R - numerical errors with analytical gradient? - r

Question

I've got the following code:
theta=0.05
n=1000
m=200
r=rnorm(2000)
#ER check function
nu=Vectorize(function(a,tau){return(abs(tau-(a<0))*a^2)})
#Selecting 10 lowest sum values (lowest10 function returns indices)
lowest10=function(x){
values=sort(x)[1:min(10,length(x))]
indices=match(values,x)
return(indices)
}
sym.expectile=function(beta,e,abs.r){return(beta[1]+beta[2]*e+beta[3]*abs.r)}
ERsum=function(beta,tau,start,end){
y=r[(start+1):end]
X1=rep(1,n-1)
X3=abs(r[start:(end-1)])
X2=c()
X2[1]=e.sym.optimal[start-m]
for (i in 2:(n-1)){
X2[i]=sym.expectile(beta,X2[i-1],X3[i-1])
}
X=matrix(c(X1,X2,X3),ncol=3)
res=y-X%*%beta
sum.nu=mean(nu(res,tau))
return(sum.nu)
}
ERsum.gr=function(beta,tau,start,end){
y=r[(start+1):end]
X1=rep(1,n-1)
X3=abs(r[start:(end-1)])
X2=c()
X2[1]=e.sym.optimal[start-m]
for (i in 2:(n-1)){
X2[i]=sym.expectile(beta,X2[i-1],X3[i-1])
}
X=matrix(c(X1,X2,X3),ncol=3)
partial.beta0=c()
for (i in 1:(n-1)){partial.beta0[i]=-(1-beta[2]^(i))/(1-beta[2])}
gr.beta0=2/T*sum(abs(tau-(y<X%*%beta))*(y-X%*%beta)*partial.beta0)/1000
partial.beta1=c()
partial.beta1[1]=-X2[1]
for (i in 2:(n-1)){partial.beta1[i]=partial.beta1[i-1]*beta[2]-X2[i]}
gr.beta1=2/T*sum(abs(tau-(y<X%*%beta))*(y-X%*%beta)*partial.beta1)/1000
partial.beta2=c()
partial.beta2[1]=-X3[1]
for (i in 2:(n-1)){partial.beta2[i]=partial.beta2[i-1]*beta[2]-X3[i]}
gr.beta2=2/T*sum(abs(tau-(y<X%*%beta))*(y-X%*%beta)*partial.beta2)/1000
c(gr.beta0,gr.beta1,gr.beta2)
}
beta=matrix(nrow=1e4,ncol=3)
beta[,1]=runif(1e4,-1,0)#beta0
beta[,2]=runif(1e4,0,1)#beta1
beta[,3]=runif(1e4,-1,0)#beta2
e.sym.optimal=c()
tau.found.sym.optim=0.02234724
library('expectreg')
e.sym.optimal[1]=expectile(r[1:m],tau.found.sym.optim)
ERsums.sym=c()
for (i in 1:nrow(beta)){
ERsums.sym[i]=ERsum(beta[i,],tau.found.sym.optim,m+1,m+n)
}
initialbeta.esym=beta[lowest10(ERsums.sym),]
intermedietebeta.esym=matrix(ncol=3,nrow=10)
for (i in 1:10){
intermedietebeta.esym[i,]=optim(initialbeta.esym[i,],ERsum,
gr=ERsum.gr,tau=tau.found.sym.optim,
start=m+1,end=m+n,
method="BFGS")$par
}
I tried to replace the optim function with optimx, but got the following error:
Error: Gradient function might be wrong - check it!
To check if my gradient is ok I tried to evaluate values of gradient function using function grad from numDeriv and directly calling my ERsum.gr function. For the sample vector
beta
[1] -0.8256490 0.7146256 -0.4945032
I obtained following results:
>grad(function(beta) ERsum(c(beta[1],beta[2],beta[3]),tau.found.sym.optim,m+1,m+n),beta)
[1] -0.6703170 2.8812666 -0.5573101
> ERsum.gr2(beta,tau.found.sym.optim,m+1,m+n)
[1] -0.6696467 2.8783853 -0.5567527
So here is my question: is it possible that these differences are just some numerical errors caused by rounding down the partial.beta0, partial.beta1, partial.beta2 which are just the components of the sum representing gradient? I think so, because if my analytical formula for gradient misses something, the discrepancies would be probably much larger, but how can I be sure? If this is a case is there any other way to obtain more accurate values of gradient?

Answer 1

You've got further problems down the line even if you solve the question of whether that is really a proper gradient, which I see as too complex to tackle. If you take out the gr argument and try to run with only optimx instead of optim, you get:
Error in intermedietebeta.esym[i, ] <- optimx(initialbeta.esym[i, ], ERsum, :
number of items to replace is not a multiple of replacement length
This probably relates to the fact that optimx does not return the same structure as is returned by optim:
> optimx(initialbeta.esym[i,],ERsum,
+ tau=tau.found.sym.optim,
+ start=m+1,end=m+n,
+ method="BFGS")$par
NULL
> optimx(initialbeta.esym[i,],ERsum,
+ tau=tau.found.sym.optim,
+ start=m+1,end=m+n,
+ method="BFGS") # leave out `$par`
p1 p2 p3 value fevals gevals niter convcode kkt1 kkt2 xtimes
BFGS -1.0325 0.2978319 0.04921863 0.09326904 102 100 NA 1 TRUE FALSE 3.366
If you disagree with the decision to allow a default gradient estimate, hten you need to narrow down your debugging to the function that throws the error:
Error: Gradient function might be wrong - check it!
> traceback()
3: stop("Gradient function might be wrong - check it! \n", call. = FALSE)
2: optimx.check(par, optcfg$ufn, optcfg$ugr, optcfg$uhess, lower,
upper, hessian, optcfg$ctrl, have.bounds = optcfg$have.bounds,
usenumDeriv = optcfg$usenumDeriv, ...)
1: optimx(initialbeta.esym[i, ], ERsum, gr = ERsum.gr, tau = tau.found.sym.optim,
start = m + 1, end = m + n, method = "BFGS")
And look at the documentation (there was no help page) and code for optimx:::optimx.check. This is the section of code that does the checking:
if (!is.null(ugr) && !usenumDeriv) {
gname <- deparse(substitute(ugr))
if (ctrl$trace > 0)
cat("Analytic gradient from function ", gname,
"\n\n")
fval <- ufn(par, ...)
gn <- grad(func = ufn, x = par, ...)
ga <- ugr(par, ...)
teps <- (.Machine$double.eps)^(1/3)
if (max(abs(gn - ga))/(1 + abs(fval)) >= teps) {
stop("Gradient function might be wrong - check it! \n",
call. = FALSE)
optchk$grbad <- TRUE
}