I am trying to calculate hypergeometric probabilities using phyper in R, and notice a strange behavior. I am looking at gene set overlap probabilities, and in one case, there are no "successful draws," so:
x=0
m=430
n=19500
k=2
Since in general I'm looking for over-enrichment I use:
phyper(0,600,19000,2,lower.tail=FALSE)
and get 0.043 which appears to be significant.
However, choosing two "n" genes should have probability of
19500/19930*19499/19929=0.957
so, shouldn't the phyper result be greater than 0.957?
Related
I am using the useful gratia package by Gavin Simpson to extract the difference in two smooths for two different levels of a factor variable. The smooths are generated by the wonderful mgcv package. For example
library(mgcv)
library(gratia)
m1 <- gam(outcome ~ s(dep_var, by = fact_var) + fact_var, data = my.data)
diff1 <- difference_smooths(m1, smooth = "s(dep_var)")
draw(diff1)
This give me a graph of the difference between the two smooths for each level of the "by" variable in the gam() call. The graph has a shaded 95% credible interval (CI) for the difference.
Statistical significance, or areas of statistical significance at the 0.05 level, is assessed by whether or where the y = 0 line crosses the CI, where the y axis represents the difference between the smooths.
Here is an example from Gavin's site where the "by" factor variable had 3 levels.
The differences are clearly statistically significant (at 0.05) over nearly all of the graphs.
Here is another example I have generated using a "by" variable with 2 levels.
The difference in my example is clearly not statistically significant anywhere.
In the mgcv package, an approximate p value is outputted for a smooth fit that tests the null hypothesis that the coefficients are all = 0, based on a chi square test.
My question is, can anyone suggest a way of calculating a p value that similarly assesses the difference between the two smooths instead of solely relying on graphical evidence?
The output from difference_smooths() is a data frame with differences between the smooth functions at 100 points in the range of the smoothed variable, the standard error for the difference and the upper and lower limits of the CI.
Here is a link to the release of gratia 0.4 that explains the difference_smooths() function
enter link description here
but gratia is now at version 0.6
enter link description here
Thanks in advance for taking the time to consider this.
Don
One way of getting a p value for the interaction between the by factor variables is to manipulate the difference_smooths() function by activating the ci_level option. Default is 0.95. The ci_level can be manipulated to find a level where the y = 0 is no longer within the CI bands. If for example this occurred when ci_level = my_level, the p value for testing the hypothesis that the difference is zero everywhere would be 1 - my_level.
This is not totally satisfactory. For example, it would take a little manual experimentation and it may be difficult to discern accurately when zero drops out of the CI. Although, a function could be written to search the accompanying data frame that is outputted with difference_smooths() as the ci_level is varied. This is not totally satisfactory either because the detection of a non-zero CI would be dependent on the 100 points chosen by difference_smooths() to assess the difference between the two curves. Then again, the standard errors are approximate for a GAM using mgcv, so that shouldn't be too much of a problem.
Here is a graph where the zero first drops out of the CI.
Zero dropped out at ci_level = 0.88 and was still in the interval at ci_level = 0.89. So an approxiamte p value would be 1 - 0.88 = 0.12.
Can anyone think of a better way?
Reply to Gavin Simpson's comments Feb 19
Thanks very much Gavin for taking the time to make your comments.
I am not sure if using the criterion, >= 0 (for negative diffs), is a good way to go. Because of the draws from the posterior, there is likely to be many diffs that meet this criterion. I am interpreting your criterion as sample the posterior distribution and count how many differences meet the criterion, calculate the percentage and that is the p value. Correct me if I have misunderstood. Using this approach, I consistently got p values at around 0.45 - 0.5 for different gam models, even when it was clear the difference in the smooths should be statistically significant, at least at p = 0.05, because the confidence band around the smooth did not contain zero at a number of points.
Instead, I was thinking perhaps it would be better to compare the means of the posterior distribution of each of the diffs. For example
# get coefficients for the by smooths
coeff.level1 <- coef(gam.model1)[31:38]
coeff.level0 <- coef(gam.model1)[23:30]
# these indices are specific to my multi-variable gam.model1
# in my case 8 coefficients per smooth
# get posterior coefficients variances for the by smooths' coefficients
vp_level1 <- gam.model1$Vp[31:38, 31:38]
vp_level0 <- gam.model1$Vp[23:30, 23:30]
#run the simulation to get the distribution of each
#difference coefficient using the joint variance
library(MASS)
no.draws = 1000
sim <- mvrnorm(n = no.draws, (coeff.level1 - coeff.level0),
(vp_level1 + vp_level0))
# sim is a no.draws X no. of coefficients (8 in my case) matrix
# put the results into a data.frame.
y.group <- data.frame(y = as.vector(sim),
group = c(rep(1,no.draws), rep(2,no.draws),
rep(3,no.draws), rep(4,no.draws),
rep(5,no.draws), rep(6,no.draws),
rep(7,no.draws), rep(8,no.draws)) )
# y has the differences sampled from their posterior distributions.
# group is just a grouping name for the 8 sets of differences,
# (one set for each difference in coefficients)
# compare means with a linear regression
lm.test <- lm(y ~ as.factor(group), data = y.group)
summary(lm.test)
# The p value for the F statistic tells you how
# compatible the data are with the null hypothesis that
# all the group means are equal to each other.
# Same F statistic and p value from
anova(lm.test)
One could argue that if all coefficients are not equal to each other then they all can't be equal to zero but that isn't what we want here.
The basis of the smooth tests of fit given by summary(mgcv::gam.model1)
is a joint test of all coefficients == 0. This would be from a type of likelihood ratio test where model fit with and without a term are compared.
I would appreciate some ideas how to do this with the difference between two smooths.
Now that I got this far, I had a rethink of your original suggestion of using the criterion, >= 0 (for negative diffs). I reinterpreted this as meaning for each simulated coefficient difference distribution (in my case 8), count when this occurs and make a table where each row (my case, 8) is for one of these distributions with two columns holding this count and (number of simulation draws minus count), Then on this table run a chi square test. When I did this, I got a very low p value when I believe I shouldn't have as 0 was well within the smooth difference CI across almost all the levels of the exposure. Maybe I am still misunderstanding your suggestion.
Follow up thought Feb 24
In a follow up thought, we could create a variable that represents the interaction between the by factor and continuous variable
library(dplyr)
my.dat <- my.dat %>% mutate(interact.var =
ifelse(factor.2levels == "yes", 1, 0)*cont.var)
Here I am assuming that factor.2levels has the levels ("no", "yes"), and "no" is the reference level. The ifelse function creates a dummy variable which is multiplied by the continuous variable to generate the interactive variable.
Then we place this interactive variable in the GAM and get the usual statistical test for fit, that is, testing all the coefficients == 0.
#GavinSimpson actually posted a method of how to get the difference between two smooths and assess its statistical significance here in 2017. Thanks to Matteo Fasiolo for pointing me in that direction.
In that approach, the by variable is converted to an ordered categorical variable which causes mgcv::gam to produce difference smooths in comparison to the reference level. Statistical significance for the difference smooths is then tested in the usual way with the summary command for the gam model.
However, and correct me if I have misunderstood, the ordered factor approach causes the smooth for the main effect to now be the smooth for the reference level of the ordered factor.
The approach I suggested, see the main post under the heading, Follow up thought Feb 24, where the interaction variable is created, gives an almost identical result for the p value for the difference smooth but does not change the smooth for the main effect. It also does not change the intercept and the linear term for the by categorical variable which also both changed with the ordered variable approach.
I've tried to contact William Revelle about this but he isn't responding.
In the psych package there is a function called cor.smoother, which determines whether or not a correlation matrix is positive definite. Its explanation is as follows:
"cor.smoother examines all of nvar minors of rank nvar-1 by systematically dropping one variable at a time and finding the eigen value decomposition. It reports those variables, which, when dropped, produce a positive definite matrix. It also reports the number of negative eigenvalues when each variable is dropped. Finally, it compares the original correlation matrix to the smoothed correlation matrix and reports those items with absolute deviations great than cut. These are all hints as to what might be wrong with a correlation matrix."
It is the really the statement in bold that I am hoping someone can interpret in a more understandable way for me?
A belated answer to your question.
Correlation matrices are said to be improper (or more accurately, not positive semi-definite) when at least one of the eigen values of the matrix is less than 0. This can happen if you have some missing data and are using pair-wise complete correlations. It is particularly likely to happen if you are doing tetrachoric or polychoric correlations based upon data sets with some or even a lot of missing data.
(A correlation matrix, R, may be decomposed into a set of eigen vectors (X) and eigen values (lambda) where R = X lambda X’. This decomposition is the basis of components analysis and factor analysis, but that is more than you want to know.)
The cor.smooth function finds the eigen values and then adjusts the negative ones by making them slightly positive (and adjusting the other ones to compensate for this change).
The cor.smoother function attempts to identify the variables that are making the matrix improper. It does this by considering all the matrices generated by dropping one variable at a time and seeing which ones of those are not positive semi-definite (i.e. have eigen values < 0.) Ideally, this will identify one variable that is messing things up.
An example of this is in the burt data set where the sorrow-tenderness correlation was probably mistyped and the .87 should be .81.
cor.smoother(burt) #identifies tenderness and sorrow as likely culprits
Hello and thanks in advance for the help!
A while back I asked a question about randomly selecting values according to a probability distribution. This is related, but I think it deserves its own post.
The vector I created in the last question was binary, now I would like to generate a weighted vector (ie with bounded integers). I am sampling from a zero-inflated or quasi-poisson distribution with a long tail, so there is a much higher probability of selecting a zero than another value, but there is a finite probability of selecting a large value (eg 63).
I can use rpois to select values from a poisson distribution and create a vector of a given length. This is similar to what I would like to do, so I will use it as an example.
e=seq(0:63)
vec<-c(0,0,0,1,1,1)
ones <- which(vec == 1L)
temp=rpois((sum(vec)),e)
vec[ones]<-temp
This works well for assigning a specific number of values selected from a poisson distribution to a vector. Is there anyway to make it quasi-poisson or zero inflated?
There's a big list of the different distributions here: http://cran.r-project.org/web/views/Distributions.html
For zero inflated poisson...
install.packages("gamlss.dist")
library(gamlss.dist)
rZIP(n, mu, sigma)
For quasi-poisson, it looks like there are some capabilities within the VGAM package with quasipoissonff, but that seems to be for fitting rather than generating. It looks like Arthur Charpentier was on to something here - but you really need to know what you're looking for to get the distribution right: http://freakonometrics.blog.free.fr/index.php?post/2010/10/21/How-to-genrerate-variables-from-a-quasi-Poisson-distribution
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