# How to sample and compute the likelihood from a Mollified Uniform distribution?

I want to draw samples from the mollified Uniform distribution presented in another Cross Validated thread, cf the answer from whuber. What is the best way to do so?

I have tried drawing $$\mu \sim U[0, 1]$$, and then drawing $$x|\mu \sim \mathcal{N}(\mu, \sigma)$$, where $$\sigma$$ is the standard deviation of the mollifier. It seems to work, cf the histogram below for 1,000,000 points with $$\sigma=0.1$$.

If true, could someone explain why this work, please? And if I get a new point, let say $$x_{new} = 1.04$$, how can I compute the likelihood of this observation?

Another way to view what you're doing is as $$\mu \sim U[0, 1] \quad \delta \sim \mathcal N(0, \sigma) \quad X = \mu + \delta .$$ The density of the sum of two variables is the convolution of their densities, which is exactly how @whuber defined the mollified uniform distribution here.
Evaluating the pdf at a single point is a little more complicated. If $$X$$ is much farther from either $$0$$ or $$1$$ than $$\sigma$$, i.e. $$\min \{ \lvert X - 0 \rvert, \lvert X - 1 \rvert \} \gg \sigma$$, then for practical purposes you can simply treat the likelihood as either 0 or 1. In your example, though, it seems like your $$\sigma$$ is fairly large. In that case, your density is the value of the convolution $$f(x) = \int_0^1 \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2\sigma^2} (x - \mu)^2} \mathrm d \mu .$$ This integral basically asks, "what's the probability density of seeing $$x$$ given that my original uniform sample was $$\mu$$", and marginalizes over all possible values of $$\mu$$.
One way to compute this integral is to simply notice that, while we defined it for $$x$$ being the normal variable, it's exactly the same formula to think of us as computing the probability that a normal random variable $$\mu \sim \mathcal N(x, \sigma)$$ is in the interval $$[0, 1]$$: $$f(x) = \Phi\left( \frac{1 - x}{\sigma} \right) - \Phi\left( \frac{-x}{\sigma} \right) .$$ Indeed, we can see that as $$\sigma \to 0$$, when $$x \in (0, 1)$$ it'll become $$\Phi(\infty) - \Phi(-\infty) = 1 - 0 = 1$$, when $$x > 1$$ it'll be $$\Phi(-\infty) - \Phi(-\infty) = 0$$, and when $$x < 0$$ it'll be $$\Phi(\infty) - \Phi(\infty) = 0$$: a uniform, like we wanted. (The exception is that right at $$x = 1$$ or $$x = 0$$ it'll be $$\tfrac12$$, but this single point doesn't really matter.)
• Oh, I see... The trick with switching the roles of $\mu$ and $x$ is really smart! Thank you @Danica. Apr 22, 2022 at 6:29