I work with fold changes as defined in the wikipedia article:
http://en.wikipedia.org/wiki/Fold_change
I work with fold changes for quite a while now but there was never really the need to calculate an average fold change across all my fold changes. When I did so yesterday I realised it's not a straight forward as it seems by just taking the mean of all fold change values. Here's an example which demonstrates my problem:
Let's consider 6 fold changes:
A = 1.1635710
B = 0.9284593
C = 1.1688855
D = 1.6400114
E = 1.2073252
F = 1.2830912
What I always do is to transform fold changes below 1 into the -1 format by dividing -1 by the fold change:
A = 1.1635710
B = -1.077053
C = 1.1688855
D = 1.6400114
E = 1.2073252
F = 1.2830912
Then to calculate the average fold change I took the mean of all fold change values which results in 0.8976386 suggesting that the average fold change is decreased even though from the values itself it is obvious that the average fold change should be increased as most of them are increased with a even higher magnitude.
Then I thought I calculate the average fold change before I transform fold changes below 1 into the minus format. Then I got an average value of 1.231891 which seems much more plausible. However, I doubt that this is correct either.
Let's simplify it even more. Let's say we have two fold changes with values of 0.8 (-1.25) and 1.25. Apparently, when looking at -1.25 and 1.25 the average fold change should be 1 so no average change. Which makes sense. But taking the average of 0.8 and 1.25 results in 1.025 so a slight increase.
Long story made short, I am not sure of how to best average a list of fold changes to get a representative average fold change.
This is a great question and I've been searching for the answer myself. Here is what I've come up with:
1) take the log of the fold changes (on the 0 to infinity scale); 2) average the log values; 3) calculate the anti-log; 4) then transform to +/- values if necessary
In your second example:
log(0.8) = -0.09691
log(1.25) = 0.09691
average = 0
10^0 = 1 (average fold change of 1)
For your first example with the 6 numbers and using this method I get your average to be 1.214425 fold.
Related
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 looked thoroughly on the internet for an answer to this question, but it seems to be too specific for an answer anywhere else. This is my last stop.
To preface, this is not a homework problem, but it is adapted from an online Coursera course, whose quiz has already passed. I got the correct answer, but it was mostly luck. Also, it is a more of a general programming question than anything related to the course, so I know for a fact that it is within my right to ask it on a public forum.
The last thing is that I'm trying to do this in MatLab; however if you have an answer that is in C++ or Python or any other high level language, that would be wonderful, as I could easily adapt those solutions to MatLab syntax.
Here it is:
I have two vectors, T and M, each with 600,000 elements/entries/integers.
T is entered as milliseconds from 1 to 600,000 in ascending order, and each element in M represents 'on' or 'off' (entered as 1 or 0 respectively) for each corresponding millisecond entry in T. So there are random 1's and 0's in M that correspond to a particular millisecond from 1 to 600,000 in T.
I need to take, starting with the 150th millisecond of T, and in 150 element/millisecond increments from there on (inclusive), the average millisecond value of those groups of 150 but ONLY of those milliseconds whose entries are 1 in M ('on'). For example, I need to look at the first 150 milliseconds in T, see which ones have a value of 1 in M, and then average them. Then I need to do it again with entries 151 to 300 in T, then 301 to 450, etc. etc. These new averages should also be stored in a new vector. The problem is, the number of corresponding 1's in M isn't going to be the same for every group of 150 milliseconds in T. (And yes, we are trying to average the actual value of the milliseconds, so the values we are using to average and the order of the entries in T will be the same).
My attempt:
It turns out there are only 53,583 random 1's in M (out of the 600,000 entries, the rest are 0). I used a 'find' operator to extract those entries from M that are a 1 into a new vector K that has the millisecond value corresponding from T. So K looks like a bunch of random numbers in ascending order, which is just a list of all the milliseconds in T who are 'on' (assigned a 1 in M).
So K looks something like [2 5 11 27 39 40 79 ...... 599,698 599,727 etc.] (all of the millisecond values who are a 1 in M).
So I have the vector K which is all of the values that I need to average in groups of 150, but the problem is that I need to go in groups of 150 based on the vector T (1 to 600,000), which means there won't always be the same number of 1's (or values in K) in every group of 150 milliseconds in T, which in turn means the number I need to divide by to get the average of each group is going to change for each group of 150. I know I need to use a loop to do the average millisecond value for every 150 entries, but how do I get the dividing number (the number of entries for each group of 150 who is assigned a 1 or 'on') to change on each iteration of the loop? Is there a way to bind T and M together so that they only use the requisite values from K whenever there is a 1 in M, and then just use a simple counter to average?
It's not a complicated problem, but it is very hard to explain. Sorry about that! I hope I explained as clearly as I could. Any help would be appreciated, although I'm sure you'll have questions first.
Thank you very much!
I think this should work OK.
sz = length(T);
n = sz / 150;
K = T.*M';
t = 1;
aver = zeros(n-1,1); % Your result vector
for i = 1:150:sz-150
aver(t) = mean(K(i:(i+150)-1));
t = t + 1;
end
-Rob
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].
I want to generate samples according to a simple categorical probability distribution, e.g.
p(A) = 0.1
p(B) = 0.5
p(C) = 0.25
p(D) = 0.15
Using rand(), which uniformly generates samples in (0,1] what is the best way to achieve this?
You could just check if the random number is less than the probability of each category, in order of increasing probability:
value = rand()
if value < p(A)
return A
if value < p(A)+p(B)
return B
if value < p(A)+p(B)+P(C)
return C
else
return D
I can't really tell you the best way to get them in order without knowing more about your code. If you have only a small number of cases that won't be changing, it might be easiest just to hard-code it once by hand as I've done above.
Edit: now that I think about it, since we're accumulating the probabilities, it doesn't really matter what order they're in. I've adjusted my code accordingly.
Edit edit: I think this is essentially how randsample works.
I have production (q) values from 4 different methods stored in the 4 matrices. Each of the 4 matrices contains q values from a different method as:
Matrix_1 = 1 row x 20 column
Matrix_2 = 100 rows x 20 columns
Matrix_3 = 100 rows x 20 columns
Matrix_4 = 100 rows x 20 columns
The number of columns indicate the number of years. 1 row would contain the production values corresponding to the 20 years. Other 99 rows for matrix 2, 3 and 4 are just the different realizations (or simulation runs). So basically the other 99 rows for matrix 2,3 and 4 are repeat cases (but not with exact values because of random numbers).
Consider Matrix_1 as the reference truth (or base case ). Now I want to compare the other 3 matrices with Matrix_1 to see which one among those three matrices (each with 100 repeats) compares best, or closely imitates, with Matrix_1.
How can this be done in Matlab?
I know, manually, that we use confidence interval (CI) by plotting the mean of Matrix_1, and drawing each distribution of mean of Matrix_2, mean of Matrix_3 and mean of Matrix_4. The largest CI among matrix 2, 3 and 4 which contains the reference truth (or mean of Matrix_1) will be the answer.
mean of Matrix_1 = (1 row x 1 column)
mean of Matrix_2 = (100 rows x 1 column)
mean of Matrix_3 = (100 rows x 1 column)
mean of Matrix_4 = (100 rows x 1 column)
I hope the question is clear and relevant to SO. Otherwise please feel free to edit/suggest anything in question. Thanks!
EDIT: My three methods I talked about are a1, a2 and a3 respectively. Here's my result:
ci_a1 =
1.0e+008 *
4.084733001497999
4.097677503988565
ci_a2 =
1.0e+008 *
5.424396063219890
5.586301025525149
ci_a3 =
1.0e+008 *
2.429145282593182
2.838897116739112
p_a1 =
8.094614835195452e-130
p_a2 =
2.824626709966993e-072
p_a3 =
3.054667629953656e-012
h_a1 = 1; h_a2 = 1; h_a3 = 1
None of my CI, from the three methods, includes the mean ( = 3.454992884900722e+008) inside it. So do we still consider p-value to choose the best result?
If I understand correctly the calculation in MATLAB is pretty strait-forward.
Steps 1-2 (mean calculation):
k1_mean = mean(k1);
k2_mean = mean(k2);
k3_mean = mean(k3);
k4_mean = mean(k4);
Step 3, use HIST to plot distribution histograms:
hist([k2_mean; k3_mean; k4_mean]')
Step 4. You can do t-test comparing your vectors 2, 3 and 4 against normal distribution with mean k1_mean and unknown variance. See TTEST for details.
[h,p,ci] = ttest(k2_mean,k1_mean);
EDIT : I misinterpreted your question. See the answer of Yuk and following comments. My answer is what you need if you want to compare distributions of two vectors instead of a vector against a single value. Apparently, the latter is the case here.
Regarding your t-tests, you should keep in mind that they test against a "true" mean. Given the number of values for each matrix and the confidence intervals it's not too difficult to guess the standard deviation on your results. This is a measure of the "spread" of your results. Now the error on your mean is calculated as the standard deviation of your results divided by the number of observations. And the confidence interval is calculated by multiplying that standard error with appx. 2.
This confidence interval contains the true mean in 95% of the cases. So if the true mean is exactly at the border of that interval, the p-value is 0.05 the further away the mean, the lower the p-value. This can be interpreted as the chance that the values you have in matrix 2, 3 or 4 come from a population with a mean as in matrix 1. If you see your p-values, these chances can be said to be non-existent.
So you see that when the number of values get high, the confidence interval becomes smaller and the t-test becomes very sensitive. What this tells you, is nothing more that the three matrices differ significantly from the mean. If you have to choose one, I'd take a look at the distributions anyway. Otherwise the one with the closest mean seems a good guess. If you want to get deeper into this, you could also ask on stats.stackexchange.com
Your question and your method aren't really clear :
Is the distribution equal in all columns? This is important, as two distributions can have the same mean, but differ significantly :
is there a reason why you don't use the Central Limit Theorem? This seems to me like a very complex way of obtaining a result that can easily be found using the fact that the distribution of a mean approaches a normal distribution where sd(mean) = sd(observations)/number of observations. Saves you quite some work -if the distributions are alike! -
Now if the question is really the comparison of distributions, you should consider looking at a qqplot for a general idea, and at a 2-sample kolmogorov-smirnov test for formal testing. But please read in on this test, as you have to understand what it does in order to interprete the results correctly.
On a sidenote : if you do this test on multiple cases, make sure you understand the problem of multiple comparisons and use the appropriate correction, eg. Bonferroni or Dunn-Sidak.