I am trying to solve the following problem.

Let $(X_n \,:\,n\geq 1)$ be a sequence of IID random variable uniformly distributed on $[0,1]$. Let $M_n:=max\{X_1,...,X_n\}$. Show that the random variable $n(1-M_n)$ converges in distribution and find the limit distribution.

The solution is straightforward: Noting that $\mathbb{P}(M_n\leq x)=x^n$, one gets $\mathbb{P}(n(1-M_n)\leq x)=1-(1-x/n)^n\to 1-e^{-x}$. Hence, by Portmanteau's theorem, the limit distribution is an exponential of parameter $1$.

However, I would like to get the same result using characteristic functions, i.e. I would like to show that $\phi_{n(1-M_n)}(t)\to \frac{1}{1-it}$ (where $\frac{1}{1-it}$ is the characteristic function of an exponential of parameter $1$). Integrating by parts I find $\phi_{M_n}(t)=\int_0^1 e^{itx} nx^{n-1}dx=(\frac{n}{it}-\frac{n(n-1)}{(it)^2}+...-\frac{n!}{(-it)^n})e^{it}+\frac{n!}{(-it)^n}$. Then $\phi_{n(1-M_n)}(t)=e^{int}\phi_{M_n}(-nt)=-\frac{n}{int}-\frac{n(n-1)}{(int)^2}+...-\frac{n!}{(int)^n}+\frac{n!}{(int)^n}e^{int}$.

However, I am not sure whether this is right since $|\frac{n!}{(int)^n}e^{int}|\to0$ and the first part of the series does not seem to converge to $\frac{1}{1-it}$. Can you spot any mistake?

Thank you.

  • $\begingroup$ On second thought, the first part of the series does converge to $\frac{1}{1-it}$. What was confusing me was that, however, this series is not the Laurent series of $\frac{1}{1-it}$ centered at infinity, which is instead $\sum_{k=1}^\infty -\frac{1}{(it)^k}$ (and I couldn't understand how this was possible since the Laurent series is unique). But the point is that in this case the coefficients are not constant (they change with $n$) and in fact they themselves converge to the coefficients of the Laurent series: for each fixed $k$, $n(n-1)...(n-k)/n^k\to 1$ as $n\to \infty$. $\endgroup$
    – Titti
    Oct 22, 2022 at 17:50

1 Answer 1


A more robust method is to study the asymptotic behavior of the integral directly:

\begin{align*} \phi_{n(1-M_n)}(t) &= \int_{0}^{1} e^{itn(1-x)} n x^{n-1} \, \mathrm{d}x \end{align*}

Substituting $x = 1 - \frac{u}{n}$, this becomes

\begin{align*} \phi_{n(1-M_n)}(t) &= \int_{0}^{n} e^{itu} \left( 1 - \frac{u}{n} \right)^{n-1} \, \mathrm{d}u \\ &= \int_{0}^{\infty} e^{itu} \left( 1 - \frac{u}{n} \right)_+^{n-1} \, \mathrm{d}u, \end{align*}

where $a_+ := \max\{0, a\}$ denotes the positive part of $a \in \mathbb{R}$. Now by noting that $ (1 - u/n)_+ \leq e^{-u/n} $ holds for all $ u \in \mathbb{R} $, it follows that, for $n \geq 2$,

$$ \left| e^{itu} \left( 1 - \frac{u}{n} \right)^{n-1} \mathbf{1}_{[0, n]}(u) \right| \leq e^{-(n-1)u/n} \leq e^{-u/2}. $$

This bound is integrable over $[0, \infty)$, hence the dominated convergence theorem is applicable and yields

\begin{align*} \lim_{n\to\infty} \phi_{n(1-M_n)}(t) &= \int_{0}^{\infty} \lim_{n\to\infty} e^{itu} \left( 1 - \frac{u}{n} \right)_+^{n-1} \, \mathrm{d}u \\ &= \int_{0}^{\infty} \lim_{n\to\infty} e^{itu} e^{-u} \, \mathrm{d}u \\ &= \frac{1}{1 - it}. \end{align*}

Therefore the limit in distribution of $n(1-M_n)$ is $\operatorname{Exp}(1)$.


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