I have a number of data points, lets say in a vector v, and lets say there are "num" of them. If I write sd = std(v) did it assume a sample i.e. it used num-1 (in the denominator) or did I get a population standard dev i.e. it used num? How can I request one or the other?
MATLAB: Simple question about Standard Deviation.
standard deviationstatistics
Related Solutions
The question you are asking is a specific instance of "propagation of uncertainty". You might want to learn more about the topic.
For this specific case, I believe the answer is
std_sum = sqrt(1^2 + 2^2 + 3^2); % error of sum = square root of (sum of squares of individual uncertainties)
std_mean = std_sum/3Here is a simulation that suggests that this is correct:
rng defaultN = 100000;x1 = 10 + 1*randn(N,1);x2 = 15 + 2*randn(N,1);x3 = 20 + 3*randn(N,1);x = [x1, x2, x3];mean_x = mean(x,2);mean(mean_x)std(mean_x,1)figuresubplot(4,1,1), histogram(x(:,1))xlim([0 30])subplot(4,1,2), histogram(x(:,2))xlim([0 30])subplot(4,1,3), histogram(x(:,3))xlim([0 30])subplot(4,1,4), histogram(mean_x)xlim([0 30])
I would encourage you to be really careful about your terminology. Often when people write x +/- dx, dx refers to the standard error, not the standard deviation. The standard deviation is a property of the parent population. The standard error (of the mean, for example) will depend on the sampling -- specifically on sample size.
I think that what I have written above probably corresponds to what you wanted to know, but be careful.
std automatically assumes you are doing this for a complete population, NOT as a sample. When you are not sure about something, the best way is to test it! So, how can we test my claim? How might you have done so, and gotten an answer 6 hours earlier?
X = rand(1,5);First, what does std do, with no weights employed?
std(X)ans = 0.32851std(X,1)ans = 0.29383So as one should expect, the two are different, by a ratio of
sqrt(5)/2ans = 1.118std(X)/std(X,1)ans = 1.118That is as expected. std(X) divides by sqrt(n-1) in the formula, but std(X,1) divides by sqrt(n).
Now, lets see what happens when we use weights. A very simple weight vector is sufficient here.
W = ones(1,5);std(X,W)ans = 0.29383This is the population standard deviation, as produced by std(X,1).
std(X,1)ans = 0.29383The point is, it makes no sense at all to talk about a sample standard deviation when you have weights. Well, relatively little sense. Given a set of weights, we can only interpret this as the entire population.
I will concede that the documentation (both doc and help) for std should have made this fact explicitly clear, even though it seems clear to me regardless, since the alternative makes no sense. If you have weights, the points are treated as a complete population.
std(X,W,0)Error using sizeDimension argument must be a positive integer scalar within indexing range.Error in var (line 109)n = size(x,dim);Error in std (line 51)y = sqrt(var(varargin{:})); std(X,0,W)Error using sizeDimension argument must be a positive integer scalar within indexing range.Error in var (line 109)n = size(x,dim);Error in std (line 51)y = sqrt(var(varargin{:}));Yep. std agrees with me.
Best Answer