# Can you spot any mistake in what I did (problem in elementary probability)

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.

• 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 $n\to \infty$. Oct 22, 2022 at 17:50

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 $$\operatorname{Exp}(1)$$.