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}
Related
Didn't know how to paraphrase the question well.
Function for example:
Data:https://www.dropbox.com/s/wr61qyhhf6ujvny/data.mat?dl=0
In this case how do I calculate that the rest point of this function is ~1? I have access to the vector that makes the plot.
I guess the mean is an approximation but in some cases it can be pretty bad.
Under the assumption that the "rest" point is the steady-state value in your data and the fact that the steady-state value happens the majority of the times in your data, you can simply bin all of the points and use each unique value as a separate bin. The bin with the highest count should correspond to the steady-state value.
You can do this by a combination of histc and unique. Assuming your data is stored in y, do this:
%// Find all unique values in your data
bins = unique(y);
%// Find the total number of occurrences per unique value
counts = histc(y, bins);
%// Figure out which bin has the largest count
[~,max_bin] = max(counts);
%// Figure out the corresponding y value
ss_value = bins(max_bin);
ss_value contains the steady-state value of your data, corresponding to the most occurring output point with the assumptions I laid out above.
A minor caveat with the above approach is that this is not friendly to floating point data whose unique values are generated by floating point values whose decimal values beyond the first few significant digits are different.
Here's an example of your data from point 2300 to 2320:
>> format long g;
>> y(2300:2320)
ans =
0.99995724232555
0.999957488454868
0.999957733165346
0.999957976465197
0.999958218362579
0.999958458865564
0.999958697982251
0.999958935720613
0.999959172088623
0.999959407094224
0.999959640745246
0.999959873049548
0.999960104014889
0.999960333649014
0.999960561959611
0.999960788954326
0.99996101464076
0.999961239026462
0.999961462118947
0.999961683925704
0.999961904454139
Therefore, what I'd recommend is to perhaps round so that the first 5 or so significant digits are maintained.
You can do this to your dataset before you continue:
num_digits = 5;
y_round = round(y*(10^num_digits))/(10^num_digits);
This will first multiply by 10^n where n is the number of digits you desire so that the decimal point is shifted over by n positions. We round this result, then divide by 10^n to bring it back to the scale that it was before. If you do this, for those points that were 0.9999... where there are n decimal places, these will get rounded to 1, and it may help in the above calculations.
However, more recent versions of MATLAB have this functionality already built-in to round, and you can just do this:
num_digits = 5;
y_round = round(y,num_digits);
Minor Note
More recent versions of MATLAB discourage the use of histc and recommend you use histcounts instead. Same function definition and expected inputs and outputs... so just replace histc with histcounts if your MATLAB version can handle it.
Using the above logic, you could also use the median too. If the majority of data is fluctuating around 1, then the median would have a high probability that the steady-state value is chosen... so try this too:
ss_value = median(y_round);
I'm looking for a function that will generate random values between 0 and 1, inclusive. I have generated 120,000 random values by using rand() function in octave, but haven't once got the values 0 or 1 as output. Does rand() ever produce such values? If not, is there any other function I can use to achieve the desired result?
If you read the documentation of rand in both Octave and MATLAB, it is an open interval between (0,1), so no, it shouldn't generate the numbers 0 or 1.
However, you can perhaps generate a set of random integers, then normalize the values so that they lie between [0,1]. So perhaps use something like randi (MATLAB docs, Octave docs) where it generates integer values from 1 up to a given maximum. With this, define this maximum number, then subtract by 1 and divide by this offset maximum to get values between [0,1] inclusive:
max_num = 10000; %// Define maximum number
N = 1000; %// Define size of vector
out = (randi(max_num, N, 1) - 1) / (max_num - 1); %// Output
If you want this to act more like rand but including 0 and 1, make the max_num variable quite large.
Mathematically, if you sample from a (continuous) uniform distribution on the closed interval [0 1], values 0 and 1 (or any value, in fact) have probability strictly zero.
Programmatically,
If you have a random generator that produces values of type double on the closed interval [0 1], the probability of getting the value 0, or 1, is not zero, but it's so small it can be neglected.
If the random generator produces values from the open interval (0, 1), the probability of getting a value 0, or 1, is strictly zero.
So the probability is either strictly zero or so small it can be neglected. Therefore, you shouldn't worry about that: in either case the probability is zero for practical purposes. Even if rand were of type (1) above, and thus could produce 0 and 1, it would produce them with probability so small that you would "never" see those values.
Does that sound strange? Well, that happens with any number. You "never" see rand ever outputting exactly 1/4, either. There are so many possible outputs, all of them equally likely, that the probability of any given output is virtually zero.
rand produces numbers from the open interval (0,1), which does not include 0 or 1, so you should never get those values.. This was more clearly documented in previous versions, but it's still stated in the help text for rand (type help rand rather than doc rand).
However, since it produces doubles, there are only a finite number of values that it will actually produce. The precise set varies depending on the RNG algorithm used. For Mersenne twister, the default algorithm, the possible values are all multiples of 2^(-53), within the open interval (0,1). (See doc RandStream.list, and then "Choosing a Random Number Generator" for info on other generators).
Note that 2^(-53) is eps/2. Therefore, it's equivalent to drawing from the closed interval [2^(-53), 1-2^(-53)], or [eps/2, 1-eps/2].
You can scale this interval to [0,1] by subtracting eps/2 and dividing by 1-eps. (Use format hex to display enough precision to check that at the bit level).
So x = (rand-eps/2)/(1-eps) should give you values on the closed interval [0,1].
But I should give a word of caution: they've put a lot of effort into making sure that output of rand gives an appropriate distribution of any given double within (0,1), and I don't think you're going to get the same nice properties on [0,1] if you apply the scaling I suggested. My knowledge of floating-point math and RNGs isn't up to explaining why, or what you might do about that.
I just tried this:
octave:1> max(rand(10000000,1))
ans = 1.00000
octave:2> min(rand(10000000,1))
ans = 3.3788e-08
Did not give me 0 strictly, so watch out for floating point operations.
Edit
Even though I said, watch out for floating point operations I did fall for that. As #eigenchris pointed out:
format long g
octave:1> a=max(rand(1000000,1))
a = 0.999999711020176
It yields a floating number close to one, not equal, as you can see now after changing the precision, as #rayryeng suggested.
Although not direct to the question here, I find it helpful to link to this SO post Octave - random generate number that has a one liner to generate 1s and 0s using r = rand > 0.5.
An issue I've come across multiple times is wanting to take two similar data sets and create histograms from them where the bins are identical, so as to easily calculate things like histogram overlap.
You can define the number of bins (obviously) using
[counts, bins] = hist(data,number_of_bins)
But there's not an obvious way (as far as I can see) to make the bin size equal for several different data sets. If remember when I initially looked finding various people who seem to have the same issue, but no good solutions.
The right, easy way
As pointed out by horchler, this can easily be achieved using either histc (which lets you define your bins vector), or vectorizing your histogram input into hist.
The wrong, stupid way
I'm leaving below as a reminder to others that even stupid questions can yield worthwhile answers
I've been using the following approach for a while, so figured it might be useful for others (or, someone can very quickly point out the correct way to do this!).
The general approach relies on the fact that MATLAB's hist function defines an equally spaced number of bins between the largest and smallest value in your sample. So, if you append a start (smallest) and end (largest) value to your various samples which is the min and max for all samples of interest, this forces the histogram range to be equal for all your data sets. You can then truncate the first and last values to recreate your original data.
For example, create the following data set
A = randn(1,2000)+7
B = randn(1,2000)+9
[counts_A, bins_A] = hist(A, 500);
[counts_B, bins_B] = hist(B, 500);
Here for my specific data sets I get the following results
bins_A(1) % 3.8127 (this is also min(A) )
bins_A(500) % 10.3081 (this is also max(A) )
bins_B(1) % 5.6310 (this is also min(B) )
bins_B(500) % 13.0254 (this is also max(B) )
To create equal bins you can simply first define a min and max value which is slightly smaller than both ranges;
topval = max([max(A) max(B)])+0.05;
bottomval = min([min(A) min(B)])-0.05;
The addition/subtraction of 0.05 is based on knowledge of the range of values - you don't want your extra bin to be too far or too close to the actual range. That being said, for this example by using the joint min/max values this code will work irrespective of the A and B values generated.
Now we re-create histogram counts and bins using (note the extra 2 bins are for our new largest and smallest value)
[counts_Ae, bins_Ae] = hist([bottomval, A, topval], 502);
[counts_Be, bins_Be] = hist([bottomval, B, topval], 502);
Finally, you truncate the first and last bin and value entries to recreate your original sample exactly
bins_A = bins_Ae(2:501)
bins_B = bins_Ae(2:501)
counts_A = counts_Ae(2:501)
counts_B = counts_Be(2:501)
Now
bins_A(1) % 3.7655
bins_A(500) % 13.0735
bins_B(1) % 3.7655
bins_B(500) % 13.0735
From this you can easily plot both histograms again
bar([bins_A;bins_B]', [counts_A;counts_B]')
And also plot the histogram overlap with ease
bar(bins_A,(counts_A+counts_B)-(abs(counts_A-counts_B)))
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.
So say, I have a = [2 7 4 9 2 4 999]
And I'd like to remove 999 from the matrix (which is an obvious outlier).
Is there a general way to remove values like this? I have a set of vectors and not all of them have extreme values like that. prctile(a,99.5) is going to output the largest number in the vector no matter how extreme (or non-extreme) it is.
There are several way to do that, but first you must define what is "extreme'? Is it above some threshold? above some number of standard deviations?
Or, if you know you have exactly n of these extreme events and that their values are larger than the rest, you can use sort and the delete the last n elements. etc...
For example a(a>threshold)=[] will take care of a threshold like definition, while a(a>mean(a)+n*std(a))=[] will take care of discarding values that are n standard deviation above the mean of a.
A completely different approach is to use the median of a, if the vector is as short as you mention, you want to look on a median value and then you can either threshold anything above some factor of that value a(a>n*median(a))=[] .
Last, a way to assess an approach to treat these spikes would be to take a histogram of the data, and work from there...
I can think of two:
Sort your matrix and remove n-elements from top and bottom.
Compute the mean and the standard deviation and discard all values that fall outside:
mean +/- (n * standard deviation)
In both cases n must be chosen by the user.
Filter your signal.
%choose the value
N = 10;
filtered = filter(ones(1,N)/N, 1, signal);
Find the noise
noise = signal - filtered;
Remove noisy elements
THRESH = 50;
signal = signal(abs(noise) < THRESH);
It is better than mean+-n*stddev approach because it looks for local changes so it won't fail on a slowly changing signal like [1 2 3 ... 998 998].