A variation of this question has been asked, but certain items remain unanswered -
I am modeling the proportion of mortality (Prop) using a single continuous predictor variable which is temperature (Temp). I have three questions.
1.) Should I be using meanBEINF for my model predicted estimates of the response? If so, how would I extract the associated standard errors? You think the way I have it specified currently would give you the response estimates, however, running predict(beinf_mod, type = "response", what = "mu") yields the same results which has me questioning.
2.) If I exponentiate the predictor variable coefficient (contained within the mu parameter) does this give me the odds between (0,1)? nu and tau currently don't have predictor variable coefficients so I'm not sure if those are to be worked in to get odds for the total domain [0,1].
3.) Is my interpretation of the odds correct in this scenario? I am familiar with a regular beta regression or logistic model, but, the uncertainties in question 2 make me wonder if this is appropriate.
Thanks in advance for the help, and it is greatly appreciated.
# generate DB
DB <- data.frame(Prop = c(0.688888889, 0.519230769, 0.378294574, 0.253644315, 0.234200744, 0.156626506,
0.191011236, 0.0625, 0.064516129, 0, 0, 0),
Temp = c(62.90857143, 62.75428571, 60.05428571, 60.23428571, 59.64285714, 57.94571429,
57.71428571, 57.14857143, 54.39714286, 51.87714286, 50.38571429, 49.1))
# beta inflated model. I understand na.omit works on the data, and that na.exclude is not really useful.
# I removed the NA's for this reproduction of the problem
beinf_mod <- gamlss(Prop ~ cs(Temp),
family=BEINF,data=na.omit(DB),na.action=na.exclude)
# obtain predictions for the estimated/expected value of y
predict(beinf_mod, type="response", se.fit=TRUE)
# get the odds of the explanatory variable. exponentiation gets us 1.47,
# so a one unit increase in temperature results in a 47% increase in the odds of mortality within the domain (0,1)
exp(coef(beinf_mod)[2])
allow me to answer my own questions
1.) yes, meanBEINF provides the estimated response values and bootstrapping would get you the errors.
2.) yes. Nu and Tau do not get worked in because one can only calculate the odds for the mu parameter (0,1) because one cannot calculate the odds for 0 and 1. You cannot divide by 0 if the odds are 1 and the odds of 0 prob to any other prob is 0.
3.) yes.
Related
I'm looking for information and guidance to help me understand the outlier test in DHARMa for negative binomial regression. Here is the diagnostic plot from DHARMa using the function simulateResiduals().
First off, The dispersion test is significant in the plot. Using testDispersion() on the model and on the residuals, I get the results of 2.495. Visually, the dots seem to aline pretty well on the QQ line. The developer stated ' If you see a dispersion parameter of 1.01, I would not worry, even if the test is significant. A significant value of 5, however, is clearly a reason to move to a model that accounts for overdispersion.' here I conclude that the deviation is within the acceptable range for the NB regression.
Second, the Outlier test is also significant. I never had this before, and I can't find much information regarding how many outliers is okay vs not okay to have. Following the recommendation of DHARMa's developer, I looked at the magnitude of the outlier to investigate this. reference. Here is the code and output:
ModelNB <- glm.nb(BUD ~ Treatment*YEAR, data=Data_Bud) simulationOutput <- simulateResiduals(fittedModel = ModelNB, plot = T) testOutliers(simulationOutput, type = "binomial")
`
DHARMa outlier test based on exact binomial test with
approximate expectations
data: simulationOutput
outliers at both margin(s) = 12, observations = 576, p-value =
0.00269
alternative hypothesis: true probability of success is not equal to 0.007968127
95 percent confidence interval:
0.01081011 0.03610864
sample estimates:
frequency of outliers (expected: 0.00796812749003984 )
0.02083333
`
**Can someone help me understand this output? ** Is having 12 outliers per 576 observations okay? In statistics classes, I was told that taking out outliers was a big No-No. What does "true probability of success is not equal to 0.007968127" mean? I can't accept H1 and need to accept H0 for the outlier???
Information on my model:
ModelNB <- glm.nb(BUD ~ Treatment*YEAR, data=Data_Bud)
BUD = The number of floral buds on a twig
Treatment = 5 different fertiliser treatment
YEAR = 2 different years (2020 and 2021)
I want to plot the posterior distribution for data sampled from gamma(2,3) with a prior distribution of gamma(3,3). I am assuming alpha=2 is known. But a graph of my posterior for different values of the rate parameter centers around 4. It should be 3. I even tried with a uniform prior to make things simpler. Can you please spot what's wrong? Thank you.
set.seed(101)
dat <- rgamma(100,shape=2,rate=3)
alpha <- 3
n <- 100
post <- function(beta_1) {
posterior<- (((beta_1^alpha)^n)/gamma(alpha)^n)*
prod(dat^(alpha-1))*exp(-beta_1*sum(dat))
return(posterior)
}
vlogl <- Vectorize(post)
curve(vlogl2,from=2,to=6)
A tricky question and possibly more related to statistics than to programming =). I initially made the same reasoning mistake as you, but subsequently realised to be more careful with the posterior and the roles of alpha and beta_1.
The prior is uniform (or flat) so the posterior distribution is proportional (not equal) to the likelihood.
The quantity you have assigned to the posterior is indeed the likelihood. Plugging in alpha=3, this evaluates to
(prod(dat^2)/(gamma(alpha)^n)) * beta_1^(3*n)*exp(-beta_1*sum(dat)).
This is the crucial step. The last two terms in the product depend on beta_1 only, so these two parts determine the shape of the posterior. The posterior distribution is thus gamma distributed with shape parameter 3*n+1 and rate parameter sum(dat). As the mode of the gamma distribution is the ratio of these two and sum(dat) is about 66 for this seed, we get a mode of 301/66 (about 4.55). This coincides perfectly with the ``posterior plot'' (again you plotted the likelihood which is not properly scaled, i.e. not properly integrating to 1) produced by your code (attached below).
I hope LifeisBetter now =).
But a graph of my posterior for different values of the rate parameter
centers around 4. It should be 3.
The mean of your data is 0.659 (~2/3). Given a gamma distribution with a shape parameter alpha = 3, we are trying to find likely values of the rate parameter, beta, that gave rise to the observed data (subject to our prior information). The mean of a gamma distribution is the shape parameter divided by the rate parameter. 100 observations should be enough to mostly overcome the somewhat informative prior (which had a mean of 1), so we should expect beta to take values somewhere in the region alpha/mean(dat), not 3.
alpha/mean(dat)
#> [1] 4.54915
I'm not going to show the derivation of the posterior distribution for beta without TeX, but it is a gamma distribution that includes the rate parameter from the prior distribution of beta (betaPrior = 3):
set.seed(101)
n <- 100
dat <- rgamma(n, 2, 3)
alpha <- 3
betaPrior <- 3
post <- function(x) dgamma(x, alpha*(n + 1), sum(dat) + betaPrior)
curve(post, 2, 6)
Notice that the mean of beta is at ~4.39 rather than ~4.55 because of the informative prior that had a mean of 1.
Unfortunately im not able to provide a reproducible example, but hopefully you get the idea regardless.
I am conducting some regression analyses where the dependent variable is a DCC of a pair of return series - two stocks. Im using dummies to represent shocks in the return series, i.e. the worst 1% of observed returns. In sum:
DCC = c + 1%Dummy
When I run the DurbinWatsonTest I get the output:
Autocorrelation: 0,9987
D-W statistic: 0
p-value: 0
HA: rho !=0
Does this just mean that its highly significant presence of autocorrelation?
I also tried dwtest, but that yields NA values for both P and DW-stat.
To correct for autocorrealtion I used the code:
spx10 = lm(bit_sp500 ~ Spx_0.1)
spx10_hc = coeftest(spx10, vcov. = vcovHC(spx10, method = "arellano",type = "HC3"))
How can I be certain that it had any effect, as I cannot run the DW-test for the spx10_hc, nor did the regression output change noteworthy. Is it common that regression analysis with 1 independent variable changes just ever so slightly when adjusting for autocorrelation?
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.
My data is in the following format and includes a particular statistic
site LRStat
1 3.580728
2 2.978038
3 5.058644
4 3.699278
5 4.349046
This is just a sample of the data.
I then obtained the null LR distribution as well by permuting random pairs of data. I used this to plot a histogram with frequency in the y-axes and LR statistic in the x-axes. How is it possible to determine the critical p-value cut-off points based on the null distribution (as shown in the below figure)?
You now have a sampling distribution of LR values. The quantile function in R will give you an estimate of whatever "critical value" you prefer. If, for instance, you decided you wanted the conventional 0.05 "p-value" you could take your dataframe, named LR_df for illustration, and issue this command:
quantile( LR_df[ , 'LRStat'] , 0.95)
If you wanted all of those "probabilities" on the figure, you would use a vector of values complementary to unity. The following code gives you the LSstat values at which a given proportion of the sample are higher than that value.
quantile( LR_df[ , 'LRStat'] , c(0.9, 0.95, 0.99, 0.999, 0.9999) )
The p-values are just a sampling distribution of a test statistic under a null hypothesis. Your null hypothesis in this case is that the LRstats are uniformly distributed. (I know it sounds strange to put it that way, but if you want to argue with the statisticians then get a copy of http://amstat.tandfonline.com/doi/pdf/10.1198/000313008X332421 .) The choice of p-value for cutoff will depend on scientific or business setting. If you were assessing an investment opportunity the cutoff might be 0.15 but if you are trying to find new scientific knowledge, I think it should be smaller (more stringent test). The field of molecular genetics has a lot of junk (i.e. fails to reproduce results) in their literature because they were not strict enough in the statistical methods.