Let $x_i$ represent samples from a Truncated Exponential distribution between $0$ and $1$, with rate parameter $\lambda$.


$\tilde x = \dfrac{\sum_{i=1}^{n}x_i}{n}$

What is the PDF of $\tilde x$?

  • 4
    $\begingroup$ Have you tried for $n=2$? Where do you get stuck? Please provide more details about your attempt so far. $\endgroup$
    – Xi'an
    Mar 22, 2017 at 17:07
  • 2
    $\begingroup$ Given that the solution is going to be an $n$-part piecewise pdf on $(0,1)$, writing it down is likely to get messy. It may be possible to obtain an excellent approximation ... it would be helpful if you could note whether (a) you are interested in small $n$ or large $n$, and (b) if you know anything about $\lambda$ (close to 0 or 1)? $\endgroup$
    – wolfies
    Mar 23, 2017 at 13:07
  • $\begingroup$ I'm without time to answer/evaluate the comments and answers now, but regarding the points in @wolfies a) I'm interested in large $n$ b) $\lambda$ should be bigger than or equal to 1. Thanks $\endgroup$ Mar 24, 2017 at 12:31
  • $\begingroup$ My answer in a closely related thread provides at least four techniques for analyzing and understanding this sum. $\endgroup$
    – whuber
    Jan 16 at 16:00

3 Answers 3


Updated answer

The solution is going to be an $n$-part piecewise pdf on (0,1). Given that the OP has noted he is interested in large $n$, expressing the exact pdf of the sample mean is likely to get messy. For large $n$ (as given), one should obtain an excellent neat simple approximation via the Central Limit Theorem.


Let $X \sim \text{TruncatedExponential}(\lambda)$ (truncated above at 1), with pdf:

$$f(x)=\frac{\lambda e^{-\lambda x}}{1-e^{-\lambda }} \quad \text{ for } 0 <x<1$$


$$\mathbb{E}[X] = \frac{1}{\lambda }+\frac{1}{1-e^{\lambda }} \quad \quad \text{and} \quad \quad \text{Var}(X) = \frac{1}{\lambda ^2}-\frac{e^{\lambda }}{\left(e^{\lambda }-1\right)^2}$$

Then if the random variables ${X_1, X_2, ...}$ are iid, by the Central Limit Theorem:

$$\bar{X}_n \;\overset{a} \sim\; N\left(\mathbb{E}[X] ,\frac{\text{Var}(X)}{n}\right)$$

All done. The following diagram compares:

  • the EXACT distribution of the sample mean (blue curve) with
  • the asymptotic Normal distribution (dashed red curve)

when the sample size is just $n = 6$:

enter image description here

Even with this tiny sample size, the simple Normal approximation already performs well in the $\lambda = 1$ case (LHS diagram). If $\lambda$ becomes larger, the distribution becomes more peaked and shifts to the left, and larger sample sizes will be needed ... but will still perform extremely well for large $n$.

For comparison, the exact pdf when $n = 6$ is:

enter image description here

Derivation of Exact PDF

To illustrate the calculation of the exact pdf, consider first two independent Truncated Exponential variables, say $X$ and $Y$ which will have joint pdf $f(x,y)$:

enter image description here

Then, the cdf of $S=X+Y$, i.e. $P(X+Y<s)$ is:

enter image description here

where I am using the Prob function from the mathStatica package for Mathematica to automate the calculation.

The pdf of $S=X+Y$is just the derivative of the cdf wrt $s$:

enter image description here

Here is a plot of the exact pdf just derived in the $n= 2$ case (here for the sample sum) when $\lambda = 1$:

enter image description here

One can derive the exact pdf of the sample sum (or sample mean) for larger $n$ in this same manner ... though for large $n$, the Central Limit Theorem will rapidly become your friend.

  • $\begingroup$ Indeed, I missed one indicator when computing $f_2$, thank you. $\endgroup$
    – Xi'an
    Mar 24, 2017 at 17:39

[There was indeed a mistake in the earlier derivation!]

If $f_n$ denotes the density of $s_n=x_1+\ldots+x_n$, it satisfies the recursion \begin{align*} f_1(s) &= \lambda e^{-\lambda s} \big/ 1-e^{-\lambda}\mathbb{I}_{(0,1)}(s)\\ f_n(s) &= \int_0^{1} f_{n-1}(s-y) \dfrac{\lambda e^{-\lambda y} }{ 1-e^{-\lambda}}\text{d}y\mathbb{I}_{(0,n)}(s)\\ \end{align*} The computation for $n=2$ leads to \begin{align*} f_2(s)&=\dfrac{\lambda^2}{(1-e^{-\lambda})^2}\int_0^{1} e^{-\lambda y}e^{-\lambda (s-y)}\mathbb{I}_{(0,1)}(s-y)\text{d}y\mathbb{I}_{(0,2)}(s)\\ &=\dfrac{\lambda^2}{(1-e^{-\lambda})^2}\int_{0\vee (s-1)}^{1\wedge s} e^{-\lambda y}e^{-\lambda (s-y)}\text{d}y\mathbb{I}_{(0,2)}(s)\\ &=\dfrac{\lambda^2e^{-\lambda s}}{(1-e^{-\lambda})^2} \left[{1\wedge s}-{0\vee (s-1)}\right]\\ &=\dfrac{\lambda^2e^{-\lambda s}}{(1-e^{-\lambda})^2}\left[s\mathbb{I}_{(0,1)}(s)+(2-s)\mathbb{I}_{(1,2)}(s)\right]\\ \end{align*}which does not show much promise for the general case.


The pdf of a $(0,1)$-trunctated exponential distribution with rate parameter $\lambda$ can be written as \begin{align} f(x)&=I_{(0,1)}(x)\frac{\lambda e^{-\lambda x}}{1-e^{-\lambda}} \\&=(1-p)I_{(0,\infty)}(x)\lambda e^{-\lambda x}+pI_{(1,\infty)}(x)\lambda e^{-\lambda(x-1)}, \end{align} where $1-p=1/(1-e^{-\lambda})$. Although one weight is negative, it follows that we can treat this as a signed mixture of two exponential distributions and proceed as if both weights were positive.

The sum of $n$ such truncated exponentials can then be seen as mixture having a total of $n+1$ components (although again, some of the associated weights are actually negative). The $i$th component is the sum of $n$ exponentials out of which $i$ are shifted one unit to the right. The $i$th component is thus a gamma distributions with shape parameter $n$ shifted $i$ to the right. The overall pdf is $$ f_n(x)=\sum_{i=0}^n {n \choose i}p^i(1-p)^{n-i}I_{(i,\infty)}(x)\frac{\lambda^n}{(n-1)!}(x-i)^{n-1}e^{-\lambda (x-i)}. $$ For $n=3$ and $\lambda=1$, the pdf has the following shape.

enter image description here

R code:

# density function of the sum
dsum01exp <- function(x, n, lambda=1) {
  p <- 1 - 1/(1 - exp(-lambda))
  d <- 0
  for (i in 0:n) {
    d <- d + choose(n, i)*p^i*(1 - p)^(n - i)*dgamma(x-i,shape=n,rate=lambda)
# random sample from (0,1)-trunctated exponential
r01exp <- function(n, lambda=1) {
# histogram of simulated sums vs. theoretical pdf
n <- 3
x <- matrix(r01exp(n*1e+5), ncol=n)
hist(apply(x,1,sum),breaks=100, prob=TRUE, xlab="sum(x)",main="")
curve(dsum01exp(x,n), 0, n, ylab=expression(f[n](x)), add=TRUE)

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