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A discrete random variable has uniform distribution with parameter $\theta$ and we have a sample $X$ of size $n$.

$$P_{\theta}(\xi=k)=\theta^{-1},\theta\in\{1,2\dots\},k\in\{1,2,\dots\theta\}$$

I've got required expectation as a function from generalized harmonic numbers.

\begin{gather*} P(X_{(n)}=k)=\\ =\sum\limits_{i=1}^nP(\text{exactly $i$ observations equals $k$ and other are less})=\\ =\sum\limits_{i=1}^n\binom{i}{n}\left(\frac{1}{\theta}\right)^i\left(\frac{k-1}{\theta}\right)^{n-i}=\left(\frac{k}{\theta}\right)^n.\\ \mathbb{E}[X_{(n)}]= \sum\limits_{k=1}^\theta k\left(\frac{k}{\theta}\right)^n= \frac{1}{\theta^n}\sum\limits_{k=1}^\theta k^{n+1}= \frac{1}{\theta^n}H_{\theta,-(n+1)}. \end{gather*}

Is there a way to represent expectation of maximum from sample $X$ with a simpler expression?

Askhat Nuriddinov
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2 Answers2

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The second line should be the CDF of $X_{(n)}$:

$$ \Pr\{X_{(n)} \leq k\} = \left(\frac {k} {\theta}\right)^n, k = 0, 1, \ldots, \theta $$

So you may use

$$ \begin{align} E[X_{(n)}] &= \sum_{k=1}^\theta k\Pr\{X_{(n)} = k\} \\ &= \sum_{k=1}^\theta \sum_{u=1}^k \Pr\{X_{(n)} = k\} \\ &= \sum_{u=1}^{\theta} \sum_{k=u}^{\theta} \Pr\{X_{(n)} = k\} \\ &= \sum_{u=1}^{\theta} \Pr\{X_{(n)} > u - 1\} \\ &= \sum_{u=1}^{\theta} [1 - \Pr\{X_{(n)} \leq u - 1\}] \\ &= \theta - \sum_{u=1}^{\theta - 1}\left(\frac {u} {\theta}\right)^n \\ \end{align} $$ and not much here you can simplify. See e.g. https://en.wikipedia.org/wiki/Faulhaber%27s_formula

BGM
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  • I believe there is disagreement with this answer: https://math.stackexchange.com/questions/852508/distribution-of-the-maximum-of-n-uniform-random-variables – Christoph Hanck Sep 22 '21 at 09:10
  • Thanks for spotting that. I have corrected the typos – BGM Sep 22 '21 at 09:37
  • As the summand is independent of $u$, we have $\sum_{u=1}^k = k$, $\sum_{u=1}^k p_{\theta} = k p_{\theta}$ – BGM Sep 22 '21 at 15:19
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If $k = \theta$, you'll get probability equals 1. That's not true. The correct formula is $ P\left(X_{(n)} = k\right) = \dfrac{k^n - (k-1)^n}{\theta^n}$

Yulia
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