I have an mle2 model that I've developed here just to demonstrate the problem. I generate values from two separate Gaussian distributions x1 and x2, combine them together to form x=c(x1,x2), and then create an MLE that attempts to re-classify x values as belonging to the left of a specific x value or the right of a specific x value via the xsplit paremeter.
The problem is that the parameters found are not ideal. Specifically, xsplit is always returned as whatever its starting value is. And if I change its starting value (e.g., as 4 or 9) there are huge differences in the log likelihood that results.
Here is the completely reproducible example:
set.seed(1001)
library(bbmle)
x1 = rnorm(n=100,mean=4,sd=0.8)
x2 = rnorm(n=100,mean=12,sd=0.4)
x = c(x1,x2)
hist(x,breaks=20)
ff = function(m1,m2,sd1,sd2,xsplit) {
outs = rep(NA,length(xvals))
for(i in seq(1,length(xvals))) {
if(xvals[i]<=xsplit) {
outs[i] = dnorm(xvals[i],mean=m1,sd=sd1,log=T)
}
else {
outs[i] = dnorm(xvals[i],mean=m2,sd=sd2,log=T)
}
}
-sum(outs)
}
# change xsplit starting value here to 9 and 4
# and realize the difference in log likelihood
# Why isn't mle finding the right value for xsplit?
mo = mle2(ff,
start=list(m1=1,m2=2,sd1=0.1,sd2=0.1,xsplit=9),
data=list(xvals=x))
#print mo to see log likelihood value
mo
#plot the result
c=coef(mo)
m1=as.numeric(c[1])
m2=as.numeric(c[2])
sd1=as.numeric(c[3])
sd2=as.numeric(c[4])
xsplit=as.numeric(c[5])
leftx = x[x<xsplit]
rightx = x[x>=xsplit]
y1=dnorm(leftx,mean=m1,sd=sd1)
y2=dnorm(rightx,mean=m2,sd=sd2)
points(leftx,y1*40,pch=20,cex=1.5,col="blue")
points(rightx,y2*90,pch=20,cex=1.5,col="red")
How can I modify my mle2 to capture the correct parameters, specifically for xsplit?
Mixture models present a lot of technical challenges (symmetry under relabeling of components, etc.); unless you have very specific needs, you might be better off using one of the large number of special-purpose mixture modeling packages that have been written for R (just library("sos"); findFn("{mixture model}") or findFn("{mixture model} Gaussian")).
However, in this case, you have a more specific problem, which is that the goodness-of-fit/likelihood surface of the xsplit parameter is "bad" (i.e. the derivative is zero almost everywhere). In particular, if you consider a pair of points x1, x2 in your data set that are neighbours, the likelihood is exactly the same for any splitting parameter between x1 and x2 (because any of those values splits the data set into the same two components). That means the likelihood surface is piecewise flat, which makes it almost impossible for any sensible optimizer -- even those such as Nelder-Mead that don't explicitly depend on derivatives. Your choices are (1) use some sort of brute-force stochastic optimizer (such as method="SANN" in optim()); (2) take xsplit out of your function and profile over it (i.e. for each possible choice of xsplit, optimize over the other four parameters); (3) smooth your splitting criterion (i.e. fit a logistic probability of belonging to one component or the other); (4) use a special-purpose mixture model fitting algorithm, as recommended above.
set.seed(1001)
library(bbmle)
x1 = rnorm(n=100,mean=4,sd=0.8)
x2 = rnorm(n=100,mean=12,sd=0.4)
x = c(x1,x2)
Your ff function can be written more compactly:
## ff can be written more compactly:
ff2 <- function(m1,m2,sd1,sd2,xsplit) {
p <- xvals<=xsplit
-sum(dnorm(xvals,mean=ifelse(p,m1,m2),
sd=ifelse(p,sd1,sd2),log=TRUE))
}
## ML estimation
mo <- mle2(ff2,
start=list(m1=1,m2=2,sd1=0.1,sd2=0.1,xsplit=9),
data=list(xvals=x))
## refit with a different starting value for xsplit
mo2 <- update(mo,start=list(m1=1,m2=2,sd1=0.1,sd2=0.1,xsplit=4))
## not used here, but maybe handy
plotfun <- function(mo,xvals=x,sizes=c(40,90)) {
c <- coef(mo)
hist(xvals,col="gray")
p <- xvals <= c["xsplit"]
y <- with(as.list(coef(mo)),
dnorm(xvals,mean=ifelse(p,m1,m2),
sd=ifelse(p,sd1,sd2))*sizes[ifelse(p,1,2)])
points(xvals,y,pch=20,cex=1.5,col=c("blue","red")[ifelse(p,1,2)])
}
plot(slice(mo),ylim=c(-0.5,10))
plot(slice(mo2),ylim=c(-0.5,10))
I cheated a little bit to extract just the xsplit parameter:
Likelihood surface around xsplit=9:
Likelihood surface around xsplit=4:
Also see p. 243 of Bolker 2008.
Update: smoothing
As I mentioned above, one solution is to make the boundary between the two mixture components smooth, or gradual, rather than sharp. I used a logistic function plogis() with midpoint at xsplit and a scale arbitrarily set to 2 (you could try to make it sharper; in principle you could make it an adjustable parameter, but if you do that you'll probably run into trouble again because the optimizer may want to make it infinite ...) In other words, rather that saying that all observations with x<xsplit are definitely in component 1 and all observations with x>xsplit are definitely in component 2, we say that observations that are equal to xsplit have a 50/50 probability of falling in either component, with the certainty of being in component 1 increasing as x decreases below xsplit. A logistic function with a very large scaling parameter approximates the sharp-split model previously attempted; generally you want to make the scaling parameter "large enough" to get a reasonable split and small enough not to run into numeric problems. (If you make the scale too large, the computed probabilities will underflow/overflow to 0 or 1 and you'll be back where you started...)
This is my second or third try; I had to do considerable fiddling (bounding values away from 0, or between 0 and 1, and fitting the standard deviations on a log scale), but the results seem reasonable. If I don't use clamp() on the logistic (plogis) function then I get 0 or 1 probabilities; if I don't use clamp() (one-sided) on the Normal probabilities then they can underflow to zero -- in either case I get infinite or NaN outcomes. Fitting the standard deviations on the log scale works better because one doesn't run into problems when the optimizer tries negative values for the standard deviation ...
## bound x values between lwr and upr
clamp <- function(x,lwr=0.001,upr=0.999) {
pmin(upr,pmax(lwr,x))
}
ff3 <- function(m1,m2,logsd1,logsd2,xsplit) {
p <- clamp(plogis(2*(xvals-xsplit)))
-sum(log((1-p)*clamp(dnorm(xvals,m1,exp(logsd1)),upr=Inf)+
p*clamp(dnorm(xvals,m2,exp(logsd2)),upr=Inf)))
}
xvals <- x
ff3(1,2,0.1,0.1,4)
mo3 <- mle2(ff3,
start=list(m1=1,m2=2,logsd1=-1,logsd2=-1,xsplit=4),
data=list(xvals=x))
## Coefficients:
## m1 m2 logsd1 logsd2 xsplit
## 3.99915532 12.00242510 -0.09344953 -1.13971551 8.43767997
The results look reasonable.
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.
How do you find the set of values for model predictors (a mixture of linear and non-linear) that yield the highest value for the response.
Example Model:
library(lme4); library(splines)
summary(lmer(formula = Solar.R ~ 1 + bs(Ozone) + Wind + Temp + (1 | Month), data = airquality, REML = F))
Here I am interested in what conditions (predictors) produce the highest solar radation (outcome).
This question seems simple, but I've failed to find a good answer using Google.
If the model was simple, I could take the derivatives to find the maximum or minimum. Someone has suggested that if the model function can be extracted, the stats::optim() function might be used. As a last resort, I could simulate all the reasonable variations of input values and plug it into the predict() function and look for the maximum value.
The last approach mentioned doesn't seem very efficient and I imagine that this is a common enough task (e.g., finding optimal customers for advertising) that someone has built some tools for handling it. Any help is appreciated.
There are some conceptual issues here.
for the simple terms (Wind and Temp), the response is a linear (and hence both monotonic and unbounded) function of the predictors. Thus if these terms have positive parameter estimates, increasing their values to infinity (Inf) will give you an infinite response (Solar.R); values should be as small as possible (negative infinite) if the coefficients are negative. Practically speaking, then, you want to set these predictors to the minimum or maximum reasonable value if the parameter estimates are respectively negative or positive.
for the bs term, I'm not sure what the properties of the B-spline are beyond the boundary knots, but I'm pretty sure that the curves go off to positive or negative infinity, so you've got the same issue. However, for the case of bs, it's also possible that there are one or more interior maxima. For this case I would probably try to extract the basis terms and evaluate the spline over the range of the data ...
Alternatively, your mentioning optim makes me think that this is a possibility:
data(airquality)
library(lme4)
library(splines)
m1 <- lmer(formula = Solar.R ~ 1 + bs(Ozone) + Wind + Temp + (1 | Month),
data = airquality, REML = FALSE)
predval <- function(x) {
newdata <- data.frame(Ozone=x[1],Wind=x[2],Temp=x[3])
## return population-averaged prediction (no Month effect)
return(predict(m1, newdata=newdata, re.form=~0))
}
aq <- na.omit(airquality)
sval <- with(aq,c(mean(Ozone),mean(Wind),mean(Temp)))
predval(sval)
opt1 <- optim(fn=predval,
par=sval,
lower=with(aq,c(min(Ozone),min(Wind),min(Temp))),
upper=with(aq,c(max(Ozone),max(Wind),max(Temp))),
method="L-BFGS-B", ## for constrained opt.
control=list(fnscale=-1)) ## for maximization
## opt1
## $par
## [1] 70.33851 20.70000 97.00000
##
## $value
## [1] 282.9784
As expected, this is intermediate in the range of Ozone(1-168), and min/max for Wind (2.3-20.7) and Temp (57-97).
This brute force solution could be made much more efficient by automatically selecting the min/max values for the simple terms and optimizing only over the complex (polynomial/spline/etc.) terms.
This question already has an answer here:
Set one or more of coefficients to a specific integer
(1 answer)
Closed 6 years ago.
In R, how can I set weights for particular variables and not observations in lm() function?
Context is as follows. I'm trying to build personal ranking system for particular products, say, for phones. I can build linear model based on price as dependent variable and other features such as screen size, memory, OS and so on as independent variables. I can then use it to predict phone real cost (as opposed to declared price), thus finding best price/goodness coefficient. This is what I have already done.
Now I want to "highlight" some features that are important for me only. For example, I may need a phone with large memory, thus I want to give it higher weight so that linear model is optimized for memory variable.
lm() function in R has weights parameter, but these are weights for observations and not variables (correct me if this is wrong). I also tried to play around with formula, but got only interpreter errors. Is there a way to incorporate weights for variables in lm()?
Of course, lm() function is not the only option. If you know how to do it with other similar solutions (e.g. glm()), this is pretty fine too.
UPD. After few comments I understood that the way I was thinking about the problem is wrong. Linear model, obtained by call to lm(), gives optimal coefficients for training examples, and there's no way (and no need) to change weights of variables, sorry for confusion I made. What I'm actually looking for is the way to change coefficients in existing linear model to manually make some parameters more important than others. Continuing previous example, let's say we've got following formula for price:
price = 300 + 30 * memory + 56 * screen_size + 12 * os_android + 9 * os_win8
This formula describes best possible linear model for dependence between price and phone parameters. However, now I want to manually change number 30 in front of memory variable to, say, 60, so it becomes:
price = 300 + 60 * memory + 56 * screen_size + 12 * os_android + 9 * os_win8
Of course, this formula doesn't reflect optimal relationship between price and phone parameters any more. Also dependent variable doesn't show actual price, just some value of goodness, taking into account that memory is twice more important for me than for average person (based on coefficients from first formula). But this value of goodness (or, more precisely, value of fraction goodness/price) is just what I need - having this I can find best (in my opinion) phone with best price.
Hope all of this makes sense. Now I have one (probably very simple) question. How can I manually set coefficients in existing linear model, obtained with lm()? That is, I'm looking for something like:
coef(model)[2] <- 60
This code doesn't work of course, but you should get the idea. Note: it is obviously possible to just double values in memory column in data frame, but I'm looking for more elegant solution, affecting model, not data.
The following code is a bit complicated because lm() minimizes residual sum of squares and with a fixed, non optimal coefficient it is no longed minimal, so that would be against what lm() is trying to do and the only way is to fix all the rest coefficients too.
To do that, we have to know coefficients of the unrestricted model first. All the adjustments have to be done by changing formula of your model, e.g. we have
price ~ memory + screen_size, and of course there is a hidden intercept. Now neither changing the data directly nor using I(c*memory) is good idea. I(c*memory) is like temporary change of data too, but to change only one coefficient by transforming the variables would be much more difficult.
So first we change price ~ memory + screen_size to price ~ offset(c1*memory) + offset(c2*screen_size). But we haven't modified the intercept, which now would try to minimize residual sum of squares and possibly become different than in original model. The final step is to remove the intercept and to add a new, fake variable, i.e. which has the same number of observations as other variables:
price ~ offset(c1*memory) + offset(c2*screen_size) + rep(c0, length(memory)) - 1
# Function to fix coefficients
setCoeffs <- function(frml, weights, len){
el <- paste0("offset(", weights[-1], "*",
unlist(strsplit(as.character(frml)[-(1:2)], " +\\+ +")), ")")
el <- c(paste0("offset(rep(", weights[1], ",", len, "))"), el)
as.formula(paste(as.character(frml)[2], "~",
paste(el, collapse = " + "), " + -1"))
}
# Example data
df <- data.frame(x1 = rnorm(10), x2 = rnorm(10, sd = 5),
y = rnorm(10, mean = 3, sd = 10))
# Writing formula explicitly
frml <- y ~ x1 + x2
# Basic model
mod <- lm(frml, data = df)
# Prime coefficients and any modifications. Note that "weights" contains
# intercept value too
weights <- mod$coef
# Setting coefficient of x1. All the rest remain the same
weights[2] <- 3
# Final model
mod2 <- update(mod, setCoeffs(frml, weights, nrow(df)))
# It is fine that mod2 returns "No coefficients"
Also, probably you are going to use mod2 only for forecasting (actually I don't know where else it could be used now) so that could be made in a simpler way, without setCoeffs:
# Data for forecasting with e.g. price unknown
df2 <- data.frame(x1 = rpois(10, 10), x2 = rpois(5, 5), y = NA)
mat <- model.matrix(frml, model.frame(frml, df2, na.action = NULL))
# Forecasts
rowSums(t(t(mat) * weights))
It looks like you are doing optimization, not model fitting (though there can be optimization within model fitting). You probably want something like the optim function or look into linear or quadratic programming (linprog and quadprog packages).
If you insist on using modeling tools like lm then use the offset argument in the formula to specify your own multiplyer rather than computing one.
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
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.