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Compute the following limit $$ I=\lim_n\int_{[0,1]^n}f(x)\,dx_1\ldots dx_n, $$ where $$ f:\mathbb{R}^n \to \mathbb{R},\ x=(x_1,\ldots,x_n) \mapsto \max\{x_1,\ldots,x_n\}. $$

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3 Answers 3

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We shall first compute $$I_n = \int_{[0,1]^n} \max\{x_1,x_2,\ldots,x_n\} dx_1 dx_2 \ldots dx_n$$ Note that $$\max\{x_1,x_2,\ldots,x_n\} = \begin{cases}x_1 & \text{ if }x_k \leq x_1 \,\,\,\,\,\,\forall k\\ x_2 & \text{ if }x_k \leq x_2 \,\,\,\,\,\,\forall k\\ x_3 & \text{ if }x_k \leq x_3 \,\,\,\,\,\,\forall k\\ \vdots & \,\,\,\,\,\,\,\,\,\,\,\,\,\vdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \vdots\\ x_n & \text{ if }x_k \leq x_n \,\,\,\,\,\,\forall k \end{cases}$$ Hence, the integral can be written as \begin{align} I_n & = \int_{x_1=0}^1 \int_{0\leq x_1,x_2,x_3,\ldots,x_n \leq x_1} x_1 dx_1 dx_2 \cdots dx_n\\ & + \int_{x_2=0}^1 \int_{0\leq x_1,x_2,x_3,\ldots,x_n \leq x_2} x_1 dx_1 dx_2 \cdots dx_n\\ & + \int_{x_3=0}^1 \int_{0\leq x_1,x_2,x_3,\ldots,x_n \leq x_3} x_1 dx_1 dx_2 \cdots dx_n\\ & + \cdots\\ & + \int_{x_n=0}^1 \int_{0\leq x_1,x_2,x_3,\ldots,x_n \leq x_n} x_1 dx_1 dx_2 \cdots dx_n \end{align} Hence, $$I_n = n \int_{x_1=0}^1 \int_{0\leq x_2,x_3,\ldots,x_n \leq x_1} x_1 dx_1 dx_2 \cdots dx_n = n \int_{x_1=0}^1 x_1^n dx_1 = \dfrac{n}{n+1}$$ $$I = \lim_{n \to \infty} I_n = 1$$

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I is the limit in first post. –  user59671 Feb 2 '13 at 3:15
    
@CutieKrait Thanks. Corrected and updated. –  user17762 Feb 2 '13 at 3:17

The integral $I_n = \int_{[0,1]^n} \max\{x_1,x_2,\ldots,x_n\} dx_1 dx_2 \ldots dx_n$ is the expected value of $M:=\max(X_1,X_2,\dots, X_n)$ where $(X_j)$ are independent, uniform(0,1) random variables. It's not hard to show that $M$ has density $f_M(x)=n x^{n-1} \,1_{[\, 0,1]}(x)$, and so $$\mathbb{E}(M)=\int_0^1 x\, f_M(x)\,dx=n\int_0^1 x^n\,dx={n\over n+1}.$$


Added: We can find the distribution function of $M$ directly. For $0\leq x\leq 1$,
$$\mathbb{P}(M\leq x) =\mathbb{P}(\cap_{i=1}^n [X_i\leq x])=\prod_{i=1}^n\mathbb{P}(X_i\leq x)=x^n.$$ Differentiating gives $f_M(x)=n x^{n-1}$.

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Very nice, but that "it is not hard to show" sounds a bit like cheating! :-) Indeed (IMHO) there lies the crux of the computation. –  Giuseppe Negro Feb 2 '13 at 3:56
1  
Fair enough. If you are willing to cheat even more, you could argue that, by symmetry, the expected values of all the order statistics are evenly spread out over ${j\over n+1}$ for $j=1,\dots, n$. Therefore the expected value of the maximum must be ${n\over n+1}$. See my answer here: math.stackexchange.com/questions/197299/… –  Byron Schmuland Feb 2 '13 at 4:50
    
Fascinating. Thank you very much! –  Giuseppe Negro Feb 2 '13 at 14:09

Here is a marginally different approach:

Let $N=\{ x \in [0,1]^n | x_i \neq x_j, \ \forall I \neq j \}$. It should be clear that $m N = m [0,1]^n$, where $m$ is the Lebesgue measure. Let $T = \{ x \in N | x_i > x_{i+1},\ \forall i \}$, and note that $N = \cup_{\pi \in \Pi_n } \pi(T)$, where $\Pi_n$ is the set of functions that permute the variables in $[0,1]^n$. Furthermore, note that $\max(x_1,...,x_n) = \max([\pi(x)]_1,...,[\pi(x)]_n)$ for all $\pi \in \Pi_n$.

Hence $I_n = n! \int_T x_1 dx = \int_{x_1=0}^1\int_{x_2=0}^{x_1} \cdots \int_{x_n=0}^{x_{n-1}} x_1 dx = n! \frac{1}{(n+1)(n-1)!} = \frac{n}{n+1}$, and $\lim_n I_n = 1$.

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