I am trying to understand the stoachstic uniform selection algorithm as described in he docs: https://se.mathworks.com/help/gads/genetic-algorithm-options.html
The ga default selection function, Stochastic uniform, lays out a line in which each parent corresponds to a section of the line of length proportional to its scaled value. The algorithm moves along the line in steps of equal size. At each step, the algorithm allocates a parent from the section it lands on. The first step is a uniform random number less than the step size.
For myself the above docs can be interpreted in two ways:
Either a random number x will be picked initially and all subsequent "steps" are simply multiple of it.
Step size: 1
Random x e.g. 0.5
Location on line: 0.5, 1, 1.5, 2, 2.5
The algorithm moves along the line in fixed steps and additionally a random x < the fixed size is added every time.
Fixed Step size: 1
Random x varies all the time but < 1
Location on line: 1.1, 2.3, 3.2, 4.5, 5.1
Number 1 faces the issue that if the random value chosen is too small only the most fit individual will be selected as we don't move along the line at all. So is the 2nd interpretation correct?
As far as I understood the scaled fitness values sum up to the count of parents that will be generated, therefore isn't the step size always 1 since we can fit exactly as many steps * parents needed on the line?
Here's a third interpretation for your consideration. The algorithm moves along the line in fixed steps. However, the starting point is less than the step size.
Fixed step size: 1
Randomly chosen start: 0.32
Locations on line: 0.32 1.32 2.32 3.32 4.32 5.32
By using a fixed step size, the algorithm knows exactly how many parents will be selected. For example, if the line is 100 units long, and the step size is 1, then exactly 100 parents will be selected. But which parents are selected is determined by the random starting point.
This assumes that there are multiple parents to choose from in each interval of length 1. And the most fit individual has a scaled length less than 1.
Related
I am wondering if anyone can explain the interpretation of the size (number of feature) in a time series? For example consider a simple script in Matlab
X= randn(2,5,2)
X(:,:,1) =
-0.5530 0.4291 0.3937 -1.2534 0.2811
-1.4926 -0.7019 -0.8305 -1.4034 1.9545
X(:,:,2) =
0.2004 0.1438 2.3655 -0.1589 0.7140
0.4905 0.2301 -0.7813 -0.6737 0.2552
Assume X is a time series with the following output
This generates 2 vectors of length 5 each has 2 rows. Can anyone tell me what is exactly the meaning of first 2 and 5?
In some websites it says a creating 5 vectors of length 5 and size 2. What does size mean here?
Is 2 like number of features and 5 is like number of time series. The reason for this confusion is because I do not understand how to interpret following sentence:
"Generate 2 vector-valued sequences of length 5; each vector has size
2."
What do size 2 and length 5 mean here?
This entirely depends on your data, and how you want to store this. If you have some 2D data over time, I find it convenient to have a data matrix with in the 1st and 2nd dimension the 2D data per time step, and in the 3rd dimension time.
Say I have a movie of 1920 by 1080 pixels with 100 frames, I'd store this as mov = rand(1080,1920,100) (1080 and 1920 swapped because of row, col order of indexing). So now mov(:,:,1) would give me the first frame etc.
BTW, your X is a normal array, not to be confused with the timeseries object.
I want to generate a matrix of random numbers (normrnd with mean == 0) that satisfy the following constraints using MATLAB (or any other language)
The sum of the absolute values in the matrix must equal X
The largest abs(single number) must equal Y
The difference between the number and its 8 neighbors (3 if in corner, 5 if on edge) must be less than Z
It would be relatively easy to satisfy one of the constraints, but I can't think of an algorithm that satisfies all of them...
Any ideas?
I am not sure whether to edit my post or to reply here, so I am editing... #MZimmerman6, you have a point. Though these constraints won't produce a unique solution, how would I obtain multiple solutions without using rand?
A very simply 3 x 3 where 5 is the max element value, 30 is the sum, and 2 is the difference
5 4 3
4 4 2
3 2 3
Rody, that actually may help...I need to think more :)
Luis ...Hmmm...why not? I can add up the absolute value of a normally distributed sample...right?
Here is an algorithm to get the 'random' numbers that you need.
Generate a valid number (for example in the middle)
Determine the feasible range for one of the numbers next to it
If there is no range, you go to step 1, otherwise generate a number and continue
Depending on your constraints it may take a while of course. You could add an other step to see if changing the existing numbers would help before going back to step 1.
How would I go about setting my tspan vector for solutions to my ode between (1,5]? I've thought of just doing >>tspan = [1:(any amount of steps):5] but is that okay?
You can't numerically integrate over a (half) open interval. Numerical integration always operates at specific numeric points, i.e. not an interval anyway, but a finite set of numbers. What you specify with the tspan argument are the smallest and largest number in that set, and both therefore are included in it. You can put more numbers into tspan to explicitly request integration results at these points, too, but however you choose those this doesn't change the fact that you don't have an interval.
If the motivation of the question is that your equations have a singularity at 1, you might specify a start point that is slightly larger, e.g. [1 + 1e-5, 5].
Seems ok, but 2 notes:
A. It should be tspan=[1:(any size of step):5];, not amount of steps. For amount of steps, you can write: tspan=linspace(1,5,(any amount of steps));
B. Those options are include '1'. If you want the interval (1,5], you shold add the size of step to '1' on each of the options. For example: tspan=[1+(size of step) : (size of step) :5];
Having read carefully the previous question
Random numbers that add to 100: Matlab
I am struggling to solve a similar but slightly more complex problem.
I would like to create an array of n elements that sums to 1, however I want an added constraint that the minimum increment (or if you like number of significant figures) for each element is fixed.
For example if I want 10 numbers that sum to 1 without any constraint the following works perfectly:
num_stocks=10;
num_simulations=100000;
temp = [zeros(num_simulations,1),sort(rand(num_simulations,num_stocks-1),2),ones(num_simulations,1)];
weights = diff(temp,[],2);
I foolishly thought that by scaling this I could add the constraint as follows
num_stocks=10;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp2 = [zeros(num_simulations,1),sort(round(rand(num_simulations,num_stocks-1)*scaling)/scaling,2),ones(num_simulations,1)];
weights2 = diff(temp2,[],2);
However though this works for small values of n & small values of increment, if for example n=1,000 & the increment is 0.1% then over a large number of trials the first and last numbers have a mean which is consistently below 0.1%.
I am sure there is a logical explanation/solution to this but I have been tearing my hair out to try & find it & wondered anybody would be so kind as to point me in the right direction. To put the problem into context create random stock portfolios (hence the sum to 1).
Thanks in advance
Thank you for the responses so far, just to clarify (as I think my initial question was perhaps badly phrased), it is the weights that have a fixed increment of 0.1% so 0%, 0.1%, 0.2% etc.
I did try using integers initially
num_stocks=1000;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp = [zeros(num_simulations,1),sort(randi([0 scaling],num_simulations,num_stocks-1),2),ones(num_simulations,1)*scaling];
weights = (diff(temp,[],2)/scaling);
test=mean(weights);
but this was worse, the mean for the 1st & last weights is well below 0.1%.....
Edit to reflect excellent answer by Floris & clarify
The original code I was using to solve this problem (before finding this forum) was
function x = monkey_weights_original(simulations,stocks)
stockmatrix=1:stocks;
base_weight=1/stocks;
r=randi(stocks,stocks,simulations);
x=histc(r,stockmatrix)*base_weight;
end
This runs very fast, which was important considering I want to run a total of 10,000,000 simulations, 10,000 simulations on 1,000 stocks takes just over 2 seconds with a single core & I am running the whole code on an 8 core machine using the parallel toolbox.
It also gives exactly the distribution I was looking for in terms of means, and I think that it is just as likely to get a portfolio that is 100% in 1 stock as it is to geta portfolio that is 0.1% in every stock (though I'm happy to be corrected).
My issue issue is that although it works for 1,000 stocks & an increment of 0.1% and I guess it works for 100 stocks & an increment of 1%, as the number of stocks decreases then each pick becomes a very large percentage (in the extreme with 2 stocks you will always get a 50/50 portfolio).
In effect I think this solution is like the binomial solution Floris suggests (but more limited)
However my question has arrisen because I would like to make my approach more flexible & have the possibility of say 3 stocks & an increment of 1% which my current code will not handle correctly, hence how I stumbled accross the original question on stackoverflow
Floris's recursive approach will get to the right answer, but the speed will be a major issue considering the scale of the problem.
An example of the original research is here
http://www.huffingtonpost.com/2013/04/05/monkeys-stocks-study_n_3021285.html
I am currently working on extending it with more flexibility on portfolio weights & numbers of stock in the index, but it appears my programming & probability theory ability are a limiting factor.......
One problem I can see is that your formula allows for numbers to be zero - when the rounding operation results in two consecutive numbers to be the same after sorting. Not sure if you consider that a problem - but I suggest you think about it (it would mean your model portfolio has fewer than N stocks in it since the contribution of one of the stocks would be zero).
The other thing to note is that the probability of getting the extreme values in your distribution is half of what you want them to be: If you have uniformly distributed numbers from 0 to 1000, and you round them, the numbers that round to 0 were in the interval [0 0.5>; the ones that round to 1 came from [0.5 1.5> - twice as big. The last number (rounding to 1000) is again from a smaller interval: [999.5 1000]. Thus you will not get the first and last number as often as you think. If instead of round you use floor I think you will get the answer you expect.
EDIT
I thought about this some more, and came up with a slow but (I think) accurate method for doing this. The basic idea is this:
Think in terms of integers; rather than dividing the interval 0 - 1 in steps of 0.001, divide the interval 0 - 1000 in integer steps
If we try to divide N into m intervals, the mean size of a step should be N / m; but being integer, we would expect the intervals to be binomially distributed
This suggests an algorithm in which we choose the first interval as a binomially distributed variate with mean (N/m) - call the first value v1; then divide the remaining interval N - v1 into m-1 steps; we can do so recursively.
The following code implements this:
% random integers adding up to a definite sum
function r = randomInt(n, limit)
% returns an array of n random integers
% whose sum is limit
% calls itself recursively; slow but accurate
if n>1
v = binomialRandom(limit, 1 / n);
r = [v randomInt(n-1, limit - v)];
else
r = limit;
end
function b = binomialRandom(N, p)
b = sum(rand(1,N)<p); % slow but direct
To get 10000 instances, you run this as follows:
tic
portfolio = zeros(10000, 10);
for ii = 1:10000
portfolio(ii,:) = randomInt(10, 1000);
end
toc
This ran in 3.8 seconds on a modest machine (single thread) - of course the method for obtaining a binomially distributed random variate is the thing slowing it down; there are statistical toolboxes with more efficient functions but I don't have one. If you increase the granularity (for example, by setting limit=10000) it will slow down more since you increase the number of random number samples that are generated; with limit = 10000 the above loop took 13.3 seconds to complete.
As a test, I found mean(portfolio)' and std(portfolio)' as follows (with limit=1000):
100.20 9.446
99.90 9.547
100.09 9.456
100.00 9.548
100.01 9.356
100.00 9.484
99.69 9.639
100.06 9.493
99.94 9.599
100.11 9.453
This looks like a pretty convincing "flat" distribution to me. We would expect the numbers to be binomially distributed with a mean of 100, and standard deviation of sqrt(p*(1-p)*n). In this case, p=0.1 so we expect s = 9.4868. The values I actually got were again quite close.
I realize that this is inefficient for large values of limit, and I made no attempt at efficiency. I find that clarity trumps speed when you develop something new. But for instance you could pre-compute the cumulative binomial distributions for p=1./(1:10), then do a random lookup; but if you are just going to do this once, for 100,000 instances, it will run in under a minute; unless you intend to do it many times, I wouldn't bother. But if anyone wants to improve this code I'd be happy to hear from them.
Eventually I have solved this problem!
I found a paper by 2 academics at John Hopkins University "Sampling Uniformly From The Unit Simplex"
http://www.cs.cmu.edu/~nasmith/papers/smith+tromble.tr04.pdf
In the paper they outline how naive algorthms don't work, in a way very similar to woodchips answer to the Random numbers that add to 100 question. They then go on to show that the method suggested by David Schwartz can also be slightly biased and propose a modified algorithm which appear to work.
If you want x numbers that sum to y
Sample uniformly x-1 random numbers from the range 1 to x+y-1 without replacement
Sort them
Add a zero at the beginning & x+y at the end
difference them & subtract 1 from each value
If you want to scale them as I do, then divide by y
It took me a while to realise why this works when the original approach didn't and it come down to the probability of getting a zero weight (as highlighted by Floris in his answer). To get a zero weight in the original version for all but the 1st or last weights your random numbers had to have 2 values the same but for the 1st & last ones then a random number of zero or the maximum number would result in a zero weight which is more likely.
In the revised algorithm, zero & the maximum number are not in the set of random choices & a zero weight occurs only if you select two consecutive numbers which is equally likely for every position.
I coded it up in Matlab as follows
function weights = unbiased_monkey_weights(num_simulations,num_stocks,min_increment)
scaling=1/min_increment;
sample=NaN(num_simulations,num_stocks-1);
for i=1:num_simulations
allcomb=randperm(scaling+num_stocks-1);
sample(i,:)=allcomb(1:num_stocks-1);
end
temp = [zeros(num_simulations,1),sort(sample,2),ones(num_simulations,1)*(scaling+num_stocks)];
weights = (diff(temp,[],2)-1)/scaling;
end
Obviously the loop is a bit clunky and as I'm using the 2009 version the randperm function only allows you to generate permutations of the whole set, however despite this I can run 10,000 simulations for 1,000 numbers in 5 seconds on my clunky laptop which is fast enough.
The mean weights are now correct & as a quick test I replicated woodchips generating 3 numbers that sum to 1 with the minimum increment being 0.01% & it also look right
Thank you all for your help and I hope this solution is useful to somebody else in the future
The simple answer is to use the schemes that work well with NO minimum increment, then transform the problem. As always, be careful. Some methods do NOT yield uniform sets of numbers.
Thus, suppose I want 11 numbers that sum to 100, with a constraint of a minimum increment of 5. I would first find 11 numbers that sum to 45, with no lower bound on the samples (other than zero.) I could use a tool from the file exchange for this. Simplest is to simply sample 10 numbers in the interval [0,45]. Sort them, then find the differences.
X = diff([0,sort(rand(1,10)),1]*45);
The vector X is a sample of numbers that sums to 45. But the vector Y sums to 100, with a minimum value of 5.
Y = X + 5;
Of course, this is trivially vectorized if you wish to find multiple sets of numbers with the given constraint.
I have 500,000 values for a variable derived from financial markets. Specifically, this variable represents distance from the mean (in standard deviations). This variable has a arbitrary distribution. I need a formula that will allow me to select a range around any value of this variable such that an equal (or close to it) amount of data points fall within that range.
This will allow me to then analyze all of the data points within a specific range and to treat them as "similar situations to the input."
From what I understand, this means that I need to convert it from arbitrary distribution to uniform distribution. I have read (but barely understood) that what I am looking for is called "probability integral transform."
Can anyone assist me with some code (Matlab preferred, but it doesn't really matter) to help me accomplish this?
Here's something I put together quickly. It's not polished and not perfect, but it does what you want to do.
clear
randList=[randn(1e4,1);2*randn(1e4,1)+5];
[xCdf,xList]=ksdensity(randList,'npoints',5e3,'function','cdf');
xRange=getInterval(5,xList,xCdf,0.1);
and the function getInterval is
function out=getInterval(yPoint,xList,xCdf,areaFraction)
yCdf=interp1(xList,xCdf,yPoint);
yCdfRange=[-areaFraction/2, areaFraction/2]+yCdf;
out=interp1(xCdf,xList,yCdfRange);
Explanation:
The CDF of the random distribution is shown below by the line in blue. You provide a point (here 5 in the input to getInterval) about which you want a range that gives you 10% of the area (input 0.1 to getInterval). The chosen point is marked by the red cross and the
interval is marked by the lines in green. You can get the corresponding points from the original list that lie within this interval as
newList=randList(randList>=xRange(1) & randList<=xRange(2));
You'll find that on an average, the number of points in this example is ~2000, which is 10% of numel(randList)
numel(newList)
ans =
2045
NOTE:
Please note that this was done quickly and I haven't made any checks to see if the chosen point is outside the range or if yCdfRange falls outside [0 1], in which case interp1 will return a NaN. This is fairly straightforward to implement, and I'll leave that to you.
Also, ksdensity is very CPU intensive. I wouldn't recommend increasing npoints to more than 1e4. I assume you're only working with a fixed list (i.e., you have a list of 5e5 points that you've obtained somehow and now you're just running tests/analyzing it). In that case, you can run ksdensity once and save the result.
I do not speak Matlab, but you need to find quantiles in your data. This is Mathematica code which would do this:
In[88]:= data = RandomVariate[SkewNormalDistribution[0, 1, 2], 10^4];
Compute quantile points:
In[91]:= q10 = Quantile[data, Range[0, 10]/10];
Now form pairs of consecutive quantiles:
In[92]:= intervals = Partition[q10, 2, 1];
In[93]:= intervals
Out[93]= {{-1.397, -0.136989}, {-0.136989, 0.123689}, {0.123689,
0.312232}, {0.312232, 0.478551}, {0.478551, 0.652482}, {0.652482,
0.829642}, {0.829642, 1.02801}, {1.02801, 1.27609}, {1.27609,
1.6237}, {1.6237, 4.04219}}
Verify that the splitting points separate data nearly evenly:
In[94]:= Table[Count[data, x_ /; i[[1]] <= x < i[[2]]], {i, intervals}]
Out[94]= {999, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000}