Calculating event probabilities in mixed, discrete/continuous distributions 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...
 A: There are two different object you might be interested in: 


*

*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).

*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 $b$. 
