Clarification of DFA average
$begingroup$
I've been reading detrended fluctiation analysis computiation on wikipedia.
https://en.wikipedia.org/wiki/Detrended_fluctuation_analysis
Just something I wanted clarified.
$$X_t = sum_{i=1}^{t}(x_i-(x))$$
where $(x)$ = mean
$$F(n) = sqrt{frac{1}{N}sum_{t=1}^{N}(X_t-Y_t)^2}$$
For example, if we had a time series of length $1000$ we would want to divide the time series into chunks/boxes to build our log log graph.
1 box of length 1000
2 boxes of length 500
4 boxes of length 250
10 boxes of length 100
20 boxes of length 50
etc....
Is the average calculated based on length of each box size?
1 box of length 1000, average of 1000
2 boxes of length 500, average of 500
4 boxes of length 250, average of 250
10 boxes of length 100, average of 100
20 boxes of length 50, average of 50
etc....
So that would mean you are essentially calculating $$X_t = sum_{i=1}^{t}(x_i-(x))$$
based on how many chunks we divide time series into based on box length. It seems this makes the most sense vs simply just calculating 1 average of 1000 and detrending based on that.
calculus probability
asked Dec 17 '18 at 1:20
ForextraderForextrader
878
$endgroup$
add a comment |
$begingroup$
I've been reading detrended fluctiation analysis computiation on wikipedia.
https://en.wikipedia.org/wiki/Detrended_fluctuation_analysis
Just something I wanted clarified.
$$X_t = sum_{i=1}^{t}(x_i-(x))$$
where $(x)$ = mean
$$F(n) = sqrt{frac{1}{N}sum_{t=1}^{N}(X_t-Y_t)^2}$$
For example, if we had a time series of length $1000$ we would want to divide the time series into chunks/boxes to build our log log graph.
1 box of length 1000
2 boxes of length 500
4 boxes of length 250
10 boxes of length 100
20 boxes of length 50
etc....
Is the average calculated based on length of each box size?
1 box of length 1000, average of 1000
2 boxes of length 500, average of 500
4 boxes of length 250, average of 250
10 boxes of length 100, average of 100
20 boxes of length 50, average of 50
etc....
So that would mean you are essentially calculating $$X_t = sum_{i=1}^{t}(x_i-(x))$$
based on how many chunks we divide time series into based on box length. It seems this makes the most sense vs simply just calculating 1 average of 1000 and detrending based on that.
calculus probability
asked Dec 17 '18 at 1:20
ForextraderForextrader
878
$endgroup$
add a comment |
$begingroup$
I've been reading detrended fluctiation analysis computiation on wikipedia.
https://en.wikipedia.org/wiki/Detrended_fluctuation_analysis
Just something I wanted clarified.
$$X_t = sum_{i=1}^{t}(x_i-(x))$$
where $(x)$ = mean
$$F(n) = sqrt{frac{1}{N}sum_{t=1}^{N}(X_t-Y_t)^2}$$
For example, if we had a time series of length $1000$ we would want to divide the time series into chunks/boxes to build our log log graph.
1 box of length 1000
2 boxes of length 500
4 boxes of length 250
10 boxes of length 100
20 boxes of length 50
etc....
Is the average calculated based on length of each box size?
1 box of length 1000, average of 1000
2 boxes of length 500, average of 500
4 boxes of length 250, average of 250
10 boxes of length 100, average of 100
20 boxes of length 50, average of 50
etc....
So that would mean you are essentially calculating $$X_t = sum_{i=1}^{t}(x_i-(x))$$
based on how many chunks we divide time series into based on box length. It seems this makes the most sense vs simply just calculating 1 average of 1000 and detrending based on that.
calculus probability
asked Dec 17 '18 at 1:20
ForextraderForextrader
878
$endgroup$
I've been reading detrended fluctiation analysis computiation on wikipedia.
https://en.wikipedia.org/wiki/Detrended_fluctuation_analysis
Just something I wanted clarified.
$$X_t = sum_{i=1}^{t}(x_i-(x))$$
where $(x)$ = mean
$$F(n) = sqrt{frac{1}{N}sum_{t=1}^{N}(X_t-Y_t)^2}$$
For example, if we had a time series of length $1000$ we would want to divide the time series into chunks/boxes to build our log log graph.
1 box of length 1000
2 boxes of length 500
4 boxes of length 250
10 boxes of length 100
20 boxes of length 50
etc....
Is the average calculated based on length of each box size?
1 box of length 1000, average of 1000
2 boxes of length 500, average of 500
4 boxes of length 250, average of 250
10 boxes of length 100, average of 100
20 boxes of length 50, average of 50
etc....
So that would mean you are essentially calculating $$X_t = sum_{i=1}^{t}(x_i-(x))$$
based on how many chunks we divide time series into based on box length. It seems this makes the most sense vs simply just calculating 1 average of 1000 and detrending based on that.
calculus probability
calculus probability
asked Dec 17 '18 at 1:20
ForextraderForextrader
878
asked Dec 17 '18 at 1:20
ForextraderForextrader
878
asked Dec 17 '18 at 1:20
ForextraderForextrader
878
asked Dec 17 '18 at 1:20
ForextraderForextrader
878
asked Dec 17 '18 at 1:20
ForextraderForextrader
878
878
add a comment |
add a comment |
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