You want a matrix that is 400x1000, where the 1000 columns are "resampled" versions of the orignal data. You say the data have an exponential distribution, and you say that you do not want to lose the exponential decay when you resample. Is there a possible confusion about the distinction between exponential distribution and exponential decay? Is there possible confusion about resampling versus reshuffling?
Matlab's resample() keeps the time order of the data the same. It samples the data faster or slower than the original sampling. I think this is not what you want, since it would lead to resampled data with more (if faster) or less (if slower) than 400 samples.
Reshuffling is used in many machine learning applications and in some statistical testing. It will preserve the number of samples and will destroy any time correlations in the data.
Neither reshuffling nor resampling will alter the distribution of the data. The distribution is independent from the time ordering. A histogram of the data reveals the distribution of the sample. @Star Strider plotted the distribution of your data, and of the best-fit lognormal distribution. His plot shows that the sample histogram looks more like a lognormal distribution than an exponential distribution. I suspect that when you said your data has an exponential distribution, you meant to say that the data shows exponential decay with time (plus random noise). This code reshuffles the data. After reshuffling, it no longer has exponential decay with time. But the distribution is unchanged.
y(:,i)=std_spk_avg(randperm(N));
If the code above works as intended, then the mean of each column of y should equal the mean of std_spk_avg, and the std.dev. of each column of y should equal the st.dev. of std_spk_avg.. Let's check and compare:
disp([mean(std_spk_avg),min(yMean),max(yMean)])
disp([std(std_spk_avg),min(yStd),max(yStd)])
Looks like it works as intended.
