Hello,
How can I generate a uniformly distributed random vector with its sum to be equal to 1?
Thank you
MATLAB: Random vector v from uniform distribution at (0,1) with sum(v)=1
probability random vector
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This is not simple, if you want to find a truly uniform set. In 24 dimensions, things get fairly complicated. And the domain the numbers can live in is actually pretty wide, thus the interval [5,800].
So, if you have all numbers at or near their mininum,
5*24ans = 120>> 800*24ans = 19200>> (5 + 800)/2*24ans = 9660The point being that it will be a quite rare event that 24 numbers, chosen randomly from that interval, will have a sum in the interval [3550,3650].
The law of large numbers tells us that the sum of that set will be effectively normally distributed. And, in fact, this gives us a solution to the problem!
Lets see. Each member of that set will be uniform, on the interval [5,800]. That means the mean of x_i will be
a = 5;b = 800;(a + b)/2ans = 402.5And the variance (of each number in the set) will be
(b-a)^2/12ans = 52668.75So now the sum of 24 of them will be fairly accurately normally distributed, with mean, variance, and standard deviation:
sumMean = (a+b)/2*24xmean = 9660sumVar = (b-a)^2/12*24xvar = 1264050sumStd = sqrt(sumVar)xstd = 1124.29978208661So, now what are the odds that you will find a sum in the range from [3550,3650]? That would be found from a cumulative normal.
sumLimits = [3550,3650];normcdf(sumLimits,sumMean,sumStd)ans = 2.74761306010529e-08 4.50716095152589e-08 diff(ans)ans = 1.7595478914206e-08Hmm. So you will probably need to generate roughly 1e8 sets of 24 uniform numbers before you find a single set that lives in the interval [3550,3650].
But! As I said, this actually gives you a near trivial solution! How? We break the problem down into two simpler problems. Hard problems are hard. Easy problems are way easier. So lets create two simpler problems.
First, can we sample from a truncated normal distributiuon? That part is easy.
norminv(sumNormProbLimits(1) + rand()*diff(sumNormProbLimits),sumMean,sumStd)ans = 3637.00633112408As you can see, that is a number that has atuncated normal distribution, with the desired parameters. So now we can just use randfixedsum. I'll put it all together here:
a = 5;b = 800;ndim = 24;sumMean = (a+b)/2*ndim;sumVar = (b-a)^2/12*ndim;sumStd = sqrt(sumVar);sumLimits = [3550,3650];sumNormProbLimits = normcdf(sumLimits,sumMean,sumStd);S = norminv(sumNormProbLimits(1) + rand()*diff(sumNormProbLimits),sumMean,sumStd);randfixedsum(24,1,S,a,b)ans = 32.6666351919248 226.171574450247 26.4486244673284 59.5273902733376 105.140425145937 198.085619814426 313.588691531256 125.527412419838 222.108869160278 78.6355991625287 116.832965448386 25.2315920517267 10.7762636314948 295.321051677379 339.225232904641 91.7004913938697 29.5858150543207 184.827862803465 37.7525635267701 115.866539562388 512.657889835195 372.544944227602 13.6087483202661 58.868132270981So, a set of 24 uniformly distributed random numbers that have their sum in the desired interval. The numbers really are uniformly distributed to lie in the interval [5,800], but the condition that the sum be so relatively small, makes it unlikely that any of them are near the maximum. On this random set, the max was as large as 512. That seems reasonable to me, under the conditional sampling that was required.
sum(ans)ans = 3592.70093432559It was really pretty easy to do. I did need randfixedsum, which you can get from the File Exchange for free download.
This is a problem of appreciating the terminology, and the double negative way things get worded. Do two negatives make a positive? Not always. But then, not, not, always either. ;-)
Lets see what happens when we throw a very non-random sequence at runstest.
runstest(cumsum(rand(1,100)))ans = 1runstest(mod(1:100,2))ans = 1The null hypothesis is that the sequence posed is truly random. Can we KNOW a sequence was a random one? Well, you can't KNOW that as fact. You can only test how random it seems to be, or, how non-random. So both cases above are clearly seen to be unlikely events, IF the sequence were truly a random one. Essentially, we reject the null hypothesis, so the a priori assumption that the sequences really are random.
But, when I do this:
runstest(rand(1,1000))ans = 0runstest is not willing to say the sequence seems to be non-random. So, we cannot reject the null hypothesis. Does that mean the sequence is random? NO!!!!!! Only that we cannot comfortably decide the sequence is probably non-random.
So, now lets look at p.
[h,p] = runstest(cumsum(rand(1,100)))h = 1p = 4.04395443273876e-29Yep. It is non-random. I'll bet some decent money on that. p is TINY, it reflects a measure of how likely we think this event might be, in context of a truly random sequence.
Now, lets try a sequence of sequences.
x = rand(1,100);[h,p] = runstest(x)h = 0p = 0.838684318956303[h,p] = runstest([x,x])h = 0p = 0.720356034109542[h,p] = runstest([x,x,x])h = 0p = 0.64038079177353[h,p] = runstest([x,x,x,x])h = 0p = 0.578177151383335See that the way I've constructed those sequences, the latter ones are seen to be somewhat less likely to be random in context of the runs seen, although runstest is not designed to detect the totally non-random way I extended the sequences. It cannot detect long lags as I extended the sequence, which in fact, make the longer versions non-random.
If I try a sequence of length 400 directly from rand, p is now larger.
[h,p] = runstest(rand(1,400))h = 0p = 0.72633274852919So h==1 says runstest has detected an event that did not seem consistent with randomness. h==0 says it did not detect significant non-randomness, but it cannot tell you to conclude a sequence is truly random, or not. p tells you the strength of what it sees.
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