I run a GLMM_adaptive model (I am doing a resource selection function) and I am using the joint_tests function (emmeans package) to compute joint tests of the terms in the model. The problem is that one of the interactions does not appear in the results.
The model is:
mod.hinc <- mixed_model(fixed = Used ~ scale(ndvi) * season * vegfactor +
scale(ndvi^2) + scale(distance^2) + scale(distance) * season,
random = ~ 1 | id, data = hin.c,
family = binomial(link="logit"))
After running the model I run the joint_tests function:
install.packages("emmeans")
library(emmeans)
joint_tests(mod.hinc)
And this is the result:
joint_tests(mod.hinc)
model term df1 df2 F.ratio p.value
ndvi 1 Inf 36.465 <.0001
season 3 Inf 22.265 <.0001
vegfactor 4 Inf 4.548 0.0011
distance 1 Inf 33.939 <.0001
ndvi:season 3 Inf 13.826 <.0001
ndvi:vegfactor 4 Inf 8.500 <.0001
season:vegfactor 12 Inf 6.544 <.0001
ndvi:season:vegfactor 12 Inf 5.165 <.0001
I cannot find the reason why the interaction scale(distance)*season does not appear in the results.
Any help on that issue is welcome. I can provide more details about the model if is required.
Thank you very much in advance.
Juan
The short answer is that distance:season is not shown because it came up with zero d.f. for the associated interaction contrasts. You could verify this by running joint_tests(mod.hinc, show0df = TRUE).
Why it has 0 d.f. is less clear. However, that is not the only problem here. You have to be extremely careful with numeric predictors when using joint_tests(); it does not do a model ANOVA; instead, as documented, it constructs a reference grid from the fitted model and performs joint tests of interaction contrasts related to the predictors. With numeric predictors, the results depend on the reference grid used.
In this particular instance, the model includes quadratic effects of ndvi and distance; however, the default reference grid is constructed using the range of the covariates -- only two distinct values. Thus, we can pick up the effects of the overall linear trends, but not the curvature effects implied by the quadratic terms. That's why only 1 d.f. of those factors' main effects are tested. There are really 2 d.f. in the effects of ndvi and distance. In order to capture all of those effects, we need to have at least three distinct values of these covariates in the reference grid. One way (not the only way) to accomplish that is to reduce the covariates to their means, plus or minus 1 SD -- which can be accomplished via this code:
meanpm1sd <- function(x)
c(mean(x) - sd(x), mean(x), mean(x) + sd(x))
joint_tests(mod.hinc, cov.reduce = meanpm1sd)
This will yield a different set of joint tests that likely will include 2-d.f. tests of ndvi and distance. But I don't know if you will still have some interactions missing due to zero-d.f. dimensionalities.
You can look directly at the estimates being tested in detail if you have any questions about what those effects are. For example, for season:distance,
### construct the needed reference grid once and for all
RG <- ref_grid(mod.h1nc, cov.reduce = meanpm1sd)
EMM <- emmeans(RG, ~ season * distance)
CON <- contrast(EMM, interaction = "consec")
EMM ### see estimates
CON ### see interaction contrasts
test(CON, joint = TRUE)
I hope this helps shed some light on what is going on.
Related
I have a mgcv::gam mixed model of the form:
m1 <- gam(Y ~ A + s(B, bs = "re"), data = dataframe, family = gaussian,
method = "REML")
The random term s(B, bs = "re") is quoted in summary(m1) as, for example,
Approximate significance of smooth terms:
# edf Ref.df F p-value
s(B) 4.486 5 97.195 6.7e-08 ***
My question is, how would I quote this result (statistic and P value) in a formal document, for example a technical report or paper?
For example, one possibility is
F[4.486,5] = 97.195, P = 6.7e-08
However, arguing against this idea, “reverse engineering” of the result using
pf(q= 97.195, df1= 4.486, df2= 5, lower.tail=FALSE)
gives an incorrect p value:
[1] 5.931567e-05
I would be very grateful for your advice. Many thanks for your help!
The F statistic in question doesn't actually follow an F with the degrees of freedom you have identified. The Ref df one is related to the test, but you'd need to read and understand Wood (2013) to fully grep how the degrees of freedom for the test are derived.
I would simply quote the statistic and the p-value and then cite Simon's paper if anyone wants to know how they were computed. I don't think you can easily get at the degrees of freedom that actually get used. (well, not without debugging the summary.gam() code and seeing how they are computed.)
References
Wood, S. N. 2013. A simple test for random effects in regression models. Biometrika 100: 1005–1010. doi:10.1093/biomet/ast038
I'm attempting to model fMRI data so I can check the efficacy of an experimental design. I have been following a couple of tutorials and have a question.
I first need to model the BOLD response by convolving a stimulus input time series with a canonical haemodynamic response function (HRF). The first tutorial I checked said that one can make an HRF that is of any amplitude as long as the 'shape' of the HRF is correct so they created the following HRF in matlab:
hrf = [ 0 0 1 5 8 9.2 9 7 4 2 0 -1 -1 -0.8 -0.7 -0.5 -0.3 -0.1 0 ]
And then convolved the HRF with the stimulus by just using 'conv' so:
hrf_convolved_with_stim_time_series = conv(input,hrf);
This is very straight forward but I want my model to eventually be as accurate as possible so I checked a more advanced tutorial and they did the following. First they created a vector of 20 timepoints then used the 'gampdf' function to create the HRF.
t = 1:1:20; % MEASUREMENTS
h = gampdf(t,6) + -.5*gampdf(t,10); % HRF MODEL
h = h/max(h); % SCALE HRF TO HAVE MAX AMPLITUDE OF 1
Is there a benefit to doing it this way over the simpler one? I suppose I have 3 specific questions.
The 'gampdf' help page is super short and only says the '6' and '10' in each function call represents 'A' which is a 'shape' parameter. What does this mean? It gives no other information. Why is it 6 in the first call and 10 in the second?
This question is directly related to the above one. This code is written for a situation where there is a TR = 1 and the stimulus is very short (like 1s). In my situation my TR = 2 and my stimulus is quite long (12s). I tried to adapt the above code to make a working HRF for my situation by doing the following:
t = 1:2:40; % 2s timestep with the 40 to try to equate total time to above
h = gampdf(t,6) + -.5*gampdf(t,10); % HRF MODEL
h = h/max(h); % SCALE HRF TO HAVE MAX AMPLITUDE OF 1
Because I have no idea what the 'gampdf' parameters mean (or what that line does, in all actuality) I'm not sure this gives me what I'm looking for. I essentially get out 20 values where 1-14 have SOME numeric value in them but 15-20 are all 0. I'm assuming there will be a response during the entire 12s stimulus period (first 6 TRs so values 1-6) with the appropriate rectification which could be the rest of the values but I'm not sure.
Final question. The other code does not 'scale' the HRF to have an amplitude of 1. Will that matter, ultimately?
The canonical HRF you choose is dependent upon where in the brain the BOLD signal is coming from. It would be inappropriate to choose just any HRF. Your best source of a model is going to come from a lit review. I've linked a paper discussing the merits of multiple HRF models. The methods section brings up some salient points.
I am currently reading "Introduction to machine learning" by Ethem Alpaydin and I came across nearest centroid classifiers and tried to implement it. I guess I have correctly implemented the classifier but I am getting only 68% accuracy . So, is the nearest centroid classifier itself is inefficient or is there some error in my implementation (below) ?
The data set contains 1372 data points each having 4 features and there are 2 output classes
My MATLAB implementation :
DATA = load("-ascii", "data.txt");
#DATA is 1372x5 matrix with 762 data points of class 0 and 610 data points of class 1
#there are 4 features of each data point
X = DATA(:,1:4); #matrix to store all features
X0 = DATA(1:762,1:4); #matrix to store the features of class 0
X1 = DATA(763:1372,1:4); #matrix to store the features of class 1
X0 = X0(1:610,:); #to make sure both datasets have same size for prior probability to be equal
Y = DATA(:,5); # to store outputs
mean0 = sum(X0)/610; #mean of features of class 0
mean1 = sum(X1)/610; #mean of featurs of class 1
count = 0;
for i = 1:1372
pre = 0;
cost1 = X(i,:)*(mean0'); #calculates the dot product of dataset with mean of features of both classes
cost2 = X(i,:)*(mean1');
if (cost1<cost2)
pre = 1;
end
if pre == Y(i)
count = count+1; #counts the number of correctly predicted values
end
end
disp("accuracy"); #calculates the accuracy
disp((count/1372)*100);
There are at least a few things here:
You are using dot product to assign similarity in the input space, this is almost never valid. The only reason to use dot product would be the assumption that all your data points have the same norm, or that the norm does not matter (nearly never true). Try using Euclidean distance instead, as even though it is very naive - it should be significantly better
Is it an inefficient classifier? Depends on the definition of efficiency. It is an extremely simple and fast one, but in terms of predictive power it is extremely bad. In fact, it is worse than Naive Bayes, which is already considered "toy model".
There is something wrong with the code too
X0 = DATA(1:762,1:4); #matrix to store the features of class 0
X1 = DATA(763:1372,1:4); #matrix to store the features of class 1
X0 = X0(1:610,:); #to make sure both datasets have same size for prior probability to be equal
Once you subsamples X0, you have 1220 training samples, yet later during "testing" you test on both training and "missing elements of X0", this does not really make sense from probabilistic perspective. First of all you should never test accuracy on the training set (as it overestimates true accuracy), second of all by subsampling your training data your are not equalizing priors. Not in the method like this one, you are simply degrading quality of your centroid estimate, nothing else. These kind of techniques (sub/over- sampling) equalize priors for models that do model priors. Your method does not (as it is basically generative model with the assumed prior of 1/2), so nothing good can happen.
I hope you can help me with this one.
I am using cointegration to discover potential pairs trading opportunities within stocks and more precisely I am utilizing the Johansen trace test for only two stocks at a time.
I have several securities, but for each test I only test two at a time.
If two stocks are found to be cointegrated using the Johansen test, the idea is to define the spread as
beta' * p(t-1) - c
where beta'=[1 beta2] and p(t-1) is the (2x1) vector of the previous stock prices. Notice that I seek a normalized first coefficient of the cointegration vector. c is a constant which is allowed within the cointegration relationship.
I am using Matlab to run the tests (jcitest), but have also tried utilizing Eviews for comparison of results. The two programs yields the same.
When I run the test and find two stocks to be cointegrated, I usually get output like
beta_1 = 12.7290
beta_2 = -35.9655
c = 121.3422
Since I want a normalized first beta coefficient, I set beta1 = 1 and obtain
beta_2 = -35.9655/12.7290 = -2.8255
c =121.3422/12.7290 = 9.5327
I can then generate the spread as beta' * p(t-1) - c. When the spread gets sufficiently low, I buy 1 share of stock 1 and short beta_2 shares of stock 2 and vice versa when the spread gets high.
~~~~~~~~~~~~~~~~ The problem ~~~~~~~~~~~~~~~~~~~~~~~
Since I am testing an awful lot of stock pairs, I obtain a lot of output. Quite often, however, I receive output where the estimated beta_1 and beta_2 are of the same sign, e.g.
beta_1= -1.4
beta_2= -3.9
When I normalize these according to beta_1, I get:
beta_1 = 1
beta_2 = 2.728
The current pairs trading literature doesn't mention any cases where the betas are of the same sign - how should it be interpreted? Since this is pairs trading, I am supposed to long one stock and short the other when the spread deviates from its long run mean. However, when the betas are of the same sign, to me it seems that I should always go long/short in both at the same time? Is this the correct interpretation? Or should I modify the way in which I normalize the coefficients?
I could really use some help...
EXTRA QUESTION:
Under some of my tests, I reject both the hypothesis of r=0 cointegration relationships and r<=1 cointegration relationships. I find this very mysterious, as I am only considering two variables at a time, and there can, at maximum, only be r=1 cointegration relationship. Can anyone tell me what this means?
I am modelling the diffusion of movies through a contact network (based on telephone data) using a zero inflated negative binomial model (package: pscl)
m1 <- zeroinfl(LENGTH_OF_DIFF ~ ., data = trainData, type = "negbin")
(variables described below.)
The next step is to evaluate the performance of the model.
My attempt has been to do multiple out-of-sample predictions and calculate the MSE.
Using
predict(m1, newdata = testData)
I received a prediction for the mean length of a diffusion chain for each datapoint, and using
predict(m1, newdata = testData, type = "prob")
I received a matrix containing the probability of each datapoint being a certain length.
Problem with the evaluation: Since I have a 0 (and 1) inflated dataset, the model would be correct most of the time if it predicted 0 for all the values. The predictions I receive are good for chains of length zero (according to the MSE), but the deviation between the predicted and the true value for chains of length 1 or larger is substantial.
My question is:
How can we assess how well our model predicts chains of non-zero length?
Is this approach the correct way to make predictions from a zero inflated negative binomial model?
If yes: how do I interpret these results?
If no: what alternative can I use?
My variables are:
Dependent variable:
length of the diffusion chain (count [0,36])
Independent variables:
movie characteristics (both dummies and continuous variables).
Thanks!
It is straightforward to evaluate RMSPE (root mean square predictive error), but is probably best to transform your counts beforehand, to ensure that the really big counts do not dominate this sum.
You may find false negative and false positive error rates (FNR and FPR) to be useful here. FNR is the chance that a chain of actual non-zero length is predicted to have zero length (i.e. absence, also known as negative). FPR is the chance that a chain of actual zero length is falsely predicted to have non-zero (i.e. positive) length. I suggest doing a Google on these terms to find a paper in your favourite quantitative journals or a chapter in a book that helps explain these simply. For ecologists I tend to go back to Fielding & Bell (1997, Environmental Conservation).
First, let's define a repeatable example, that anyone can use (not sure where your trainData comes from). This is from help on zeroinfl function in the pscl library:
# an example from help on zeroinfl function in pscl library
library(pscl)
fm_zinb2 <- zeroinfl(art ~ . | ., data = bioChemists, dist = "negbin")
There are several packages in R that calculate these. But here's the by hand approach. First calculate observed and predicted values.
# store observed values, and determine how many are nonzero
obs <- bioChemists$art
obs.nonzero <- obs > 0
table(obs)
table(obs.nonzero)
# calculate predicted counts, and check their distribution
preds.count <- predict(fm_zinb2, type="response")
plot(density(preds.count))
# also the predicted probability that each item is nonzero
preds <- 1-predict(fm_zinb2, type = "prob")[,1]
preds.nonzero <- preds > 0.5
plot(density(preds))
table(preds.nonzero)
Then get the confusion matrix (basis of FNR, FPR)
# the confusion matrix is obtained by tabulating the dichotomized observations and predictions
confusion.matrix <- table(preds.nonzero, obs.nonzero)
FNR <- confusion.matrix[2,1] / sum(confusion.matrix[,1])
FNR
In terms of calibration we can do it visually or via calibration
# let's look at how well the counts are being predicted
library(ggplot2)
output <- as.data.frame(list(preds.count=preds.count, obs=obs))
ggplot(aes(x=obs, y=preds.count), data=output) + geom_point(alpha=0.3) + geom_smooth(col="aqua")
Transforming the counts to "see" what is going on:
output$log.obs <- log(output$obs)
output$log.preds.count <- log(output$preds.count)
ggplot(aes(x=log.obs, y=log.preds.count), data=output[!is.na(output$log.obs) & !is.na(output$log.preds.count),]) + geom_jitter(alpha=0.3, width=.15, size=2) + geom_smooth(col="blue") + labs(x="Observed count (non-zero, natural logarithm)", y="Predicted count (non-zero, natural logarithm)")
In your case you could also evaluate the correlations, between the predicted counts and the actual counts, either including or excluding the zeros.
So you could fit a regression as a kind of calibration to evaluate this!
However, since the predictions are not necessarily counts, we can't use a poisson
regression, so instead we can use a lognormal, by regressing the log
prediction against the log observed, assuming a Normal response.
calibrate <- lm(log(preds.count) ~ log(obs), data=output[output$obs!=0 & output$preds.count!=0,])
summary(calibrate)
sigma <- summary(calibrate)$sigma
sigma
There are more fancy ways of assessing calibration I suppose, as in any modelling exercise ... but this is a start.
For a more advanced assessment of zero-inflated models, check out the ways in which the log likelihood can be used, in the references provided for the zeroinfl function. This requires a bit of finesse.