I have tried to find the limiting distribution of $X_n\sim\text{Exponential}(\lambda/n)$ by finding the cdf and taking the limit. I got:

\begin{align*} F_{X_n}(X)) = \int_{0}^{X} \frac{\lambda}{n} e^{\frac{-\lambda x}{n}} dx &= \left. - e^{\frac{-\lambda x}{n}}\right|_0^{X}\\ &= 1 - e^{\frac{-\lambda X}{n}}\\ \lim_{n \rightarrow \infty} F_{X_n}(X_n)) =0 \end{align*}

Somehow this does not really look right to me...? Did I make an algebra mistake? My goal here would ultimately be to find the limiting distribution of $x - \lfloor x \rfloor$ given $X_n\sim\text{Exponential}(\lambda/n)$.


1 Answer 1


By definition, the law of $X_n - \lfloor X_n\rfloor$ is

$$F_n(x) = \Pr(X_n - \lfloor X_n\rfloor \le x)$$

for $0 \le x \lt 1$. The event $E^{(n)}: X_n - \lfloor X_n\rfloor \le x$ is a countable union of the disjoint events $E^{(n)}_i: i \le X_n \le i + x$ for $i=0, 1, 2, \ldots$. Therefore (because probability is countably summable)

$$F_n(x) = \Pr(E^{(n)})= \sum_{i=0}^\infty \Pr(E^{(n)}_i).$$

When $X_n$ has an Exponential$(\lambda/n)$ distribution,

$$\Pr(E^{(n)}_i) = \Pr( i \le X_n \le i + x) = e^{-\lambda i/n} - e^{-\lambda (i+x)/n} = \left(1 - e^{-\lambda x / n}\right)e^{-\lambda i/n},$$


$$F_n(x) = \left(1 - e^{-\lambda x / n}\right)\sum_{i=0}^\infty e^{-\lambda i/n}.$$

The last term sums a geometric series with initial term $1$ and common ratio $e^{-\lambda/n}$, immediately simplifying the whole expression to

$$F_n(x) = \frac{1 - e^{-\lambda x/n}}{1 - e^{-\lambda/n}}.$$

The limiting value as $n\to \infty$ is most easily obtained with L'Hopital's Rule,

$$\lim_{n\to\infty} F_n(x) = \lim_{n\to\infty} \frac{\lambda x e^{-\lambda x/n}}{\lambda e^{-\lambda/n}} = x\lim_{n\to\infty} e^{\lambda/n(1-x)} = x.$$

This is the law of the Uniform distribution on $[0, 1)$.

  • $\begingroup$ I used a similar analysis to derive the eighth method at stats.stackexchange.com/a/117711. There, $X_n$ has a Normal distribution. The sum is a theta function in that case. $\endgroup$
    – whuber
    Feb 9, 2015 at 18:17
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    $\begingroup$ I had a hunch that the result would be related to uniform distribution considering the difference $x - \lfloor x \rfloor$ is always between 0 and 1, but wasn't quite sure how to proceed with the proof. Thanks so much for the clear explanation! $\endgroup$
    – datatista
    Feb 9, 2015 at 18:22
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    $\begingroup$ The result is true whenever $F_n$ is a given continuous distribution (with bounded density) scaled by $\sigma_n$ and $\sigma_n\to \infty$. $\endgroup$
    – whuber
    Feb 9, 2015 at 18:34

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