I have two vectors a and b with same length. The vectors contains number of times a game has been played. So for example game 1 has been played 265350 in group a while it has been played 52516 in group b.
a <- c(265350, 89148, 243182, 208991, 113090, 124698, 146574, 33649, 276435, 9320, 58630, 20139, 26178, 7837, 6405, 399)
b <- c(52516, 42840, 60571, 58355, 46975, 47262, 58197, 42074, 50090, 27198, 45491, 43048, 44512, 27266, 43519, 28766)
I want to use Pearsons Chi square test to test Independence between the two vector. In R I type
chisq.test(a,b)
and I get a p-value 0.2348 meaning that the two vectors are independent (H is true).
But when I run pairwise.prop.test(a,b) and get all the pairwise p-values and almost all of them are very low, meaning that there are pairwise dependence between the two vectors but this is in contrast to the first result. How can that be ?
The pairwise.prop.test is not the correct test for your case.
As it mentions in the documentation:
Calculate pairwise comparisons between pairs of proportions with correction for multiple testing
And also:
x (first argument).
Vector of counts of successes or a matrix with 2 columns giving the counts of successes and failures, respectively.
And
n (second argument).
Vector of counts of trials; ignored if x is a matrix.
So, x in the number of successes out of n which is the trials, i.e. x <= (less than or equal) to each pair in n. And this is why pairwise.prop.test is used for proportions. As an example imagine tossing a coin 1000 times getting heads in 550. x would be 550 and n would be 1000. In your case you do not have something similar, you just have counts of a game in two groups.
The correct hypothesis test for testing independence is the chisq.test(a,b) that you have already used and I would trust that.
Related
So we are given a set of integers from 0 to n. This is then randomized. The goal is to calculate the number of expected integers which remain in the same position in both lists. I have tried to set up two indicator variables for each integer and then mapping it to the two different sets, but I don't really know how to go from there.
The random variable X, representing the number of your integers which remain in the same position after randomisation, follows the binomial distribution with n+1 trials and a probability of 1/(n+1), therefore the expected number of integers remaining in place is 1.
My reasoning is:
Each integer can move to any other position in the list after randomisation, with equal probability. So whether an integer remains in place can be considered a Bernoulli distribution, with probability 1/(n+1), since there are n+1 possible position it could move to, and only 1 position for it to have remained in place.
There are therefore n+1 Bernoulli distributions, all with the same probability, and all independent of each other. (A Bernoulli distribution represents a yes / no outcome where the yes has a given probability.)
The binomial distribution is defined as the number of successes in a given number of identical independent trials, or (equivalently) the number of "yes" outcomes in a given number of independent Bernoulli distributions with the same probability.
The number of your integers which remain in place after randomisation is therefore a bimonial distribution, probability 1/(n+1) and with n+1 trials.
The mean of a binomial distribution with n trials with probability p is np, therefore in your case the expected number of integers remaining in place is (n+1) . (1/(n+1)) which is 1.
For more info on the binomial distribution, see wikipedia.
I would like to measure the hamming sequence similarity in which the substitution costs are not based on the substitution rates in the observed sequences but based on the spatial autocorrelation within the study area of the different states (states are thus not related to DNA but something else).
I divided my study area in grid cells of equal size (e.g. 1000m) and measured how often the same "state" is observed in a neighboring cell (Rook-case). Consequently the weight matrix indicates that from state A to A (to move within the same states) has a much higher probability than to go from A to B or B to C or A to C. This already indicates that states have a high spatial autocorrelation.
The problem is, if you want to measure sequence similarity the substitution matrix should be 0 at the diagonal. Therefore I was wondering whether there is a kind of transformation to go from an "autocorrelation matrix" to a substitution matrix, with 0 values along the diagonal. By means of this we would like to account for spatial autocorrelation in the study area in our sequence similarity measure. To do my analysis I am using the package TraMineR.
Example matrix in R for sequences consisting out of four states (A,B,C,D):
Sequence example: AAAAAABBBBCCCCCCCCCCCCDDDDDDDDDDDDDDDDDDDDDDDAAAAAAAAA
Autocorrelation matrix:
A = c(17.50,3.00,1.00,0.05)
B = c(3.00,10.00,2.00,1.00)
C = c(1.00,2.00,30.00,3.00)
D = c(0.05,1.00,3.00,20.00)
subm = rbind(A,B,C,D)
colnames(subm) = c("A","B","C","D")
how to transform this matrix to a substitution matrix?
First, TraMineR computes the Hamming distance, i.e., a dissimilarity, not a similarity.
The simple Hamming distance is just the count of mismatches between two sequences. For example, the Hamming distance between AABBCC and ABBBAC is 2, and between AAAAAA and AAAAAA it is 0 since there are no mismatches.
Generalized Hamming allows to weighting mismatches (not matches!) with substitution costs. For example if the substitution cost between A and B is 1.5, and is 2 between B and C, then the distance would be the weighted sum of mismatches, i.e., 3.5 between the first two sequences. It would still be zero between one sequence and itself.
From what I understand, the shown matrix is not the matrix of substitution costs. It is the matrix of what you call 'spatial autocorrelations', and you look for how you can turn this information into substitutions costs.
The idea is to assign high substitution cost (mismatch weight) when the autocorrelation (a rate in your case) is low, i.e., when there is a low probability to find say state B in the neighborhood of state A, and to assign a low substitution cost when the probability is high. Since your probability matrix is symmetric, a simple solution is to use $1 - p(A|B)$ for all off diagonal terms, and leave 0 on the diagonal for the reason explained above.
sm <- 1 - subm/100
diag(sm) <- 0
sm
For non symmetric probabilities, you could use a similar formula to the one used for deriving the costs from transition rates, i.e., $2 - p(A|B) - p(B|A)$.
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.
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.