R: Left-censoring a probablity distribution function

Question

I would like to left-censor (at zero) a probablity distribution function, but I just can't find a way to implement this in R. I have reviewed previous questions about censoring, but none have provided a solution to my problem. I already have a CDF with $\mu$ and $\sigma$ parameters and this CDF has to be censored at zero.

Answer 1

if you try to find conditional distribution of X s.to X >0, then this will be p(X |X >0) = p1(X) / (1-F(0)), p1(X) is density of X * I(x>0). It will integrate to 1 as p1(X) will integrate to 1 - F(0). The answer above just need to clarify that truncated density for normal distribution will be 2 * Normal density at x >0, and 0 for x<0, so integrating over all values is 2 * 1/2 =1 .

Answer 2

I think you mean a truncated distribution. A truncated distribution and one which arises from censoring both conceptually involve "cutting" a distribution past a certain threshold. However, censoring refers to the unobserved event time falling past a certain timepoint.

To "implement" a probability distribution in R - in other words to calculate the density function - you just make use of the following result:

Let $\theta$ be the "left" truncated value, meaning no observation less than $\theta$ can be observed. Let $X$ be a random variable distributed according to the "non-truncated" distribution function.

$$ f_Y(y) = \left\{ \begin{array}{cccc} \text{if} & y > \theta & \text{then} & \frac{f_X(y)} {1-F_X(\theta)} \\ &\text{else} && 0 \end{array} \right. $$

In other words, we just normalize the density according to the cumulative probability of the remaining distribution that we haven't cut.

In R this is trivial. Suppose we want the probability density of $y=1$ in a standard normal density that we have cut at 0. We simply call:

dnorm(1, 0, 1) / pnorm(0, 0, 1)

Which is exactly equal to dnorm(1,0,1)*2 which is what we expect: a doubling of the normal density since we have cut it in half.