Say, I have N observations stored in an array, X = [x_1,x_2,...,x_N]. What is the meaning of E[\sum_{i=1}^N{x_i}]/N? This to me appears an average operation. But not sure. Can somebody please help what is the meaning of this operator with the help of any example in Matlab?
In general, (1) what is E[x]~ wherexis a random variable,
(2)E[x^2], and
(3)E[d]whered = 1/N(sum_{i=1}^N x_i)`
Your notation is still pretty bad to the point it's hard to understand, BUT there's a limit to what you can do on this site since it does not have MathJax enabled...
The expectation value is a generalized average in the sense that it should be weighted by the distribution from which X is drawn. If X is uniformly distributed then you'll get what you would call the "average" and what I think your first formula is giving. If X is distributed by some other distribution, then you will get something else.
If the distribution is discrete, then in general
E[f(x)] = sum_{i=1}^N [f(x_i) p(x_i)],
where p(x) is the distribution for the random variable.
I have a dataset with 274 samples (9 months) of the daily energy (Watts.hour) used on a residential household. I'm not sure if i'm applying the lpc function correctly.
My code is the following:
filename='9-months.csv';
energy = csvread(filename);
C=zeros(5,1);
counter=0;
N=3;
for n=274:-1:31
w2=energy(1:n-1,1);
a=lpc(w2,N);
energy_estimated=0;
for X = 1:N
energy_estimated = energy_estimated + (-a(X+1)*energy(n-X));
end
w_real=energy(n);
error2=abs(w_real-energy_estimated);
counter=counter+1;
C(counter,1)=error2;
end
mean_error=round(mean(C));
Being "n" the sample on analysis, I will use the energy array's values, from 1 to n-1, to calculate the lpc coefficientes (with N=3).
After that, it will apply the calculated coefficients on the "for" cycle presented, in order to calculate the estimated energy.
Finally, error2 outputs the error between the real energy and estimated value.
On the example presented ( http://www.mathworks.com/help/signal/ref/lpc.html ) some filters are used. Do I need to apply any filter to it? Is my methodology correct?
Thank you very much in advance!
The lpc seems to be used correctly, but there are a few other things about your code. I am adressign the part at he "for n" :
for n=31:274 %for me it would seem more logically to go forward in time
w2=energy(1:n-1,1);
a=lpc(w2,N);
energy_estimate=filter([0 -a(2:end)],1,w2);
energy_estimate=energy_estimate(end);
estimates(n)=energy_estimate;
end
error=energy(31:274)-estimates(31:274)';
meanerror=mean(error); %you dont really round mean errors
filter is exactly what you are trying to do with the X=1:N loop. but this will perform the calculation for the entire w2 vector. If you just want the last value take the (end) command as well.
Now there is no reason to calculate the error for every single value and then add them to a vector you can do that faster after the calculation.
Now if your trying to estimate future values with a lpc it could work like that, but you are implying that every value is only dependend on the last 3 values. Have you tried something like a polynominal approach? i would think that this would be closer to reality.
I'm working on a time series forecasting problem and I would like to confirm if it makes sense to compute the standard deviation of the root mean squared error. If so, is this the correct way?
STD_test = std(sqrt((y_real-y_pred).^2))
Also, imagine that the output of the model is 100, the RMSE 20 and the STD 10. This means that the real value is between [70,120] ?
The term y_real-y_pred is the vector of errors. The expression squares each element of it, and then sqrts each element of it, thus having the effect of abs(). Then std() is run on the vector of errors. Thus, this is computing the S.D. of the (absolute) error. That is a meaningful metric, but unlikely to be what you are after. Try:
e = y_real-y_pred;
MSE = mean(e.^2);
RMSE = sqrt(MSE);
sd = std(RMSE);
That will compute what you want. However, since RMSE is a scalar value, the value sd will be zero, so to answer the first part of your question, no it is not meaningful. What is meaningful is to look at the s.d. of the error itself:
sd = std(e);
RMSE and s.d. are somewhat related but they are distinct.
Your RMSE is fine; but the final conclusion is not! A std of 10 means there's a roughly 68% chance that your output lies within +- std. You can refer to this wiki link to learn more about the rule.
To apply the combination of SVD perturbation:
I = imread('image.jpg');
Ibw = single(im2double(I));
[U S V] = svd(Ibw);
% calculate derviced image
P = U * power(S, i) * V'; % where i is between 1 and 2
%To compute the combined image of SVD perturbations:
J = (single(I) + (alpha*P))/(1+alpha); % where alpha is between 0 and 1
I applied this method to a specific face recognition model and I noticed the accuracy was highly increased!! So it is very efficient!. Interestingly, I used the value i=3/4 and alpha=0.25 according to a paper that was published in a journal in 2012 in which the authors used i=3/4 and alpha=0.25. But I didn't make attention that i must be between 1 and 2! (I don't know if the authors make an error of dictation or they in fact used the value 3/4). So I tried to change the value of i to a value greater than 1, the accuracy decreased!!. So can I use the value 3/4 ? If yes, how can I argument therefore my approach?
The paper that I read is entitled "Enhanced SVD based face recognition". In page 3, they used the value i=3/4.
(http://www.oalib.com/paper/2050079)
Kindly I need your help and opinions. Any help will be very appreciated!
The idea to have the value between one and two is to magnify the singular values to make them invariant to illumination changes.
Refer to this paper: A New Face Recognition Method based on SVD Perturbation for Single Example Image per Person: Daoqiang Zhang,Songcan Chen,and Zhi-Hua Zhou
Note that when n equals to 1, the derived image P is equivalent to the original image I . If we
choose n>1, then the singular values satisfying s_i > 1 will be magnified. Thus the reconstructed
image P emphasizes the contribution of the large singular values, while restraining that of the
small ones. So by integrating P into I , we get a combined image J which keeps the main
information of the original image and is expected to work better against minor changes of
expression, illumination and occlusions.
My take:
When you scale the singular values in the exponent, you are basically introducing a non-linearity, so its possible that for a specific dataset, scaling down the singular values may be beneficial. Its like adjusting the gamma correction factor in a monitor.
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.