Random samples obeying the exponential distribution can be generated by the inverse sampling technique by using the quantile function of the exponential distribution:

x=F−1(u)=−1λln(u)

where u is a sample drawn from the uniform distribution on the unit interval (0,1).

In OpenFOAM software, a distribution model called

`exponential`

(here) can be used to generate exponential-distribution random samples, and its users can, supposedly,choose a minimum and maximum valuefor the exponential-distribution samplesprior tothe random-number generation.The governing expression implemented into this software is as follows:

x=−1λln[exp(−λtmin)+u{exp(−λtmax)−exp(−λtmin)}]

where tmin and tmax are user-defined minimum and maximum values, respectively.

Two (rather broad) questions arise:

- Have you ever come across the above expression (or similar one) in the literature for generating exponential-distribution random samples with a given min-max? Or do you think this expression looks like (or is) a heuristic solution?
- If heuristic, would you suggest a way to carry out verification tests on this expression to test whether the expression produces samples obeying the exponential distribution within [min,max]? (Plotting normalised histogram (i.e. counts in a bin divided by the number of observations times the bin width) and comparing it with the analytical exponential distribution seem to be problematic due to the min/max limits).

**Answer**

You describe truncation to an interval. I will elaborate.

Suppose X is any random variable (such as an exponential variable) and let FX be its distribution function,

FX(x)=Pr

For an interval [a,b], the *truncation* limits X to that interval. That lops off some probability from X, namely the chance that X either is less than a or greater than b. The chance that is left is

\Pr(X\in[a,b]) = \Pr(X\le b) – \Pr(X\le a) + \Pr(X=a) = F_X(b) – F_X(a) + \Pr(X=a).

Thus, to make the total probability come out to 1, the distribution function for the truncated X must be zero when x\lt a, 1 when x\ge b, and otherwise is

F_X(x;a,b) = \frac{\Pr(X\in[a,x])}{\Pr(X\in[a,b])}= \frac{F_X(x) – F_X(a) + \Pr(X=a)}{F_X(b) – F_X(a) + \Pr(X=a)}.

When you can compute the inverse of the distribution function–which almost always means X is a continuous variable–it’s straightforward to generate samples: draw a uniform random probability U (from the interval [0,1], of course) and find a number x for which F_X(x) = U. This value is written

x = F^{-1}_X(U).

F_X^{-1} is called the “percentage point function” or “inverse distribution function.”

For example, when X has an Exponential distribution with rate \lambda \gt 0,

U = F_X(x) = 1 – \exp(-\lambda x),

which we can solve to obtain

F_X^{-1}(U) = -\frac{1}{\lambda}\log(U).

This is called “inverting the distribution” or “applying the percentage point function.”

It turns out–and this is the point of this post–that *when you can invert F_X, you can also invert the truncated distribution.* Given U, this amounts to solving

U = F_X(x;a,b) = \frac{F_X(x)-F_X(a)}{F_X(b) – F_X(a)},

because (since we are now assuming X is continuous) the terms \Pr(X=a)=0 drop out. The solution is

x = F_X^{-1}(U;a,b) = F_X^{-1}\left(F_X(a)+\left[F_X(b) – F_X(a)\right]U\right).

That is, the *only* change is that after drawing U, you must rescale and shift it to make its value lie between F_X(a) and F_X(b), and *then* you invert it.

This yields the second formula in the question.

An equivalent procedure is to draw a uniform value V from the interval [F_X(a),F_X(b)] and compute F_X^{-1}(V). This works because the scaled and shifted version of U has a uniform distribution in this interval. I use this method in the code below.

The figure illustrates the results of this algorithm with \lambda=1/2 and truncation to the interval [2,7]. I think it alone is a pretty good verification of the procedure.

The `R`

code is general-purpose: replace `ff`

(which implements F_X) and `f.inv`

(which implements F^{-1}_X) with the corresponding functions for *any* continuous random variable.

```
#
# Provide a CDF and its percentage point function.
#
lambda <- 1/2
ff <- function(x) pexp(x, lambda)
f.inv <- function(q) qexp(q, lambda)
#
# Specify the interval of truncation.
#
a <- 2
b <- 7
#
# Simulate data and truncated data.
#
n <- 1e6
x <- f.inv(runif(n))
x.trunc <- f.inv(runif(n, ff(a), ff(b)))
#
# Draw histograms.
#
dx <- (b - a) / 25
bins <- seq(a - ceiling((a - min(x))/dx)*dx, max(x)+dx, by=dx)
h <- hist(x.trunc, breaks=bins, plot=FALSE)
hist(x, breaks=bins, freq=FALSE, ylim=c(0, max(h$density)), col="#e0e0e0",
xlab="Value",
main="Histogram of X and its truncated version")
plot(h, add=TRUE, freq=FALSE, col="#2020ff40")
abline(v = c(a,b), lty=3, lwd=2)
mtext(c(expression(a), expression(b)), at = c(a, b), side=1, line=0.25)
```

**Attribution***Source : Link , Question Author : Herpes Free Engineer , Answer Author : whuber*