Show Convergence in Distribution

Question

Consider the maximum ${U_{(n)}}$ of $n$ simulated uniform (0,1) i.i.d. random variables ${U_1},...,{U_n}$.

Show that $n\left( {1 - {U_{(n)}}} \right)$ converges in distribution to a standard exponential distribution with distribution function $F\left( y \right) = 1 - {e^{ - y}}$ for $y>0$, as $n \to \infty $. (Hint: Compute $P\left( {n\left( {1 - {U_{\left( n \right)}}} \right) > y} \right)$ and take the limit for $n \to \infty $.)

How would you adapt this result if the random variables were uniformly distributed on $(0, a)$ for some $a > 0$. [Hint: think how to transform the given case to the uniform $(0,1)$ case.]

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I'm kind of starting slow in this new chapter and I don't know how to procede in this type of exercise. What does it mean when he says "maximum ${U_{(n)}}$"? To follow his hint, do I solve for ${U_{(n)}}$ or for $y$? Should I treat $n\left( {1 - {U_{(n)}}} \right)$ as a completely new variable, as in a transformation?

Thank you immensely for any insights you may have!

Answer 1

Let $U_1,U_2,...,U_n$ be i.i.d. with uniform distribution on $(0,a)$. Let $V_i=\frac {U_i} a$. Then $V_1,V_2,...,V_n$ are i.i.d. with uniform distribution on $(0,1)$. Also $V_{(n)}=\max \{V_1,V_2,...,V_n\}=\frac {U_{(n)}} a$. Since $n(1-V_{(n)})$ converges to an exponential distribution by the first part we see that $n(1-a^{-n}U_{(n)})$ converges to an exponential distribution.

Answer 2

Recall that if $X_1,\ldots,X_n$ are i.i.d. random variables then the distribution of $X_{(n)}$ is given by \begin{align} \mathbb P(X_{(n)}\leqslant x) &= \mathbb P\left(\bigcap_{i=1}^n \{X_i\leqslant x\}\right)\\ &=\prod_{i=1}^n \mathbb P(X_i\leqslant x)\\ &=\mathbb P(X_1\leqslant x)^n. \end{align} In the case where $U_1\sim\mathsf U(0,1)$, we have $\mathbb P(U_1\leqslant x) = x$ for $x\in[0,1]$, and hence $\mathbb P(U_{(n)}\leqslant x)=x^n$. It follows that $$ \mathbb P(n(1-U_{(n)})>y) = \mathbb P\left(U_{(n)}< 1-\frac yn\right) = \left(1-\frac yn\right)^n. $$ It is well known that $$ \lim_{n\to\infty}\left(1-\frac yn\right)^n = e^{-y}, $$ and therefore $n(1-U_{(n)})$ converges to a standard exponential distribution as $n\to\infty$.