Calculating Sn: solve Qn=Cn*{|Xi−Xj|;i<j}(k) using {1, 2, 3, 5, 6, 7}

[harrison]

How would one solve the following equation using inputs {1,2,3,5,6,7}

[FONT=MathJax_Math-italic]Q[FONT=MathJax_Math-italic]n[FONT=MathJax_Main]=[FONT=MathJax_Math-italic]C[FONT=MathJax_Math-italic]n*[FONT=MathJax_Main]{[FONT=MathJax_Main]|[FONT=MathJax_Math-italic]X[FONT=MathJax_Math-italic]i[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]X[FONT=MathJax_Math-italic]j[/FONT][/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Main];[/FONT][FONT=MathJax_Math-italic]i[/FONT][FONT=MathJax_Main]<[/FONT][FONT=MathJax_Math-italic]j[/FONT][FONT=MathJax_Main]}[FONT=MathJax_Main]([FONT=MathJax_Math-italic]k[FONT=MathJax_Main])[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]

Where Cn is a constant = 2.2219

Where (k) = kth order statistic

For reference, please see (3.3) of https://www.researchgate.net/publication/221996720_Alternatives_to_Median_Absolute_Deviation

So far I have:

-- 7 6 5 3 2 1
1 6 5 4 2 1
2 5 4 3 1
3 4 3 2
5 2 1
6 1
7

If anyone can help that would be great!

 

[stapel]

How would one solve the following equation using inputs {1, 2, 3, 5, 6, 7}?

. . . . .\(\displaystyle Q_n\, =\, C_n\, \cdot\, \bigg{\{}\big| X_i\, -\, X_j\big|\, \mbox{ for }\, i\, <\, j\bigg{\}}_{(k)}\)

Where C_n is a constant equal to 2.2219, and where (k) denotes the k^th-order statistic

For reference, please see (3.3) of https://www.researchgate.net/publication/221996720_Alternatives_to_Median_Absolute_Deviation

So far I have:

-- 7 6 5 3 2 1
1 6 5 4 2 1
2 5 4 3 1
3 4 3 2
5 2 1
6 1
7

How did you obtain these strings of numbers? How does this relate to the exercise?

Please be complete. Thank you!

 

[harrison]

How did you obtain these strings of numbers? How does this relate to the exercise?

Please be complete. Thank you!

The string of numbers is the kth order statistic of pairwise differences.

 

[stapel]

The string of numbers is the kth order statistic of pairwise differences.

Okay.... Could you please show your work, so we can see what's going on? Thank you!

 

[harrison]

Okay.... Could you please show your work, so we can see what's going on? Thank you!

That is all that I have so far . I really wish I had more and I have tried searching all over the place for ways to calculate the formula.

 

[stapel]

That is all that I have so far . I really wish I had more and I have tried searching all over the place for ways to calculate the formula.

Well, you arrived at those strings of numbers by doing... something. Could you please show us what that "something" was? Thank you!

 

[harrison]

Well, you arrived at those strings of numbers by doing... something. Could you please show us what that "something" was? Thank you!

Honestly, I am confused as to what you are asking for. I thought what I did was arrange data according to the rules of the formula. If we are finding the kth order statistic then the data is arranged in that order buy subtracting Xi from Xj.

k = (h/2) where h = [n/2] +1. Therefore since n = 6, then h = 4, so k = (4/2) = 2.

I am just struggling to put it all together. I get that people in this forum answering questions are reluctant to show step by step solutions as it seems "lazy" by the person asking for that, but sometimes it helps as really I am lost.

 

[stapel]

Honestly, I am confused as to what you are asking for. I thought what I did was arrange data according to the rules of the formula. If we are finding the kth order statistic then the data is arranged in that order buy subtracting Xi from Xj.

k = (h/2) where h = [n/2] +1. Therefore since n = 6, then h = 4, so k = (4/2) = 2.

I am just struggling to put it all together. I get that people in this forum answering questions are reluctant to show step by step solutions as it seems "lazy" by the person asking for that, but sometimes it helps as really I am lost.

Perhaps the reason nobody else is trying to help is that they're as lost as I am, regarding what's going on here, and what you've done. Now I'm wondering if somebody gave you the string "7 6 5 3 2 1", because you're not able to show any work as to how they're derived.

Until you understand the material well enough to be able to explain at least how that first step was completed, it's possible that nobody will be able to assist. Sorry.

 

[Dr.Peterson]

Honestly, I am confused as to what you are asking for. I thought what I did was arrange data according to the rules of the formula. If we are finding the kth order statistic then the data is arranged in that order buy subtracting Xi from Xj.

k = (h/2) where h = [n/2] +1. Therefore since n = 6, then h = 4, so k = (4/2) = 2.

I am just struggling to put it all together. I get that people in this forum answering questions are reluctant to show step by step solutions as it seems "lazy" by the person asking for that, but sometimes it helps as really I am lost.

I think in this case it's not a matter of keeping you from being lazy, but just of wanting to work with you, and therefore having to ask you to explain your thinking more fully. What you wrote (below) as your work is a list of numbers with no explanation. You've added some in what you just said (above). I'll try to express what I think you are saying, being as complete as I think appropriate, and then the discussion may open up a bit.

How would one solve the following equation using inputs {1,2,3,5,6,7}

[FONT=MathJax_Math-italic]Q[FONT=MathJax_Math-italic]n[FONT=MathJax_Main]=[FONT=MathJax_Math-italic]C[FONT=MathJax_Math-italic]n*[FONT=MathJax_Main]{[FONT=MathJax_Main]|[FONT=MathJax_Math-italic]X[FONT=MathJax_Math-italic]i[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]X[FONT=MathJax_Math-italic]j[/FONT][/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Main];[/FONT][FONT=MathJax_Math-italic]i[/FONT][FONT=MathJax_Main]<[/FONT][FONT=MathJax_Math-italic]j[/FONT][FONT=MathJax_Main]}[FONT=MathJax_Main]([FONT=MathJax_Math-italic]k[FONT=MathJax_Main])[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]

Where Cn is a constant = 2.2219

Where (k) = kth order statistic

For reference, please see (3.3) of https://www.researchgate.net/publication/221996720_Alternatives_to_Median_Absolute_Deviation

So far I have:

-- 7 6 5 3 2 1
1 6 5 4 2 1
2 5 4 3 1
3 4 3 2
5 2 1
6 1
7

If anyone can help that would be great!

So here's what I think you're saying (with my comments in [...]):

You want to understand formula (3.3) in the reference, by applying it to the arbitrary set {1, 2, 3, 5, 6, 7} which you have chosen just as an example. Here n=6, so h = [n/2] +1 = [6/2] +1 = 4, and \(\displaystyle k=\binom{h}{2}=\binom{4}{2}=\frac{4!}{2!2!}=6\). [Note that you seem to have misread that. I don't know how it affects your work, since you didn't get that far.] The article also says that k is approximately \(\displaystyle k=\dfrac{\binom{n}{2}}{4}=\binom{6}{2}/4=\frac{1}{4}\frac{6!}{2!4!}=\dfrac{15}{4}\). [This seems a little low; maybe it works better for large n. I haven't explored that.]

Anyway, the formula is Q_n=d{|x_i - x_j|; i<j\}_(k). This means, since k = 6, that we are listing the 15 absolute differences in order and choosing the 6th of them. You've made a table to organize these differences,

	7	6	5	3	2	1
1	6	5	4	2	1
2	5	4	3	1
3	4	3	2
5	2	1
6	1
7

So the set of differences, in order, is {1,1,1,1,2,2,2,3,3,4,4,4,5,5,6}, and the 6th element is 2. So Q_n=d * 2.