I am working in R in package lme4 and in MPlus and have a following situation:
I want to predict variable B (which is dichotomous) from variable A (continous) controlling for random effects on the level of a) Subjects; b) Tasks.
A -> B (1)
The problem is that when I use model to predict the values of B from A, values below probability of 0.5 get predicted, and in my case that doesn´t make sense, because, if you guess at random, the probability of correct answer on B would be 0.5.
I want to know how I can constrain the model (1) in R or in MPlus so that it doesn´t predict values lower than 0.5 in variable B.
Thank you!
I found a solution to the question thanks to Mr Kenneth Knoblauch. Basically, you need the psyphy package to use mafc.logit function.
For example, the code then looks like this:
mod <- glm(B ~ A, data = df, family = binomial(mafc.logit(.m = 2)))
It then involves the guessing parameter for (.m = 2) - two-choice tasks.
Cheers!
Related
Long story short:
I need to run a multinomial logit regression with both individual and time fixed effects in R.
I thought I could use the packages mlogit and survival to this purpose, but I am cannot find a way to include fixed effects.
Now the long story:
I have found many questions on this topic on various stack-related websites, none of them were able to provide an answer. Also, I have noticed a lot of confusion regarding what a multinomial logit regression with fixed effects is (people use different names) and about the R packages implementing this function.
So I think it would be beneficial to provide some background before getting to the point.
Consider the following.
In a multiple choice question, each respondent take one choice.
Respondents are asked the same question every year. There is no apriori on the extent to which choice at time t is affected by the choice at t-1.
Now imagine to have a panel data recording these choices. The data, would look like this:
set.seed(123)
# number of observations
n <- 100
# number of possible choice
possible_choice <- letters[1:4]
# number of years
years <- 3
# individual characteristics
x1 <- runif(n * 3, 5.0, 70.5)
x2 <- sample(1:n^2, n * 3, replace = F)
# actual choice at time 1
actual_choice_year_1 <- possible_choice[sample(1:4, n, replace = T, prob = rep(1/4, 4))]
actual_choice_year_2 <- possible_choice[sample(1:4, n, replace = T, prob = c(0.4, 0.3, 0.2, 0.1))]
actual_choice_year_3 <- possible_choice[sample(1:4, n, replace = T, prob = c(0.2, 0.5, 0.2, 0.1))]
# create long dataset
df <- data.frame(choice = c(actual_choice_year_1, actual_choice_year_2, actual_choice_year_3),
x1 = x1, x2 = x2,
individual_fixed_effect = as.character(rep(1:n, years)),
time_fixed_effect = as.character(rep(1:years, each = n)),
stringsAsFactors = F)
I am new to this kind of analysis. But if I understand correctly, if I want to estimate the effects of respondents' characteristics on their choice, I may use a multinomial logit regression.
In order to take advantage of the longitudinal structure of the data, I want to include in my specification individual and time fixed effects.
To the best of my knowledge, the multinomial logit regression with fixed effects was first proposed by Chamberlain (1980, Review of Economic Studies 47: 225–238). Recently, Stata users have been provided with the routines to implement this model (femlogit).
In the vignette of the femlogit package, the author refers to the R function clogit, in the survival package.
According to the help page, clogit requires data to be rearranged in a different format:
library(mlogit)
# create wide dataset
data_mlogit <- mlogit.data(df, id.var = "individual_fixed_effect",
group.var = "time_fixed_effect",
choice = "choice",
shape = "wide")
Now, if I understand correctly how clogit works, fixed effects can be passed through the function strata (see for additional details this tutorial). However, I am afraid that it is not clear to me how to use this function, as no coefficient values are returned for the individual characteristic variables (i.e. I get only NAs).
library(survival)
fit <- clogit(formula("choice ~ alt + x1 + x2 + strata(individual_fixed_effect, time_fixed_effect)"), as.data.frame(data_mlogit))
summary(fit)
Since I was not able to find a reason for this (there must be something that I am missing on the way these functions are estimated), I have looked for a solution using other packages in R: e.g., glmnet, VGAM, nnet, globaltest, and mlogit.
Only the latter seems to be able to explicitly deal with panel structures using appropriate estimation strategy. For this reason, I have decided to give it a try. However, I was only able to run a multinomial logit regression without fixed effects.
# state formula
formula_mlogit <- formula("choice ~ 1| x1 + x2")
# run multinomial regression
fit <- mlogit(formula_mlogit, data_mlogit)
summary(fit)
If I understand correctly how mlogit works, here's what I have done.
By using the function mlogit.data, I have created a dataset compatible with the function mlogit. Here, I have also specified the id of each individual (id.var = individual_fixed_effect) and the group to which individuals belongs to (group.var = "time_fixed_effect"). In my case, the group represents the observations registered in the same year.
My formula specifies that there are no variables correlated with a specific choice, and which are randomly distributed among individuals (i.e., the variables before the |). By contrast, choices are only motivated by individual characteristics (i.e., x1 and x2).
In the help of the function mlogit, it is specified that one can use the argument panel to use panel techniques. To set panel = TRUE is what I am after here.
The problem is that panel can be set to TRUE only if another argument of mlogit, i.e. rpar, is not NULL.
The argument rpar is used to specify the distribution of the random variables: i.e. the variables before the |.
The problem is that, since these variables does not exist in my case, I can't use the argument rpar and then set panel = TRUE.
An interesting question related to this is here. A few suggestions were given, and one seems to go in my direction. Unfortunately, no examples that I can replicate are provided, and I do not understand how to follow this strategy to solve my problem.
Moreover, I am not particularly interested in using mlogit, any efficient way to perform this task would be fine for me (e.g., I am ok with survival or other packages).
Do you know any solution to this problem?
Two caveats for those interested in answering:
I am interested in fixed effects, not in random effects. However, if you believe there is no other way to take advantage of the longitudinal structure of my data in R (there is indeed in Stata but I don't want to use it), please feel free to share your code.
I am not interested in going Bayesian. So if possible, please do not suggest this approach.
So I am wanting to create a logistic regression that simultaneously satisfies two constraints.
The link here, outlines how to use the Excel solver to maximize the value of Log-Likelihood value of a logistic regression, but I am wanting to implement a similar function in R
What I am trying to create in the end is an injury risk function. These take an S-shape function.
As we see, the risk curves are calculated from the following equation
Lets take some dummy data to begin with
set.seed(112233)
A <- rbinom(153, 1, 0.6)
B <- rnorm(153, mean = 50, sd = 5)
C <- rnorm(153, mean = 100, sd = 15)
df1 <- data.frame(A,B,C)
Lets assume A indicates if a bone was broken, B is the bone density and C is the force applied.
So we can form a logistic regression model that uses B and C are capable of explaining the outcome variable A. A simple example of the regression may be:
Or
glm(A ~ B + C, data=df1, family=binomial())
Now we want to make the first assumption that at zero force, we should have zero risk. This is further explained as A1. on pg.124 here
Here we set our A1=0.05 and solve the equation
A1 = 1 - (1-P(0))^n
where P(0) is the probability of injury when the injury related parameter is zero and n is the sample size.
We have our sample size and can solve for P(0). We get 3.4E-4. Such that:
The second assumption is that we should maximize the log-likelihood function of the regression
We want to maximize the following equation
Where pi is estimated from the above equation and yi is the observed value for non-break for each interval
My what i understand, I have to use one of the two functions in R to define a function for max'ing LL. I can use mle from base R or the mle2 from bbmle package.
I guess I need to write a function along these lines
log.likelihood.sum <- function(sequence, p) {
log.likelihood <- sum(log(p)*(sequence==1)) + sum(log(1-p)*(sequence==0))
}
But I am not sure where I should account for the first assumption. Ie. am I best to build it into the above code, and if so, how? Or will it be more effiecient to write a secondary finction to combine the two assumptions. Any advice would be great, as I have very limited experience in writing and understanding functions
I am very new to R (mostly program in SQL) but was faced with a problem that SQL couldn't help me with. I'll try to simplify the problem below.
Assume I have a set of data with 100 rows where each row has a different weight associated with it. Out of those 100 rows of data, 5 have an X value that is top-coded at 1000. Also assume that X can be represented by the linear equation X ~ Y + Z + U + 0 (want a positive value so I don't want a Y-intercept).
Now, without taking the weights of each row of data into consideration, the formula I used in R was:
fit = censReg(X ~ Y + Z + U + 0, left = -Inf, right = 1000, data = dataset)
If I computed summary(fit) I would get 0 left-censored values, 95 uncensored values, and 5 right censored values which is exactly what I want, minus the fact that the weights haven't been sufficiently added into the mix. I checked the reference manual on the censReg function and it doesn't seem like it accepts a weight argument.
Is there something I'm missing about the censReg function or is there another function that would be of better use to me? My end goal is to estimate X in the cases where it is censored (i.e. the 5 cases where it is 1000).
You should use Tobit regression for this situation, it is designed specifically to linearly model latent variables such as the one you describe.
The regression accounts for your weights and the censored observations, which can be seen in the derivation of the log-likelihood function for the Type I Tobit (upper and lower bounded).
Tobit regression can be found in the VGAM package using the vglm function with a tobit control parameter. An excellent example can be found here:
http://www.ats.ucla.edu/stat/r/dae/tobit.htm
So I want to plot this:
lmfit = lm (y ~ a + b)
but, "b" only has the values of zero and one. So, I want to plot two separate regression lines, that are paralel to one one another to show the difference that b makes to the y-intercept. So after plotting this:
plot(b,y)
I want to then use abline(lmfit,col="red",lwd=2) twice, once with the x value of b set to zero, and once with it set to one. So once without the term included, and once where b is just 1b.
To restate: b is categorical, 0 or 1. a is continuous with a slight linear trend.
Thank you.
Example:
You might want to consider using predict(...) with b=0 and b=1, as follows. Since you didn't provide any data, I'm using the built-in mtcars dataset.
lmfit <- lm(mpg~wt+cyl,mtcars)
plot(mpg~wt,mtcars,col=mtcars$cyl,pch=20)
curve(predict(lmfit,newdata=data.frame(wt=x,cyl=4)),col=4,add=T)
curve(predict(lmfit,newdata=data.frame(wt=x,cyl=6)),col=6,add=T)
curve(predict(lmfit,newdata=data.frame(wt=x,cyl=8)),col=8,add=T)
Given you have an additive lm model to begin with, drawing the lines is pretty straightforward, even though not completely intuitive. I tested it with the following simulated data:
y <- rnorm(30)
a <- rep(1:10,times=3)
b <- rep(c(1,0),each=15)
LM <- lm(y~a+b)
You have to access the coefficient values in the lm. Its is:
LM$coefficients
Here comes the tricky part, you have to assign the coefficients for each line.
The first one is easy:
abline(LM$coef[1],LM$coef[2])
The other one is a bit more complicated, given R works with additive coefficients, so for the second line you have:
abline(LM$coef[1]+LM$coef[3],LM$coef[2])
I hope this is what you was expecting
Unless I've misunderstood the question, all you have to do is run abline again but on a model without the b term.
abline(lm(y~a),col="red",lwd=2)
Using predict() one can obtain the predicted value of the dependent variable (y) for a certain value of the independent variable (x) for a given model. Is there any function that predicts x for a given y?
For example:
kalythos <- data.frame(x = c(20,35,45,55,70),
n = rep(50,5), y = c(6,17,26,37,44))
kalythos$Ymat <- cbind(kalythos$y, kalythos$n - kalythos$y)
model <- glm(Ymat ~ x, family = binomial, data = kalythos)
If we want to know the predicted value of the model for x=50:
predict(model, data.frame(x=50), type = "response")
I want to know which x makes y=30, for example.
Saw the previous answer is deleted. In your case, given n=50 and the model is binomial, you would calculate x given y using:
f <- function (y,m) {
(logit(y/50) - coef(m)[["(Intercept)"]]) / coef(m)[["x"]]
}
> f(30,model)
[1] 48.59833
But when doing so, you better consult a statistician to show you how to calculate the inverse prediction interval. And please, take VitoshKa's considerations into account.
Came across this old thread but thought I would add some other info. Package MASS has function dose.p for logit/probit models. SE is via delta method.
> dose.p(model,p=.6)
Dose SE
p = 0.6: 48.59833 1.944772
Fitting the inverse model (x~y) would not makes sense here because, as #VitoshKa says, we assume x is fixed and y (the 0/1 response) is random. Besides, if the data weren’t grouped you’d have only 2 values of the explanatory variable: 0 and 1. But even though we assume x is fixed it still makes sense to calculate a confidence interval for the dose x for a given p, contrary to what #VitoshKa says. Just as we can reparameterize the model in terms of ED50, we can do so for ED60 or any other quantile. Parameters are fixed, but we still calculate CI's for them.
The chemcal package has an inverse.predict() function, which works for fits of the form y ~ x and y ~ x - 1
You just have to rearrange the regression equation, but as the comments above state this may prove tricky and not necessarily have a meaningful interpretation.
However, for the case you presented you can use:
(1/coef(model)[2])*(model$family$linkfun(30/50)-coef(model)[1])
Note I did the division by the x coefficient first to allow the name attribute to be correct.
For just a quick view (without intervals and considering additional issues) you could use the TkPredict function in the TeachingDemos package. It does not do this directly, but allows you to dynamically change the x value(s) and see what the predicted y-value is, so it would be fairly simple to move x until the desired Y is found (for given values of additional x's), this will also show possibly problems with multiple x's that would work for the same y.