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Assume we have a random numbers $\sim U(0,100)$.

Then the expected value of that number will be: $\int_{0}^{100} \frac{x}{100}$ = 50.5

Now assume we have 5 random numbers $\sim U(0,100)$.

How can I calculate what would be the expected value of the maximal number?

Thanks.

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    $\begingroup$ why 50.5? Shouldn't it be 50? $\endgroup$
    – Arash
    Jan 13, 2014 at 12:03
  • $\begingroup$ you are right 50.5 is for the integral from 1 to 100 $\endgroup$
    – user121014
    Jan 13, 2014 at 12:06
  • $\begingroup$ @user121014 No it is not ... $\endgroup$
    – wolfies
    Jan 13, 2014 at 15:54

2 Answers 2

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You need to learn about order statistics: https://en.wikipedia.org/wiki/Order_statistics

The maximum of five independent observations is the fifth order statistic (of that sample). In your case, that will have a certain (scaled) beta distribution. You can find the datails in wikipedia above. In your case it will be (100 multiplied by) a beta(5,1)-variable, with expectation $100 \cdot \frac{5}{6}$.

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  • $\begingroup$ thanks @kjetil, can you please explain a bit more and give some intuition? $\endgroup$
    – user121014
    Jan 13, 2014 at 12:29
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Suppose $X_1, X_2, \dots, X_n$ form a random sample of size $n$ from some distribution with p.d.f. $f$, and c.d.f. $F$, then let:

$$ Y = max\{X_1, X_2, \dots, X_n\} $$

Then the c.d.f. of $Y$ is given by

\begin{align*} G(y) & = \text{Pr}(Y \le y) = \text{Pr}(X_1 \le y, X_2 \le y, \dots, X_n \le y)\\ & = \text{Pr}(X_1 \le y)\text{Pr}(X_2 \le y)\dots\text{Pr}(X_n \le y) \\ & = F(y)F(y)\dots F(y) = [F(y)]^n \end{align*}

The first line says $y$ must be greater or equal to all $n$ samples. The second line follows from the independence of each sample. Then the p.d.f. of $Y$ is

$$ g(y) = n[F(y)]^{n - 1}f(y) $$

which is just the derivative of $F$.

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    $\begingroup$ @user121014: Note that in your uniform case, $F(y)=\frac{y}{100}$. Thus $g(y)=5\frac{y^4}{100^5}$. Now calculate $\int_0^{100} yg(y)\,dy$. We get $\frac{5}{6}(100)$. $\endgroup$ Jan 13, 2014 at 16:37

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