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I'm trying to get the shape and scale parameters for this data using the optim function in R.

incomeData = data.frame(L = c(850,rep(1000,24),rep(2001,112),rep(3001,267),rep(4001,598),rep(5001,1146)),
       U = c(999,rep(2000,24),rep(3000,112),rep(4000,267),rep(5000,598),rep(10000,1146)),
       Interval = c(1,rep(2,24),rep(3,112),rep(4,267),rep(5,598),rep(6,1146)))

The maximum likelihood function is defined as this:

MaxLikelihoodFunction

F is the cumulative gamma function evaluated in the upper and lower bound of the income interval with shape = shape and scale = scale.

And for the initial values of the parameters I'm using the methods of moments:

mu: mean of the middle points of the invervals.

phi

Where s^2 is the variance of the middle points of the intervals.

The initial parameters were calculated using the method of moments

incomeData$middle = (incomeData$U+incomeData$L)/2 # middle point of the interval

middlePointMean = mean(incomeData$middle) # mean of the middle points
middlePointVar = var(incomeData$middle) # variance of the middle points

initialPar1 = middlePointVar/(middlePointMean^2) # initial shape parameter (this was suggested)
initialPar2 = initialPar1/middlePointMean # initial scale parameter

This is the code I used to run the optimization

 # The likelihood function for this problem is defined by the product of the difference between the 
 # cumulative gamma evaluated in the upper bound of the interval - the cumulative gamma evaluated in 
 # the lower bound of the interval. 
 
 logLikelihood = function(par){

 ub = incomeData$U 
 lb = incomeData$L 

  # I'm applying sum instead of prod since the log of a product would be the sum
 
 logLike = sum(pgamma(ub,shape = par[1],scale = par[2]) - 
 pgamma(lb,shape = par[1],scale = par[2]))

return(-logLike)

}

optim(par = c(initialPar1,initialPar2),fn = logLikelihood,
method = "L-BFGS-B",lower = 0.00001,upper=.99999)

I get these results

$par
[1] 1.014180e-01 1.737418e-05

$value
[1] 0

$counts
function gradient 
  1       1 

$convergence
[1] 0

$message
[1] "CONVERGENCE: NORM OF PROJECTED GRADIENT <= PGTOL"

When I test the results with those parameters the values are too low and I can't plot the distribution nor the likelihood function and it doesn't make sense to me. This is supposed to give the proability of falling in a particular income interval. What I'm doing wrong?

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  • $\begingroup$ Your code uses the sum of the likelihoods (not the loglikelihoods) and calls it logLike. $\endgroup$ May 22, 2021 at 23:15

1 Answer 1

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It's a bit strange that your data don't include any intervals at the ends, say $<850$ or $>10000$. I modified your approach and got some sensible results, I think. Instead of the individual data, I used the grouped data with the frequencies of each interval. Further, I used the log of the shape and rate of the gamma distribution for fitting for numerical purposes. Lastly, your code doesn't calculate the log-likelihood, just the likelihood so I changed that as well.

Here's the code:

# Grouped data
dat <- data.frame(
  U = c(999, 2000, 3000, 4000, 5000, 10000)
  , L = c(850, 1000, 2001, 3001, 4001, 5001)
  , f = c(1, 24, 112, 267, 598, 1146)
)

# Log-likelihood
logLikelihood = function(par, data){
  df <- function(f, low, up, par1, par2) {
    f*log(pgamma(up, par1, par2) - pgamma(low, par1, par2))
  }  
  shape = exp(par[1])
  rate = exp(par[2])  
  -sum(df(dat$f, dat$L, dat$U, shape, rate))
}

# Fit
res <- optim(par = c(3, -6), fn = logLikelihood, data = dat, control = list(reltol = 1e-15))

# Results
exp(res$par)
[1] 11.264619015  0.002143716

So the fitted gamma distribution has shape $11.265$ and rate $0.00214$. The density looks like this:

Gamma_PDF

The mean of the gamma distribution is $11.265/0.00214=5254.7$ which is not too far from the mean of the grouped data ($5837.3$).

Here is the code to produce the graph:

par(mar = c(5.5, 5.5, 0.1, 0.5), mgp=c(4.5,1,0))
curve(dgamma(x, exp(res$par)[1], exp(res$par)[2]), from = 0, to = 15000, n = 1000L, lwd = 2, col = "steelblue", ylab = "Density", xlab = "Income", xaxt = "n", las = 1)
axis(1, at = seq(0, 15000, 2500))
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  • $\begingroup$ As you said, I also think that the grouped data works better. I was checking the code and if I remove the exponential from the shape and rate and the control parameter in the optim function, the results don't change that much. This is not a big deal is it, or there might be some implications? Also, sorry for my inexperience with R graphs, but how did you plot the density? $\endgroup$
    – Seb
    May 23, 2021 at 0:57
  • $\begingroup$ @Seb I added the R code I used to produce the density plot. The reason I used the log-parameters for optimization is that it removes any boundaries and hopefully makes the fitting easier and more stable. As you said: The fit also works on the scale of the original parameters. $\endgroup$ May 23, 2021 at 6:28

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