[R] problem with convergence in mle2/optim function
Adam Zeilinger
zeil0006 at umn.edu
Wed Oct 10 00:21:42 CEST 2012
Dear R help,
Thanks again for the responses. I increased the lower constraint to:
lower = list(p1 = 0.0001, p2 = 0.0001, mu1 = 0.0001, mu2 = 0.0001).
I also included an upper box constraint of:
upper = list(p1 = Inf, p2 = Inf, mu1 = p1t, mu2 = p2t).
Making these changes improved the rate of convergence among stochastic
simulation runs, but I still had convergence problems.
I found success in switching from mle2/optim to spg (BB package). So
far, spg has produced similarly precise estimates as L-BFGS-B and
consistently provides parameter estimates.
If anyone is interested, here is the new objective function and spg
call, instead of my previous objective function and mle2 call. All
other parts of my reproducible code are the same as I've previously
supplied:
######################################################################
library(BB)
# Objective function for spg()
NLL2 <- function(par, y){
p1 <- par[1]
p2 <- par[2]
mu1 <- par[3]
mu2 <- par[4]
t <- y$tv
n1 <- y$n1
n2 <- y$n2
n3 <- y$n3
P1 <- (p1*((-1 + exp(sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2)))*t))*((-mu2)*(mu2 - p1 + p2) +
mu1*(mu2 + 2*p2)) - mu2*sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))) -
exp(sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))*t)*
mu2*sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2))) +
2*exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))))*t)*mu2*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))))/
exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))))*t)/(2*(mu2*p1 + mu1*(mu2 + p2))*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2))))
P2 <- (p2*((-1 + exp(sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2)))*t))*(-mu1^2 + 2*mu2*p1 +
mu1*(mu2 - p1 + p2)) - mu1*sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))) -
exp(sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))*t)*
mu1*sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2))) +
2*exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))))*t)*mu1*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))))/
exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))))*t)/(2*(mu2*p1 + mu1*(mu2 + p2))*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2))))
P3 <- 1 - P1 - P2
p.all <- c(P1, P2, P3)
#cat("NLL.free p.all {P1,P2,P3}\n")
#print(matrix(p.all, ncol=3))
-sum(dmnom2(c(n1, n2, n3), prob = p.all, log = TRUE))
}
par <- c(p1t, p2t, mu1t, mu2t)
spg.fit <- spg(par = par, fn = NLL2, y = yt,
lower = c(0.001, 0.001, 0.001, 0.001),
control = list(maxit = 5000))
########################################################################
My next problem is that spg takes about twice as long as L-BFGS-B to
converge. The spg help file strongly suggests the use of an exact
gradient function to improve speed. But I am having trouble writing a
gradient function. Here is what I have so far:
I derived the gradient function by taking the derivative of my NLL
equation with respect to each parameter. My NLL equation is the
probability mass function of the trinomial distribution. Here is some
reproducible code:
#########################################################################
library(Ryacas)
p1 <- Sym("p1"); p2 <- Sym("p2"); mu1 <- Sym("mu1"); mu2 <- Sym("mu2")
t <- Sym("t"); n1 <- Sym("n1"); n2 <- Sym("n2"); n3 <- Sym("n3")
P1.symb <- ((p1*((-1 + exp(sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2)))*t))*((-mu2)*(mu2 - p1 + p2) +
mu1*(mu2 + 2*p2)) - mu2*sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))) -
exp(sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))*t)*
mu2*sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2))) +
2*exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))))*t)*mu2*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))))/
exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))))*t)/(2*(mu2*p1 + mu1*(mu2 + p2))*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))))
P2.symb <- ((p2*((-1 + exp(sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2)))*t))*(-mu1^2 + 2*mu2*p1 +
mu1*(mu2 - p1 + p2)) - mu1*sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))) -
exp(sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))*t)*
mu1*sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2))) +
2*exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))))*t)*mu1*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))))/
exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))))*t)/(2*(mu2*p1 + mu1*(mu2 + p2))*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))))
P3.symb <- 1 - P1.symb - P2.symb
# gradient equation with respect to parameter p1 for probability mass
# function of the trinomial distribution with probabilities P1, P2, P3
gr.p1 <- deriv(log(P1.symb^n1), p1) + deriv(log(P2.symb^n2), p1) +
deriv(log(P3.symb^n3), p1)
######################################################################
gr.p1 is very large equation, which I won't reproduce here. Let's say
that the four gradient equations for the four parameters are defined as
gr.p1, gr.p2, gr.mu1, gr.mu2, and all are derived as described above for
gr.p1. These gradient equations are functions of p1, p2, mu1, mu2, t,
n1, n2, and n3. My current gradient function is:
grr <- function(par, y){
p1 <- par[1]
p2 <- par[2]
mu1 <- par[3]
mu2 <- par[4]
t <- y[,1]
n1 <- y[,2]
n2 <- y[,3]
n3 <- y[,4]
gr.p1 <- ....
gr.p2 <- ....
gr.mu1 <- ....
gr.mu2 <- ....
c(gr.p1, gr.p2, gr.mu1, gr.mu2)
}
The problem is that I need to supply values for t, n1, n2, and n3 to the
gradient function, which are from the dataset yt, above. When I supply
the dataset yt, the grr function produces a vector of length 4*nrow(yt)
= 80. However, spg requires a gradient function that returns a vector
of length(par) = 4. When I include this gradient function in my spg
function, I get an error that the gradient function is incorrect.
Does anyone have any suggestions on how to write my gradient function?
Am I calculating the gradient equation, gr.p1 incorrectly? As always,
any help would be much appreciated.
Adam Zeilinger
On 10/8/2012 3:44 AM, Berend Hasselman wrote:
>
> See below.
>
> On 08-10-2012, at 07:39, Adam Zeilinger wrote:
>
>> Dear R Help,
>>
>> Thank you to those who responded to my questions about mle2/optim convergence. A few responders pointed out that the optim error seems to arise when either one of the probabilities P1, P2, or P3 become negative or infinite. One suggested examining the exponential terms within the P1 and P2 equations.
>>
>> I may have made some progress along these lines. The exponential terms in the equations for P1 and P2 go to infinity at certain (large) values of t. The exponential terms can be found in lines 1, 5, and 7 in the P1 and P2 expressions below. Here is some example code:
>>
>> ###########################################################################
>> # P1 and P2 equations
>> P1 <- expression((p1*((-1 + exp(sqrt((mu1 + mu2 + p1 + p2)^2 -
>> 4*(mu2*p1 + mu1*(mu2 + p2)))*t))*((-mu2)*(mu2 - p1 + p2) +
>> mu1*(mu2 + 2*p2)) - mu2*sqrt((mu1 + mu2 + p1 + p2)^2 -
>> 4*(mu2*p1 + mu1*(mu2 + p2))) -
>> exp(sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))*t)*
>> mu2*sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2))) +
>> 2*exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
>> 4*(mu2*p1 + mu1*(mu2 + p2))))*t)*mu2*
>> sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))))/
>> exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
>> 4*(mu2*p1 + mu1*(mu2 + p2))))*t)/(2*(mu2*p1 + mu1*(mu2 + p2))*
>> sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))))
>> P2 <- expression((p2*((-1 + exp(sqrt((mu1 + mu2 + p1 + p2)^2 -
>> 4*(mu2*p1 + mu1*(mu2 + p2)))*t))*(-mu1^2 + 2*mu2*p1 +
>> mu1*(mu2 - p1 + p2)) - mu1*sqrt((mu1 + mu2 + p1 + p2)^2 -
>> 4*(mu2*p1 + mu1*(mu2 + p2))) -
>> exp(sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))*t)*
>> mu1*sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2))) +
>> 2*exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
>> 4*(mu2*p1 + mu1*(mu2 + p2))))*t)*mu1*
>> sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))))/
>> exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
>> 4*(mu2*p1 + mu1*(mu2 + p2))))*t)/(2*(mu2*p1 + mu1*(mu2 + p2))*
>> sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))))
>>
>> # Vector of t values
>> tv <- c(1:200)
>>
>> # 'true' parameter values
>> p1t = 2; p2t = 2; mu1t = 0.001; mu2t = 0.001
>>
>> # Function to calculate probabilities from 'true' parameter values
>> psim <- function(x){
>> params <- list(p1 = p1t, p2 = p2t, mu1 = mu1t, mu2 = mu2t, t = x)
>> eval.P1 <- eval(P1, params)
>> eval.P2 <- eval(P2, params)
>> P3 <- 1 - eval.P1 - eval.P2
>> c(x, matrix(c(eval.P1, eval.P2, P3), ncol = 3))
>> }
>> pdat <- sapply(tv, psim, simplify = TRUE)
>> Pdat <- as.data.frame(t(pdat))
>> names(Pdat) <- c("time", "P1", "P2", "P3")
>>
>> matplot(Pdat[,-1], type = "l", xlab = "time", ylab = "Probability",
>> col = c("black", "brown", "blue"),
>> lty = c(1:3), lwd = 2, ylim = c(0,1))
>> legend("topright", c("P1", "P2", "P3"),
>> col = c("black", "brown", "blue"),
>> lty = c(1:3), lwd = 2)
>> Pdat[160:180,] # psim function begins to return "NaN" at t = 178
>>
>> # exponential terms in P1 and P2 expressions are problematic
>> params <- list(p1 = p1t, p2 = p2t, mu1 = mu1t, mu2 = mu2t, t = 178)
>>
>> exp1 <- expression(exp(sqrt((mu1 + mu2 + p1 + p2)^2 -
>> 4*(mu2*p1 + mu1*(mu2 + p2)))*t))
>> eval(exp1, params) # returns Inf at t = 178
>>
>> exp2 <- expression(exp((1/2)*(mu1 + mu2 + p1 + p2 +
>> sqrt((mu1 + mu2 + p1 + p2)^2 -
>> 4*(mu2*p1 + mu1*(mu2 + p2))))*t))
>> eval(exp2, params) # also returns Inf at t = 178
>> ##########################################################################
>>
>> The time step at which the exponential terms go to infinity depends on the values of the parameters p1, p2, mu1, mu2. It seems that the convergence problems may be due to these exponential terms going to infinity. Thus my convergence problem appears to be an overflow problem?
>>
>
> It is quite possible that letting t becoming larger and larger is going to get you into trouble here,
> But the maximum value of t in your initial post was 20.
> So it's unlikely that the value of t was causing the problems with the calculation of the gradient (error message "non-finite finite-difference value [3]").
>
> If you change the lower bounds for mu1 and mu2 to something slightly larger than 0 mle2 converges. Because now optim has enough leeway to calculate a finite difference. At least I think that is the cause of your initial problem.
>
>> # mle2 call
>> mle.fit <- mle2(NLL.func, data = list(y = yt),
> + start = list(p1 = p1t, p2 = p2t, mu1 = mu1t, mu2 = mu2t),
> + control = list(trace=1,maxit = 5000, factr = 1e-5, lmm = 17),
> + method = "L-BFGS-B", skip.hessian = TRUE,
> + lower = list(p1 = 0, p2 = 0, mu1 = 0.0001, mu2 = 0.0001))
> iter 0 value -1110.031664
> iter 10 value -1110.315437
> iter 20 value -1110.608553
> final value -1110.609914
> converged
>
>> mle.fit
>
> Call:
> mle2(minuslogl = NLL.func, start = list(p1 = p1t, p2 = p2t, mu1 = mu1t,
> mu2 = mu2t), method = "L-BFGS-B", data = list(y = yt), skip.hessian = TRUE,
> control = list(trace = 1, maxit = 5000, factr = 1e-05, lmm = 17),
> lower = list(p1 = 0, p2 = 0, mu1 = 1e-04, mu2 = 1e-04))
>
> Coefficients:
> p1 p2 mu1 mu2
> 1.471146 1.618261 0.000100 0.000100
>
> Log-likelihood: 1110.61
>
>
>> Unfortunately, I'm not sure where to go from here. Due to the complexity of the P1 and P2 equations, there's no clear way to rearrange the equations to eliminate t from the exponential terms.
>>
>> Does anyone have any suggestions on how to address this problem? Perhaps there is a way to bound p1, p2, mu1, and mu2 to avoid the exponential terms going to infinity? Or bound P1 and P2?
>>
>
> If you were to transform P1, P2 and P3 to something between 0 and 1 you would be changing the model.
>
> Berend
>
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>
--
Adam Zeilinger
Post Doctoral Scholar
Department of Entomology
University of California Riverside
www.linkedin.com/in/adamzeilinger
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