know how likely it is to pick a point in $[r_{new}-\delta, r_{new}+\delta]$.
You could choose a random point in annulus with uniform distribution and use its distance to the center as the new radius. Outer region is wider and it is more likely that the new radius is closer to the outer radius.
If x is a uniform random variable between $0$ and $1$, the new radius can be chosen as $r_{new} = \sqrt{x R^2 + (1-x) r^2}$. The probability of choosing a new radius in $[r'-\delta,r'+\delta]$ is $\frac{4r'\delta}{R^2 - r^2}$. Note that the probability grows linearly with respect to $r'$.