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Below is an example from my textbook. In the book, the conclusion is simply given as an example on how integration works over sets. I want to prove the conclusion for my own sanity. I think my problem is translating the set notation to the integrals...

I am specifically wondering why $dx_1$ appears but $\int_{0}^{x_1}$ does not. Also, why are all the lower bounds zero instead of the previous $x_i$?

Given $$C=\{(x_1,\dots ,x_n)\mid 0 \leq x_1 \leq x_2 \leq \dots \leq x_n \leq 1\}$$

and $$Q(C)=\int \dots \int_{C}dx_1 dx_2 \dots dx_n$$

prove $$Q(C)=\int_{0}^{1} \int_{0}^{x_n}\dots \int_{0}^{x_2}dx_1 dx_2\dots dx_n= \frac{1}{n!}$$

If you would like to see the example it is in Chapter 1, page 8 of Introduction to Mathematical Statistics, 7th addition, Hogg, McKean, Craig

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    $\begingroup$ Your definition of $Q(C)$ contains an unspecified and undefined $f$. $\endgroup$ – md2perpe Jun 27 '17 at 20:13
  • $\begingroup$ @md2perpe corrected. Thank you. $\endgroup$ – rocksNwaves Jun 28 '17 at 12:41
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$x_n$ takes value between $0$ and $1$.

Upon fixing $x_n$, $x_{n-1}$ takes value between $0$ and $x_n$.

Similarly, upon fixing $x_{n-1}$, $x_{n-2}$ takes value between $0$ and $x_{n-1}$.

Once we determine $x_2$, $x_1$ takes value between $0$ and $x_2$.

We have then considered all the constraints.

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  • $\begingroup$ Very concise and to the point. fixing the constraints helped me understand. $\endgroup$ – rocksNwaves Jun 28 '17 at 12:49
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$0 \leq x_{1} \leq x_{2} \leq .... \leq x_{n}$ are the constraint.

If you do the calculus, you will get that $$Q_{c} = \int_{0}^{1} \int_{0}^{x_{n}}...\int_{0} ^{x_{3}} \int_{0}^{x_{2}} dx_{1}dx_{2}...dx_{n-1}dx_{n}$$

How to get this? Because of the integral, if you integrate $x_{1}$ with $ 0 \leq x_{1} \leq x_{2} $ that you will get new constraints $C^{'} = 0 \leq x_{2} \leq .... \leq x_{n}$.

$$Q_{c} = \int_{0}^{1} \int_{0}^{x_{n}}...\int_{0} ^{x_{3}} \int_{0}^{x_{2}} dx_{1}dx_{2}...d_{x_{n-1}}d_{x_{n}} = \int_{0}^{1} \int_{0}^{x_{n}}...\int_{0} ^{x_{3}} x_{2}dx_{2}...dx_{n-1}dx_{n}=... = \frac{1}{n!}$$

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For a permutation $\pi$ of $\{ 1, \ldots, n \}$, let $$C_\pi=\{(x_1,\dots ,x_n) \mid 0 < x_{\pi(1)} < x_{\pi(2)} < \dots < x_{\pi(n)} < 1\}.$$

It's easy to see that if $\pi_1$ and $\pi_2$ are two different permutations then $C_{\pi_1} \cap C_{\pi_2} = \emptyset$ but that $C_{\pi_1}$ and $C_{\pi_2}$ are congruent as geometrical objects, $C_{\pi_1} \cong C_{\pi_2}$, so they have the same $n$-volume. Also, $C_{\pi} \subset [0, 1]^n$ and $\bigcup_\pi C_\pi = [0, 1]^n \setminus \bigcup_\pi \partial C_\pi$.

Since there are $n!$ permutations it follows that the $n$-volume of $C_\pi$ is $\frac{1}{n!}$.

Your set $C = \overline{C_{\operatorname{id}}}$, where $\operatorname{id}$ is the identity permutation $\operatorname{id}(k) = k$. Your function $Q(C)$ is the $n$-volume of $C$. Thus $Q(C) = \frac{1}{n!}$.

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