Why are Truncated Probability Distributions important in statistics?

Recently, I was reading about "Truncated" Probability Distributions. As the name suggests, a Truncated Probability Distributions is created by taking some Probability Distribution Function and restricting (i.e. "truncating") the range that the random variable can take. However, instead of simply defining the original Probability Distribution Function over a "truncated range" - we end up creating a new Probability Distribution Function over this truncated range (i.e. the Truncated Probability Distribution Function):

enter image description here

My Question: Why is it necessary to create a "Truncated Probability Distribution Function" from the original Probability Distribution Function? Why can't we just restrict the range of the original Probability Distribution Function and perform all our inferences on the original Probability Distribution Function - or is this because doing so would result the in the original Probability Distribution Function not integrating to "1"? Are there any real benefits of using the Truncated Probability Distribution compared to the original Probability Distribution Function? Are there any applications or instances in the real world where it is absolutely necessary to use the Truncated Probability Distribution Function compared to the original Probability Distribution Function?



  • 2
    $\begingroup$ A lot of practical situations imply the support of the observations to be bounded. When this bound, eg an upper bound $\eta$ say, is known, the statistical modelling should account for it, as eg $\mathbb P(X>\eta)$ should be zero. Using standard distributions and restricting their support is a natural approach to such modelling. $\endgroup$
    – Xi'an
    Commented Nov 8, 2021 at 7:30
  • $\begingroup$ @ Xi'an : thank you for your reply! $\endgroup$
    – stats_noob
    Commented Nov 8, 2021 at 17:29
  • $\begingroup$ Perhaps google truncation vs. censoring just so you can avoid the trap of confusing the two concepts. (Not saying you have, want to prevent it.) $\endgroup$
    – BruceET
    Commented Nov 9, 2021 at 15:24

2 Answers 2


As you seem to have (almost) guessed, the trucated distribution comes about from imposing the restriction on the support and then multiplying by a scaling constant to make the restricted density integrate/sum to one. That is all we are doing when we create a truncated version of an initial distribution.

As to when this is useful, it is useful anytime we want to condition on a restricted range for the observable random variable. This occurs in conditional probability problems when we specify an initial distribution and then condition on the value being in some restricted part of the allowable range. It also occurs in cases where we use an approximating distribution to approximate another distribution on a smaller support. Finally, it also occurs in problems with censored data, when we condition on the non-censored part of the data range.


@Xi'an has mentioned the case in which it makes sense to have $P(X > \eta) = 0,$ so that $X$ has a natural maximum value.

In many applications of the normal distribution it is natural to have $P(X < 0) = 0.$ Heights (in inches) of female students might be modeled as $X\sim\mathsf{Norm}(\mu = 65, \sigma=3)$. Obviously, there can be no such thing as a negative height, so we require $P(X < 0) = 0.$ But this is often left without formal truncation because $P(X < 0) \approx 0,$ without any adjustment. The value $0$ is more than twenty standard deviations below the mean, and the probability more than four standard deviations below the mean is already negligibly small (about $.000032.)$ In R:

pnorm(0, 65, 3)
[1] 2.116566e-104
pnorm(65 - 12, 65, 3)
[1] 3.167124e-05

However, if we try to model some inherently positive distribution as $\mathsf{Norm}(\mu=5, \sigma = 3),$ we need to deal with an implied impossible probability of negative values amounting to almost $0.05,$ which could lead to noticeable errors.

pnorm(0, 5, 3) 
[1] 0.04779035

The cure is to increase the density function so that its positive probability will be $1.$ So we inflate the density $f(x)$ of $\mathsf{Norm}(5, 3)$ to be $g(x) = f(x)/0.9522,$ for $x>0,$ and all computations using the density function $g(x)$ will be accurate to at least four places.

1 - pnorm(0, 5, 3)
[1] 0.9522096

You have a figure in your Question along these lines. Here are the plots of densities $f(x)$ (brown) and $g(x)$ (blue), mentioned above.

![enter image description here

R code for figure.

hdr = "Densities of NORM(5,3) and its Truncated Version (blue)"
curve(dnorm(x,5,3), -3, 15, ylim = c(0,.14), lwd=2, 
  col="brown", ylab="Density", main=hdr)
 curve(dnorm(x,5,3)/(1-pnorm(0,5,3)), 0, 14, add=T, lwd=2, col="blue")
 abline(h=0, col="green2")
 abline(v=0, col="green2")

Thus, using the truncated version $P(4 \le X \le 5) = 0.1371$ (blue density), not $0.1306$ (brown).

diff(pnorm(c(4,5), 5, 3))
[1] 0.1305587
diff(pnorm(c(4,5), 5, 3))/0.9522
[1] 0.1371126

Note: In practice is may be worthwhile trying to find a distribution that naturally has support on the positive real line (e.g, a gamma distribution), instead of using a truncated normal distribution.


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