Suppose I have a following formula for a mixed-effects model:
Precipitation ~ s(month,bs="cc")+s(time)+ humidity,random= ~(humidity|year)
and I know that humidity will only have positive effects on precipitation. So I want to specify a log-normal prior or other non-negative prior for humidity in rstanarm. Suppose the coeffienct for humidity is beta.I would specify the prior as log(beta) ~ normal(0,10^4) How should I do it? I don't think the default exponential distribution is a proper prior.
That is not among the priors that are supported by rstanarm. That same syntax would work with brms, although you would need to specify a the prior like
my_prior <- prior(lognormal(0, 1e4), coef = "humidity")
However, a lognormal prior with 10^4 as the standard deviation of the logarithm is preposterous because it puts considerable probability on values that would overflow to infinity on a 64 bit computer.
Related
I am trying to create a GLMM in R. I want to find out how the emergence time of bats depends on different factors. Here I take the time difference between the departure of the respective bat and the sunset of the day as dependent variable (metric). As fixed factors I would like to include different weather data (metric) as well as the reproductive state (categorical) of the bats. Additionally, there is the transponder number (individual identification code) as a random factor to exclude inter-individual differences between the bats.
I first worked in R with a linear mixed model (package lme4), but the QQ plot of the residuals deviates very strongly from the normal distribution. Also a histogram of the data rather indicates a gamma distribution. As a result, I implemented a GLMM with a gamma distribution. Here is an example with one weather parameter:
model <- glmer(formula = difference_in_min ~ repro + precipitation + (1+repro|transponder number), data = trip, control=ctrl, family=gamma(link = log))
However, since there was no change in the QQ plot this way, I looked at the residual diagnostics of the DHARMa package. But the distribution assumption still doesn't seem to be correct, because the data in the QQ plot deviates very much here, too.
Residual diagnostics from DHARMa
But if the data also do not correspond to a gamma distribution, what alternative is there? Or maybe the problem lies somewhere else entirely.
Does anyone have an idea where the error might lie?
But if the data also do not correspond to a gamma distribution, what alternative is there?
One alternative is called the lognormal distribution (https://en.wikipedia.org/wiki/Log-normal_distribution)
Gaussian (or normal) distributions are typically used for data that are normally distributed around zero, which sounds like you do not have. But the lognormal distribution does not have the same requirements. Following your previous code, you would fit it like this:
model <- glmer(formula = log(difference_in_min) ~ repro + precipitation + (1+repro|transponder number), data = trip, control=ctrl, family=gaussian(link = identity))
or instead of glmer you can just call lmer directly where you don't need to specify the distribution (which it may tell you to do in a warning message anyway:
model <- lmer(formula = log(difference_in_min) ~ repro + precipitation + (1+repro|transponder number), data = trip, control=ctrl)
I have a model of exponential decay in the form Y = exp{a + bX + cW}. In R, I represent this as a generalized linear model (GLM) using a gamma random component with log link function.
fitted <- glm(Y ~ X + W, family=Gamma(link='log'))
I know from this post that for the standard errors to really represent an exponential rather than gamma random component, I need to specify the dispersion parameter as being 1 when I call summary.
summary(fitted, dispersion=1)
summary(fitted) # not the same!
Now, I want to find the 95% confidence intervals for my estimates of a, b, c. However, there seems to be no way to specify the dispersion parameter for the confint, even though I know it should affect the confidence interval (because it affects the standard error).
confint(fitted)
confint(fitted, dispersion=1) # same as the last confint :(
So, in order to get the confidence intervals corresponding to an exponential rather than gamma random component, how do I specify the dispersion parameter when computing the confidence interval for a GLM?
I would like to fit a (very) high order regression to a set of data in R, however the poly() function has a limit of order 25.
For this application I need an order on the range of 100 to 120.
model <- lm(noisy.y ~ poly(q,50))
# Error in poly(q, 50) : 'degree' must be less than number of unique points
model <- lm(noisy.y ~ poly(q,30))
# Error in poly(q, 30) : 'degree' must be less than number of unique points
model <- lm(noisy.y ~ poly(q,25))
# OK
Polynomials and orthogonal polynomials
poly(x) has no hard-coded limit for degree. However, there are two numerical constraints in practice.
Basis functions are constructed on unique location of x values. A polynomial of degree k has k + 1 basis and coefficients. poly generates basis without the intercept term, so degree = k implies k basis and k coefficients. If there are n unique x values, it must be satisfied that k <= n, otherwise there is simply insufficient information to construct a polynomial. Inside poly(), the following line checks this condition:
if (degree >= length(unique(x)))
stop("'degree' must be less than number of unique points")
Correlation between x ^ k and x ^ (k+1) is getting closer and closer to 1 as k increases. Such approaching speed is of course dependent on x values. poly first generates ordinary polynomial basis, then performs QR factorization to find orthogonal span. If numerical rank-deficiency occurs between x ^ k and x ^ (k+1), poly will also stop and complain:
if (QR$rank < degree)
stop("'degree' must be less than number of unique points")
But the error message is not informative in this case. Furthermore, this does not have to be an error; it can be a warning then poly can reset degree to rank to proceed. Maybe R core can improve on this bit??
Your trial-and-error shows that you can't construct a polynomial of more than 25 degree. You can first check length(unique(q)). If you have a degree smaller than this but still triggering error, you know for sure it is due to numerical rank-deficiency.
But what I want to say is that a polynomial of more than 3-5 degree is never useful! The critical reason is the Runge's phenomenon. In terms of statistical terminology: a high-order polynomial always badly overfits data!. Don't naively think that because orthogonal polynomials are numerically more stable than raw polynomials, Runge's effect can be eliminated. No, a polynomial of degree k forms a vector space, so whatever basis you use for representation, they have the same span!
Splines: piecewise cubic polynomials and its use in regression
Polynomial regression is indeed helpful, but we often want piecewise polynomials. The most popular choice is cubic spline. Like that there are different representation for polynomials, there are plenty of representation for splines:
truncated power basis
natural cubic spline basis
B-spline basis
B-spline basis is the most numerically stable, as it has compact support. As a result, the covariance matrix X'X is banded, thus solving normal equations (X'X) b = (X'y) are very stable.
In R, we can use bs function from splines package (one of R base packages) to get B-spline basis. For bs(x), The only numerical constraint on degree of freedom df is that we can't have more basis than length(unique(x)).
I am not sure of what your data look like, but perhaps you can try
library(splines)
model <- lm(noisy.y ~ bs(q, df = 10))
Penalized smoothing / regression splines
Regression spline is still likely to overfit your data, if you keep increasing the degree of freedom. The best model seems to be about choosing the best degree of freedom.
A great approach is using penalized smoothing spline or penalized regression spline, so that model estimation and selection of degree of freedom (i.e., "smoothness") are integrated.
The smooth.spline function in stats package can do both. Unlike what its name seems to suggest, for most of time it is just fitting a penalized regression spline rather than smoothing spline. Read ?smooth.spline for more. For your data, you may try
fit <- smooth.spline(q, noisy.y)
(Note, smooth.spline has no formula interface.)
Additive penalized splines and Generalized Additive Models (GAM)
Once we have more than one covariates, we need additive models to overcome the "curse of dimensionality" while being sensible. Depending on representation of smooth functions, GAM can come in various forms. The most popular, in my opinion, is the mgcv package, recommended by R.
You can still fit a univariate penalized regression spline with mgcv:
library(mgcv)
fit <- gam(noisy.y ~ s(q, bs = "cr", k = 10))
I'm fitting a linear model using OLS and have scaled my regressors with the function scale in R because of the different units of measure between variables. Then, I fit the model using the lm command and get the coefficients of the fitted model. As far as I know the coefficients of the fitted model are not in the same units of the original regressors variables and therefore must be scaled back before they can be interpreted. I have been searching for a direct way to do it by couldn't find anything. Does anyone know how to do that?
Please have a look to the code, could you please help me implementing what you proposed?
library(zoo)
filename="DataReg4.csv"
filepath=paste("C:/Reg/",filename, sep="")
separator=";"
readfile=read.zoo(filepath, sep=separator, header=T, format = "%m/%d/%Y", dec=".")
readfile=as.data.frame(readfile)
str(readfile)
DF=readfile
DF=as.data.frame(scale(DF))
fm=lm(USD_EUR~diff_int+GDP_US+Net.exports.Eur,data=DF)
summary(fm)
plot(fm)
I'm sorry this is the data.
http://www.mediafire.com/?hmcp7urt0ag8187
If you used the scale function with default arguments then your regressors will be centered (subtracting their mean) and divided by their standard deviations. You can interpret the coefficients without transforming them back to the original units:
Holding everything else constant, on average, a one standard deviation change in one of the regressors is associated with a change in the dependent variable corresponding to the coefficient of that regressor.
If you have included an intercept term in your model keep in mind that the interpretation of the intercept will change. The estimated intercept now represents the average level of the dependent variable when all of the regressors are at their average levels. This is a result of subtracting the mean from each variable.
To interpret the coefficients in non-standard deviation terms, just calculate the standard deviation of each regressor and multiple that by the coefficient.
To de-scale or back-transform regression coefficients from a regression done with scaled predictor variable(s) and non-scaled response variable the intercept and slope should be calculated as:
A = As - Bs*Xmean/sdx
B = Bs/sdx
thus the regression is,
Y = As - Bs*Xmean/sdx + Bs/sdx * X
where
As = intercept from the scaled regression
Bs = slope from the scaled regression
Xmean = the mean of the scaled predictor variable
sdx = the standard deviation of the predictor variable
This can be adjusted if Y was also scaled but it appears you decided not to do that ultimately with your dataset.
If I understand your description (that is unfortunately at the moment code-free), you are getting standardized regression coefficients for Y ~ As + Bs*Xs where all those "s" items are scaled variables. The coefficients then are the predicted change on a std deviation scale of Y associated with a change in X of one standard deviation of X. The scale function would have recorded the means and standard deviations in attributes for hte scaled object. If not, then you will have those estimates somewhere in your console log. The estimated change in dY for a change dX in X should be: dY*(1/sdY) = Bs*dX*(1/sdX). Predictions should be something along these lines:
Yest = As*(sdX) + Xmn + Bs*(Xs)*(sdX)
You probably should not have needed to standardize the Y values, and I'm hoping that you didn't because it makes dealing with the adjustment for the means of the X's easier. Put some code and example data in if you want implemented and checked answers. I think #DanielGerlance is correct in saying to multiply rather than divide by the SD's.
I need to fit Y_ij ~ NegBin(m_ij,k), hence a negative binomial distribution to a count. However, the data I have observed are censored, I know the value of y_ij, but it could be more than that value. Writting down the loglikelihood going with this problem is:
ll = \sum_{i=1}^n w_i (c_i log(P(Y_ij=y_ij|X_ij)) + (1- c_i) log(1- \sum_{k=1}^32 P(Y_ij = k|X_ij)))
Where X_ij represent the design matrix (with the covariates of interest), w_i is the weight for each observation, y_ij is the response variable and P(Y_ij=y_ij|Xij) is the negative binomial distribution where the m_ij=exp(X_ij \beta) and \alpha is the overdispersion parameter.
Does someone knows if there exist a build-in code in R that could be used to obtain this?
Check this paper out: Regression Models for Count Data in R