I am having some trouble with a textbook problem.

I am given that $U_1, ... U_n$ are independent random variables with uniform distribution over $[0,1]$ and that $U_{(n)}$ is the maximum.

I have to show that the CDF of the standardized $U_{(n)}$ approaches a limiting value.

I know that the CDF of $U_{(n)}$ is $F(U)=(u)^n$ and that the standardized $U_{(n)}$ is $U_{(n)}^*=\frac{U_{(n)}-\frac{n}{n+1}}{\sqrt{\frac{n}{(n+1)^2(n+2)}}}$, where $E[U_{(n)}] = \frac{n}{n+1}$ and $var(U_{(n)})=\frac{n}{(n+1)^2(n+2)}$.

However, I am unsure on how to find the CDF of $U_{(n)}^*$ and then show the limiting value.

Any help is appreciated, thank you in advance.

  • $\begingroup$ This comes down to the problem of computing the cdf of a random variable $Z=(X-\mu)/\sigma$ where the cdf of $X$ is known and $\mu$ and $\sigma$ are given numbers. Does that help? $\endgroup$ – whuber Oct 26 '17 at 15:16

Thanks for whuber's comments, I hope my answer is right now.

You already show that $F_{U_{(n)}}(u)=u^n, 0 \le u \le 1$


$\sigma=\frac{n}{(n+1)^2(n+2)}$ here

Now we derivate $CDF$ of $$U_{(n)}^*=\frac{U_{(n)}-\mu}{\sigma} \tag{1}$$

$F_{U_{(n)}^*}(t)=P(U_{(n)}^*\le t)=P(\frac{U_{(n)}-\mu}{\sigma} \le t)=P(U_{(n)} \le \sigma t+\mu)=F_{U_{(n)}}(\sigma t+\mu)=(\sigma t+\mu)^n$

This is the $CDF$ of $U_{(n)}^*$

Next we need to find the limit of $F_{U_{(n)}^*}(t)$, I think it should be in term of $t$

From $(1)$, the range of $t$ is $\frac{-\mu}{\sigma} < U_{(n)}^* < \frac{1-\mu}{\sigma}$


$Lim_{t \to \frac{-\mu}{\sigma}}F_{U_{(n)}^*}(t)=0$


$Lim_{t \to \frac{1-\mu}{\sigma}}F_{U_{(n)}^*}(t)=1$

******I don't feel comfortable about this answer, hope some one can help especailly in term of extreme-value distribution

  • 1
    $\begingroup$ I believe you have misinterpreted the question. $U_{(n)}$, being the maximum of a set of independent uniform values, is far from uniform--look at the CDF given by the OP!--and the limiting distribution will be an extreme-value distribution. $\endgroup$ – whuber Oct 26 '17 at 15:15
  • 1
    $\begingroup$ @whuber, yes you're right, I will modify the answer $\endgroup$ – Deep North Oct 26 '17 at 20:48

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