I’m trying to calculate a ”standard deviation series” with a formula that involves a cumulative sum, but my matlab skills apparently aren’t sufficient to figure out how to type the formula in matlab.
The equation I want to write can be seen in step 5. Here: http://en.wikipedia.org/wiki/Rescaled_range
I have a vector “X” and a vector “u” and if I just use the cumulative sum command matlab doesn’t interpret the sum in the way I want it. This is how I want the calculation to look like:
Y(1) = ( X(1)-u(1) )^2
Y(2) = ( X(1)-u(2) )^2 + ( X(2)-u(2) )^2
Y(3) = ( X(1)-u(3) )^2 + ( X(2)-u(3) )^2+ ( X(3)-u(3) )^2
Y(4) = ( X(1)-u(4) )^2 + ( X(2)-u(4) )^2+ ( X(3)-u(4) )^2+ ( X(4)-u(4) )^2
….and so on
So in other words the calculation is increasing in size for each step and keeps X(1), X(2), X(3) and so on but only uses the most current value for “u” in each of the calculations.
This should be pretty simple right? How do I make it happen in matlab, any ideas?
Thanks

3 Kommentare

Adam
Adam am 23 Sep. 2014
It may just be because it's 9pm here and I don't have my work head on, but the formula you put up there looks considerably more complicated than the one you are pointing us to which appears to just be a standard n-dimensional standard deviation.
Star Strider
Star Strider am 23 Sep. 2014
It looks like the summation inside the radical in ‘Step 5’.
Peta
Peta am 24 Sep. 2014
Yes exactly, I just wrote the (x(i)-u)^2 summation part, then it should be multiplied with 1/t and square rooted. Or do you think I interpreted the formula wrong?

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Roger Stafford
Roger Stafford am 24 Sep. 2014
Bearbeitet: Roger Stafford am 24 Sep. 2014

2 Stimmen

To use 'cumsum' you need to get the u's out of the squared expression. Let's assume X and u are row vectors.
Y = cumsum(X.^2)-2*cumsum(X).*u+(1:size(u,2)).*u.^2;

1 Kommentar

Peta
Peta am 24 Sep. 2014
Aah, beautiful! That did exactly what I wanted it to! I would never have figured that out myself, thanks!

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Star Strider
Star Strider am 23 Sep. 2014

2 Stimmen

This seems to do what you want:
X = randi(20,1,10); % Create ‘X’
u = 1:10; % Create ‘u’
for k1 = 1:10
Y(k1) = sum((X(1:k1)-u(k1)).^2);
end
At least it produces the correct result (checked with manual calculation).

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