I have a problem using fisher’s exact test in R with a simulated p-value, but I don’t know if it’s a caused by “the technique” ( R ) or if it is (statistically) intended to work that way.
One of the datasets I want to work with:
matrix(c(103,0,2,1,0,0,1,0,3,0,0,3,0,0,0,0,0,0,19,3,57,11,2,87,1,2,0,869,4,2,8,1,4,3,18,16,5,60,60,42,1,1,1,1,21,704,40,759,404,151,1491,9,40,144),ncol=2,nrow=27)
The resulting p-value is always the same, no matter how often I repeat the test:
p = 1 / (B+1)
(B = number of replicates used in the Monte Carlo test)
When I shorten the matrix it works if the number of rows is lower than 19. Nevertheless it is not a matter of number of cells in the matrix. After transforming it into a matrix with 3 columns it still does not work, although it does when using the same numbers in just two columns.
Varying simulated p-values:
>a <- matrix(c(103,0,2,1,0,0,1,0,3,0,0,3,0,0,0,0,0,0,869,4,2,8,1,4,3,18,16,5,60,60,42,1,1,1,1,21),ncol=2,nrow=18)
>b <- matrix(c(103,0,2,1,0,0,1,0,3,0,0,3,0,0,0,0,0,0,19,869,4,2,8,1,4,3,18,16,5,60,60,42,1,1,1,1,21,704),ncol=2,nrow=19)
>c <- matrix(c(103,0,2,1,0,0,1,0,3,0,0,3,0,0,0,0,0,0,869,4,2,8,1,4,3,18,16,5,60,60,42,1,1,1,1,21),ncol=3,nrow=12)
>fisher.test(a,simulate.p.value=TRUE)$p.value
Number of cells in a and b are the same, but the simulation only works with matrix a.
Does anyone know if it is a statistical issue or a R issue and, if so, how it could be solved?
Thanks for your suggestions
I think that you are just seeing a very significant result. The p-value is being computed as the number of simulated (and the original) matrices that are as extreme or more extreme than the original. If none of the randomly generated matrices are as or more extreme then the p-value will just be 1 (the original matrix is as extreme as itself) divided by the total number of matrices which is $B+1$ (the B simulated and the 1 original matrix). If you run the function with enough samples (high enough B) then you will start to see some of the random matrices as or more extreme and therefor varying p-values, but the time to do so is probably not reasonable.
Related
With R I can try to find the probability that the Age vector below resulted from random sampling. I used the runs test (from randtests package) with resulted in p-value = 0.2892. Other colleagues used the rle functune (run length encoding in R) or others to simulate whether the probabilities of random allocation generating the observed sequences. Their result shows p < 0.00000001 that this sequence is the result of random sampling. I am trying to find the R code to replicate their findings. any help is highly appreciated on how to simulate to replicate their findings.
Update: I received advice from statistician that I can do this using non-parametric bootstrap. However, I still do not know how this can be done. I appreciate your help.
example:
Age <-c(68,71,72,69,80,78,80,81,84,82,67,73,65,68,66,70,69,72,74,73,68,75,70,72,75,73,69,75,74,79,80,78,80,81,79,82,69,73,67,66,70,72,69,72,75,80,68,69,71,77,70,73) ;
randtests::runs.test(Age);
X <- rle(Age);X$lengths
What was initially presented isn't the whole story. If one looks at the supplement where these numbers are from, the reported p-value is for comparing two vectors. OP only provides one, and hence the task is not well-defined.
The full assertion of the research article is that
group1 <- c(68,71,72,69,80,78,80,81,84,82,67,73,65,68,66,70,69,72,74,73,68,75,70,72,75,73)
group2 <- c(69,75,74,79,80,78,80,81,79,82,69,73,67,66,70,72,69,72,75,80,68,69,71,77,70,73)
being two independent random samples has a p-value < 0.00000001.
Even checking identity along position (10 entries in original) with permutations within a group, I'm seeing only 2 or 3 draws per million that have a similar number of identical values. I.e., something like:
set.seed(123)
mean(replicate(1e6, sum(sample(group1, length(group1)) == group2)) >= 10)
# 2e-06
Testing correlations and/or bootstrapping could easily be in the p-value range that is reported (nothing as extreme in 100 million simulations).
In this post https://stats.stackexchange.com/questions/94909/course-of-action-for-2x2-tables-with-0s-in-cell-and-low-cell-counts, OP said that s/he got a p-value 0.5152 while conducted a Fisher's exact test for the following data:
Control Cases
A 8 0
B 14 0
But I am getting p-value=1 and odds ratio=0 for the data. My R codes are:
a <- matrix(c(8,14,0,0),2,2)
(res <- fisher.test(a))
Where am I doing mistake?
Good afternoon :)
https://en.wikipedia.org/wiki/Fisher%27s_exact_test
Haven't used these in a while, but I'm assuming its your column of two 0's:
p = choose(14, 14) * choose(8, 8)/ choose(22, 22)
which is 1.0. For odds ratio, read here: https://en.wikipedia.org/wiki/Odds_ratio
The 0's are either the numerators or the denominators. I think this makes sense, as a column of 0's effectively mean you have a group with no observations in.
You get the strange p-value=1 and OR=0 because one or more of your counts is 0. It should not be computed by the chi-square equation, which through multiplication yields chi-values of 0 for these respective cells:
Chi square equation, cell-by-cell.
Instead, you should use the Fisher's exact test ("fisher.test()") which to some extent can correct for the very low cell counts (normally you should use Fisher's for whenever you have at least 20% of cells with a count of <5). Source: https://www.ncbi.nlm.nih.gov/pubmed/23894860 Using the chi-square analysis will require you to correct using the Yates' correction, (e.g.: chisq.test(matrix, correct = T)).
I have an R matrix which dimensions are ~20,000,000 rows by 1,000 columns. The first column represents counts and the rest of the columns represent the probabilities of a multinomial distribution of these counts. So in other words, in each row the first column is n and the rest of the k columns are the probabilities of the k categories. Another point is that the matrix is sparse, meaning that in each row there are many columns with value of 0.
Here's a toy matrix I created:
mat=rbind(c(5,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1),c(2,0.2,0.2,0.2,0.2,0.2,0,0,0,0,0),c(22,0.4,0.6,0,0,0,0,0,0,0,0),c(5,0.5,0.2,0,0.1,0.2,0,0,0,0,0),c(4,0.4,0.15,0.15,0.15,0.15,0,0,0,0,0),c(10,0.6,0.1,0.1,0.1,0.1,0,0,0,0,0))
What I'd like to do is obtain an empirical measure of the variance of the counts for each category. The natural thing that comes to mind is to obtain random draws and then compute the variances over them. Something like:
draws = apply(mat,1,function(x) rmultinom(samples,x[1],x[2:ncol(mat)]))
Where say samples=100000
Then I can run an apply over draws to compute the variances.
However, for my real data dimensions this will become prohibitive at least in terms of RAM. Is whether a more efficient solution in R to this problem?
If all you need is the variance of the counts, just compute it immediately instead of returning the intermediate simulated draws.
draws = apply(mat,1,function(x) var(rmultinom(samples,x[1],x[2:ncol(mat)])))
I am new to R and cointegration so please have patience with me as I try to explain what it is that I am trying to do. I am trying to find cointegrated variables among 1500-2000 voltage variables in the west power system in Canada/US. THe frequency is hourly (common in power) and cointegrated combinations can be as few as N variables and a maximum of M variables.
I tried to use ca.jo but here are issues that I ran into:
1) ca.jo (Johansen) has a limit to the number of variables it can work with
2) ca.jo appears to force the first variable in the y(t) vector to be the dependent variable (see below).
Eigenvectors, normalised to first column: (These are the cointegration relations)
V1.l2 V2.l2 V3.l2
V1.l2 1.0000000 1.0000000 1.0000000
V2.l2 -0.2597057 -2.3888060 -0.4181294
V3.l2 -0.6443270 -0.6901678 0.5429844
As you can see ca.jo tries to find linear combinations of the 3 variables but by forcing the coefficient on the first variable (in this case V1) to be 1 (i.e. the dependent variable). My understanding was that ca.jo would try to find all combinations such that every variable is selected as a dependent variable. You can see the same treatment in the examples given in the documentation for ca.jo.
3) ca.jo does not appear to find linear combinations of fewer than the number of variables in the y(t) vector. So if there were 5 variables and 3 of them are cointegrated (i.e. V1 ~ V2 + V3) then ca.jo fails to find this combination. Perhaps I am not using ca.jo correctly but my expectation was that a cointegrated combination where V1 ~ V2 + V3 is the same as V1 ~ V2 + V3 + 0 x V4 + 0 x V5. In other words the coefficient of the variable that are NOT cointegrated should be zero and ca.jo should find this type of combination.
I would greatly appreciate some further insight as I am fairly new to R and cointegration and have spent the past 2 months teaching myself.
Thank you.
I have also posted on nabble:
http://r.789695.n4.nabble.com/ca-jo-cointegration-multivariate-case-tc3469210.html
I'm not an expert, but since no one is responding, I'm going to try to take a stab at this one.. EDIT: I noticed that I just answered to a 4 year old question. Hopefully it might still be useful to others in the future.
Your general understanding is correct. I'm not going to go in great detail about the whole procedure but will try to give some general insight. The first thing that the Johansen procedure does is create a VECM out of the VAR model that best corresponds to the data (This is why you need the lag length for the VAR as input to the procedure as well). The procedure will then investigate the non-lagged component matrix of the VECM by looking at its rank: If the variables are not cointegrated then the rank of the matrix will not be significantly different from 0. A more intuitive way of understanding the johansen VECM equations is to notice the comparibility with the ADF procedure for each distinct row of the model.
Furthermore, The rank of the matrix is equal to the number of its eigenvalues (characteristic roots) that are different from zero. Each eigenvalue is associated with a different cointegrating vector, which
is equal to its corresponding eigenvector. Hence, An eigenvalue significantly different
from zero indicates a significant cointegrating vector. Significance of the vectors can be tested with two distinct statistics: The max statistic or the trace statistic. The trace test tests the null hypothesis of less than or equal to r cointegrating vectors against the alternative of more than r cointegrating vectors. In contrast, The maximum eigenvalue test tests the null hypothesis of r cointegrating vectors against the alternative of r + 1 cointegrating vectors.
Now for an example,
# We fit data to a VAR to obtain the optimal VAR length. Use SC information criterion to find optimal model.
varest <- VAR(yourData,p=1,type="const",lag.max=24, ic="SC")
# obtain lag length of VAR that best fits the data
lagLength <- max(2,varest$p)
# Perform Johansen procedure for cointegration
# Allow intercepts in the cointegrating vector: data without zero mean
# Use trace statistic (null hypothesis: number of cointegrating vectors <= r)
res <- ca.jo(yourData,type="trace",ecdet="const",K=lagLength,spec="longrun")
testStatistics <- res#teststat
criticalValues <- res#criticalValues
# chi^2. If testStatic for r<= 0 is greater than the corresponding criticalValue, then r<=0 is rejected and we have at least one cointegrating vector
# We use 90% confidence level to make our decision
if(testStatistics[length(testStatistics)] >= criticalValues[dim(criticalValues)[1],1])
{
# Return eigenvector that has maximum eigenvalue. Note: we throw away the constant!!
return(res#V[1:ncol(yourData),which.max(res#lambda)])
}
This piece of code checks if there is at least one cointegrating vector (r<=0) and then returns the vector with the highest cointegrating properties or in other words, the vector with the highest eigenvalue (lamda).
Regarding your question: the procedure does not "force" anything. It checks all combinations, that is why you have your 3 different vectors. It is my understanding that the method just scales/normalizes the vector to the first variable.
Regarding your other question: The procedure will calculate the vectors for which the residual has the strongest mean reverting / stationarity properties. If one or more of your variables does not contribute further to these properties then the component for this variable in the vector will indeed be 0. However, if the component value is not 0 then it means that "stronger" cointegration was found by including the extra variable in the model.
Furthermore, you can test test significance of your components. Johansen allows a researcher to test a hypothesis about one or more
coefficients in the cointegrating relationship by viewing the hypothesis as
a restriction on the non-lagged component matrix in the VECM. If there exist r cointegrating vectors, only these linear combinations or linear transformations of them, or combinations of the cointegrating vectors, will be stationary. However, I'm not aware on how to perform these extra checks in R.
Probably, the best way for you to proceed is to first test the combinations that contain a smaller number of variables. You then have the option to not add extra variables to these cointegrating subsets if you don't want to. But as already mentioned, adding other variables can potentially increase the cointegrating properties / stationarity of your residuals. It will depend on your requirements whether or not this is the behaviour you want.
I've been searching for an answer to this and I think I found one so I'm sharing with you hoping it's the right solution.
By using the johansen test you test for the ranks (number of cointegration vectors), and it also returns the eigenvectors, and the alphas and betas do build said vectors.
In theory if you reject r=0 and accept r=1 (value of r=0 > critical value and r=1 < critical value) you would search for the highest eigenvalue and from that build your vector. On this case, if the highest eigenvalue was the first, it would be V1*1+V2*(-0.26)+V3*(-0.64).
This would generate the cointegration residuals for these variables.
Again, I'm not 100%, but preety sure the above is how it works.
Nonetheless, you can always use the cajools function from the urca package to create a VECM automatically. You only need to feed it a cajo object and define the number of ranks (https://cran.r-project.org/web/packages/urca/urca.pdf).
If someone could confirm / correct this, it would be appreciated.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
how to generate pseudo-random positive definite matrix with constraints on the off-diagonal elements?
The user wants to impose a unique, non-trivial, upper/lower bound on the correlation between every pair of variable in a var/covar matrix.
For example: I want a variance matrix in which all variables have 0.9 > |rho(x_i,x_j)| > 0.6, rho(x_i,x_j) being the correlation between variables x_i and x_j.
Thanks.
There are MANY issues here.
First of all, are the pseudo-random deviates assumed to be normally distributed? I'll assume they are, as any discussion of correlation matrices gets nasty if we diverge into non-normal distributions.
Next, it is rather simple to generate pseudo-random normal deviates, given a covariance matrix. Generate standard normal (independent) deviates, and then transform by multiplying by the Cholesky factor of the covariance matrix. Add in the mean at the end if the mean was not zero.
And, a covariance matrix is also rather simple to generate given a correlation matrix. Just pre and post multiply the correlation matrix by a diagonal matrix composed of the standard deviations. This scales a correlation matrix into a covariance matrix.
I'm still not sure where the problem lies in this question, since it would seem easy enough to generate a "random" correlation matrix, with elements uniformly distributed in the desired range.
So all of the above is rather trivial by any reasonable standards, and there are many tools out there to generate pseudo-random normal deviates given the above information.
Perhaps the issue is the user insists that the resulting random matrix of deviates must have correlations in the specified range. You must recognize that a set of random numbers will only have the desired distribution parameters in an asymptotic sense. Thus, as the sample size goes to infinity, you should expect to see the specified distribution parameters. But any small sample set will not necessarily have the desired parameters, in the desired ranges.
For example, (in MATLAB) here is a simple positive definite 3x3 matrix. As such, it makes a very nice covariance matrix.
S = randn(3);
S = S'*S
S =
0.78863 0.01123 -0.27879
0.01123 4.9316 3.5732
-0.27879 3.5732 2.7872
I'll convert S into a correlation matrix.
s = sqrt(diag(S));
C = diag(1./s)*S*diag(1./s)
C =
1 0.0056945 -0.18804
0.0056945 1 0.96377
-0.18804 0.96377 1
Now, I can sample from a normal distribution using the statistics toolbox (mvnrnd should do the trick.) As easy is to use a Cholesky factor.
L = chol(S)
L =
0.88805 0.012646 -0.31394
0 2.2207 1.6108
0 0 0.30643
Now, generate pseudo-random deviates, then transform them as desired.
X = randn(20,3)*L;
cov(X)
ans =
0.79069 -0.14297 -0.45032
-0.14297 6.0607 4.5459
-0.45032 4.5459 3.6549
corr(X)
ans =
1 -0.06531 -0.2649
-0.06531 1 0.96587
-0.2649 0.96587 1
If your desire was that the correlations must ALWAYS be greater than -0.188, then this sampling technique has failed, since the numbers are pseudo-random. In fact, that goal will be a difficult one to achieve unless your sample size is large enough.
You might employ a simple rejection scheme, whereby you do the sampling, then redo it repeatedly until the sample has the desired properties, with the correlations in the desired ranges. This may get tiring.
An approach that might work (but one that I've not totally thought out at this point) is to use the standard scheme as above to generate a random sample. Compute the correlations. I they fail to lie in the proper ranges, then identify the perturbation one would need to make to the actual (measured) covariance matrix of your data, so that the correlations would be as desired. Now, find a zero mean random perturbation to your sampled data that would move the sample covariance matrix in the desired direction.
This might work, but unless I knew that this is actually the question at hand, I won't bother to go any more deeply into it. (Edit: I've thought some more about this problem, and it appears to be a quadratic programming problem, with quadratic constraints, to find the smallest perturbation to a matrix X, such that the resulting covariance (or correlation) matrix has the desired properties.)
This is not a complete answer, but a suggestion of a possible constructive method:
Looking at the characterizations of the positive definite matrices (http://en.wikipedia.org/wiki/Positive-definite_matrix) I think one of the most affordable approaches could be using the Sylvester criterion.
You can start with a trivial 1x1 random matrix with positive determinant and expand it in one row and column step by step while ensuring that the new matrix has also a positive determinant (how to achieve that is up to you ^_^).
Woodship,
"First of all, are the pseudo-random deviates assumed to be normally distributed?"
yes.
"Perhaps the issue is the user insists that the resulting random matrix of deviates must have correlations in the specified range."
Yes, that's the whole difficulty
"You must recognize that a set of random numbers will only have the desired distribution parameters in an asymptotic sense."
True, but this is not the problem here: your strategy works for p=2, but fails for p>2, regardless of sample size.
"If your desire was that the correlations must ALWAYS be greater than -0.188, then this sampling technique has failed, since the numbers are pseudo-random. In fact, that goal will be a difficult one to achieve unless your sample size is large enough."
It is not a sample size issue b/c with p>2 you do not even observe convergence to the right range for the correlations, as sample size growths: i tried the technique you suggest before posting here, it obviously is flawed.
"You might employ a simple rejection scheme, whereby you do the sampling, then redo it repeatedly until the sample has the desired properties, with the correlations in the desired ranges. This may get tiring."
Not an option, for p large (say larger than 10) this option is intractable.
"Compute the correlations. I they fail to lie in the proper ranges, then identify the perturbation one would need to make to the actual (measured) covariance matrix of your data, so that the correlations would be as desired."
Ditto
As for the QP, i understand the constraints, but i'm not sure about the way you define the objective function; by using the "smallest perturbation" off some initial matrix, you will always end up getting the same (solution) matrix: all the off diagonal entries will be exactly equal to either one of the two bounds (e.g. not pseudo random); plus it is kind of an overkill isn't it ?
Come on people, there must be something simpler