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How to describe the set $$\{ 1 \leq n \leq N: \alpha_n \in(a,b)\}$$ when $(a,b) \subset [0,1)$ and you have following information:

a sequence of numbers $(\alpha_1,\alpha_2, \alpha_3,...)$, where $a_j \in [0,1)$, $j \in \mathbb{N}$, is equidistributed, if for every interval $(a,b) \subset [0,1)$ $$\lim_{N \rightarrow \infty} \frac{ \# \{1\leq n \leq N: \alpha_n \in (a,b) \}}{N}=b-a$$ holds, where $\# A$ is the cardinality of set $A$ (number of elements in the set).

I think $$A=\{ 1 \leq n \leq N: \alpha_n \in(a,b)\} = \{1,2,3,4, \dots, N \}\Rightarrow\#A = N\;,$$ but then how do you get $$\lim_{N \rightarrow \infty} \frac{ \# \{1\leq n \leq N: \alpha_n \in (a,b) \}}{N}=b-a\;?$$

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I have made a major edit to your post. I had trouble reading your original question, and I hope that I have maintained the meaning of your post. Please edit the post if I have inadvertently changed the meaning of your question. –  JavaMan Oct 29 '12 at 17:44
    
Sorry but at the moment there is no question at all here. –  Did Oct 29 '12 at 17:50
    
After the edit the task "Help me prove the formula above!" makes no sense. There are only definitions, no claims. –  Hagen von Eitzen Oct 29 '12 at 18:11
    
I have restored the OP’s original post, merely improving the formatting and making a very few minor changes in the wording. The previous edits were far too extreme and distorted the post beyond recognition. I agree with @did that it’s not at all clear what is being asked. –  Brian M. Scott Oct 29 '12 at 18:22

2 Answers 2

up vote 1 down vote accepted

You defined $$A_{\alpha,N}(a,b)=\{1 \leq n \leq N : \alpha_n \in ]a,b[\}$$

This set contains exactly the indices $n \in [|1,N|]$ for which $a<\alpha_n<b$. Since $0<a\leq b < 1$ and $\alpha_n\in[0,1[$, there is no reason why your statement would be true in general.

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Was my interpretation of set correct: $$A=\{ 1 \leq n \leq N: \alpha_n \in(a,b)\} = \{1,2,3,4, \dots, N \}\Rightarrow\#A = N\;,$$? If not, then what is right answer? –  laovultai Oct 29 '12 at 18:32
    
The implication is correct, but it has little to do with the definition of the set. –  Wok Oct 29 '12 at 19:00
    
I meant that is $$A=\{ 1 \leq n \leq N: \alpha_n \in(a,b)\} = \{1,2,3,4, \dots, N \}$$ correct? –  laovultai Oct 29 '12 at 19:04
    
No. Have a look at my answer. –  Wok Oct 29 '12 at 19:06
    
Please consider accepting the answer after satisfaction is provided. –  Wok Oct 29 '12 at 19:11

There's one mistake : your statement

$$A=\{ 1 \leq n \leq N: \alpha_n \in(a,b)\} = \{1,2,3,4, \dots, N \}$$

is false.

You can think at this problem as a probability distribution problem : Let's consider some probability distribution $\mathcal{P}$. You draw $n$ samples in $[0,1)$ called $\alpha_1,\dots,\alpha_n$. For some arbitrary interval $(a,b)$, your set $A=\{ 1 \leq n \leq N: \alpha_n \in(a,b)\}$ are the set of points that are drawn inside $(a,b)$. Of course, you won't get all the $N$ points inside $(a,b)$. The probability that a point is inside $(a,b)$ can be written as :

$$\mathbb{P}(\alpha \in (a,b) | \alpha \sim \mathcal{P}) = \lim_{N \rightarrow \infty} \frac{ \# \{1\leq n \leq N: \alpha_n \in (a,b) \}}{N}$$

(when you are drawing an infinite quantity of points, this is the proportion of points that are $(a,b)$).

You can say that $\mathcal{P}$ is an uniform law if this probability is $\frac{measure (a,b)}{measure(0,1)}= b-a$. You will say that the sample $\alpha_1,\dots,\alpha_n$ is equidistributed if it comes from an uniform law.

(I know that these explanations aren't really rigorous, sorry for that but I hope this "sketch" explanation can help)

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