lnormMLE: Maximum Likelihood Parameter Estimation of a Log Normal Model...

Description Usage Arguments Details Value Note Author(s) References See Also Examples

View source: R/durationDist.R

Description

Estimate log normal model parameters by the maximum likelihood method using possibly censored data.

Usage

1
2
lnormMLE(yi, ni = numeric(length(yi)) + 1,
         si = numeric(length(yi)) + 1)

Arguments

yi

vector of (possibly binned) observations or a spikeTrain object.

ni

vector of counts for each value of yi; default: numeric(length(yi))+1.

si

vector of counts of uncensored observations for each value of yi; default: numeric(length(yi))+1.

Details

In the absence of censored data the ML estimates are available in closed form together with the Hessian matrix at the MLE. In presence of censored data an initial guess for the parameters is obtained using the uncensored data before maximizing the likelihood function to the full data set using optim with the BFGS method.

In order to ensure good behavior of the numerical optimization routines, optimization is performed on the log of parameter sdlog.

Standard errors are obtained from the inverse of the observed information matrix at the MLE. They are transformed to go from the log scale used by the optimization routine, when the latter is used (ie, for censored data) to the parameterization requested.

Value

A list of class durationFit with the following components:

estimate

the estimated parameters, a named vector.

se

the standard errors, a named vector.

logLik

the log likelihood at maximum.

r

a function returning the log of the relative likelihood function.

mll

a function returning the opposite of the log likelihood function using the log of parameter sdlog.

call

the matched call.

Note

The returned standard errors (component se) are valid in the asymptotic limit. You should plot contours using function r in the returned list and check that the contours are reasonably close to ellipses.

Author(s)

Christophe Pouzat [email protected]

References

Lindsey, J.K. (2004) Introduction to Applied Statistics: A Modelling Approach. OUP.

See Also

Lognormal,invgaussMLE

Examples

  1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
## Simulate sample of size 100 from a log normal
## distribution
set.seed(1102006,"Mersenne-Twister")
sampleSize <- 100
meanlog.true <- -2.4
sdlog.true <- 0.4
sampLN <- rlnorm(sampleSize,meanlog.true,sdlog.true)
sampLNmleLN <- lnormMLE(sampLN)
rbind(est = sampLNmleLN$estimate,se = sampLNmleLN$se,true = c(meanlog.true,sdlog.true))
## In the absence of censoring the MLE of the log normal is available in a
## closed form together with its variance (ie, the observed information matrix)
## we can check that we did not screw up at that stage by comparing the observed
## information matrix obtained numerically with the analytical one. To do that we
## use the MINUS log likelihood function returned by lnormMLE to get a numerical
## estimate
detailedFit <- optim(fn=sampLNmleLN$mll,
                     par=as.vector(c(sampLNmleLN$estimate[1],log(sampLNmleLN$estimate[2]))),
                     method="BFGS",
                     hessian=TRUE)
## We should not forget that the "mll" function uses the log of the sdlog parameter while
## the "se" component of sampLNmleLN list is expressed on the linear scale we must therefore
## transform one into the other as follows (Kalbfleisch, 1985, p71):
## if x = u and y = exp(v) and if we have the information matrix in term of
## u and v (that's the hessian component of list detailedFit above), we create matrix:
##      du/dx du/dy
## Q =
##      dv/dx dv/dy
## and we get I in term of x and y by the following matrix product:
## I(x,y) <- t(Q) %*% I(u,v) %*% Q
## In the present case:
##  du/dx = 1, du/dy = 0, dv/dx = 0, dv/dy = 1/exp(v)
## Therefore:
Q <- diag(c(1,1/exp(detailedFit$par[2])))
numericalI <- t(Q) %*% detailedFit$hessian %*% Q
seComp <- rbind(sampLNmleLN$se, sqrt(diag(solve(numericalI))))
colnames(seComp) <- c("meanlog","sdlog")
rownames(seComp) <- c("analytical", "numerical")
seComp
## We can check the relative differences between the 2
apply(seComp,2,function(x) abs(diff(x))/x[1])

## Not run: 
## Estimate the log relative likelihood on a grid to plot contours
MeanLog <- seq(sampLNmleLN$estimate[1]-4*sampLNmleLN$se[1],
               sampLNmleLN$estimate[1]+4*sampLNmleLN$se[1],
               sampLNmleLN$se[1]/10)
SdLog <- seq(sampLNmleLN$estimate[2]-4*sampLNmleLN$se[2],
             sampLNmleLN$estimate[2]+4*sampLNmleLN$se[2],
             sampLNmleLN$se[2]/10)
sampLNmleLNcontour <- sapply(MeanLog, function(mu) sapply(SdLog, function(s) sampLNmleLN$r(mu,s)))
## plot contours using a linear scale for the parameters
## draw four contours corresponding to the following likelihood ratios:
##  0.5, 0.1, Chi2 with 2 df and p values of 0.95 and 0.99
X11(width=12,height=6)
layout(matrix(1:2,ncol=2))
contour(MeanLog,SdLog,t(sampLNmleLNcontour),
        levels=c(log(c(0.5,0.1)),-0.5*qchisq(c(0.95,0.99),df=2)),
        labels=c("log(0.5)",
          "log(0.1)",
          "-1/2*P(Chi2=0.95)",
          "-1/2*P(Chi2=0.99)"),
        xlab=expression(mu),ylab=expression(sigma),
        main="Log Relative Likelihood Contours"
        )
points(sampLNmleLN$estimate[1],sampLNmleLN$estimate[2],pch=3)
points(meanlog.true,sdlog.true,pch=16,col=2)
## The contours are not really symmetrical about the MLE we can try to
## replot them using a log scale for the parameters to see if that improves
## the situation
contour(MeanLog,log(SdLog),t(sampLNmleLNcontour),
        levels=c(log(c(0.5,0.1)),-0.5*qchisq(c(0.95,0.99),df=2)),
        labels="",
        xlab=expression(mu),ylab=expression(log(sigma)),
        main="Log Relative Likelihood Contours",
        sub=expression(paste("log scale for parameter: ",sigma)))
points(sampLNmleLN$estimate[1],log(sampLNmleLN$estimate[2]),pch=3)
points(meanlog.true,log(sdlog.true),pch=16,col=2)

## make a parametric boostrap to check the distribution of the deviance
nbReplicate <- 10000
sampleSize <- 100
system.time(
            devianceLN100 <- replicate(nbReplicate,{
                             sampLN <- rlnorm(sampleSize,meanlog=meanlog.true,sdlog=sdlog.true)
                             sampLNmleLN <- lnormMLE(sampLN)
                             -2*sampLNmleLN$r(meanlog.true,sdlog.true)
                           }
                                       )
            )[3]

## Get 95 and 99% confidence intervals for the QQ plot
ci <- sapply(1:nbReplicate,
                 function(idx) qchisq(qbeta(c(0.005,0.025,0.975,0.995),
                                            idx,
                                            nbReplicate-idx+1),
                                      df=2)
             )
## make QQ plot
X <- qchisq(ppoints(nbReplicate),df=2)
Y <- sort(devianceLN100)
X11()
plot(X,Y,type="n",
     xlab=expression(paste(chi[2]^2," quantiles")),
     ylab="MC quantiles",
     main="Deviance with true parameters after ML fit of logNorm data",
     sub=paste("sample size:", sampleSize,"MC replicates:", nbReplicate)
     )
abline(a=0,b=1)
lines(X,ci[1,],lty=2)
lines(X,ci[2,],lty=2)
lines(X,ci[3,],lty=2)
lines(X,ci[4,],lty=2)
lines(X,Y,col=2)

## End(Not run)

STAR documentation built on May 31, 2017, 2:28 a.m.