Calculating event probabilities in mixed, discrete/continuous distributions

Question

This is a simple question. I am dealing with a "clipped" normal distribution -- say, $N(0,0.5)$ clipped between $[-1,1]$. I would like to calculate the "probability" of a sample, but I know that in $(-1,1)$, the probability of a single event is 0. However, the probability of 1 is $1-\text{cdf}(1,0,0.5)$ and -1 is $\text{cdf}(-1,0,0.5)$. How does one typically reconcile these sorts of differences?

My intuition would be to use epsilon balls within $(-1,1)$, but it's not quite the RIGHT thing to do...

Answer 1

There are two different object you might be interested in:

1. the truncated normal distribution, which is a Gaussian whose range is restricted to lie within $[a,b]$. This is a density, so it's continuous (no need for epsilon balls).
2. a random variable $Z$ that results from drawing a Gaussian random variable $X$ and then "thresholding" it, setting $Z=a$ if $X<a$ and to $Z=b$ if $X>b$, to $Z=X$ if $a\leq X \leq b$.

I'm guessing it's the latter object you're interested in. As you've noted, the probability of getting a sample on the border is non-zero: $P(Z=a) = \Phi((a-\mu)/\sigma)$ and $P(Z=b)=\Phi((\mu-b)/\sigma)$, where $\Phi$ is the standard Gaussian cdf, $\mu$ is the mean, and $\sigma$ the standard deviation.

However, there is zero probability of getting any particular sample within the range $(a,b)$, but the probability density within this range is simply that of the original Gaussian. (The probability of getting a sample from somewhere within this range is simply $1-P(Z=a)-P(Z=b)$.

The resulting object is therefore defined by "point masses" at $a$ and $b$ and a continuous density between $a$ and $b$.