R/lorenzcurve.R

Defines functions lorenzcurve

Documented in lorenzcurve

#' @title The Lorenz curve
#' @description This function returns the Lorenz curve of a rule for a claims problem.
#' @param E The endowment.
#' @param d The vector of claims.
#' @param Rules The rules: AA, APRO, CE, CEA, CEL, DT, MO, PIN, PRO, RA, Talmud.
#' @param col The colors. If col=NULL then the sequence of default colors is:
#' c("red", "blue", "green", "yellow", "pink", "coral4", "darkgray", "burlywood3", "black", "darkorange", "darkviolet").
#' @param legend A logical value. The color legend is shown if legend=TRUE.
#' @return The graphical representation of the Lorenz curve of a rule (or several rules) for a claims problem.
#' @details Let \eqn{E> 0} be the endowment to be divided and \eqn{d\in \mathcal{R}^n}{d} the vector of claims
#' with \eqn{d\ge 0} and such that the sum of claims \eqn{D=\sum_{i=1}^{n} d_i\ge E}{D=\sum di} exceeds the endowment.
#'
#' Rearrange the claims from small to large, \eqn{0 \le d_1 \le...\le d_n}{%
#' 0 \le d1 \le...\le dn}. The Lorenz curve represents the proportion of the awards given to each subset of claimants by a specific rule \eqn{R}  as a function of the
#' cumulative distribution of population.
#'
#' The Lorenz curve of a rule \eqn{R} for the claims problem \eqn{(E,d)} is the polygonal path connecting the \eqn{n+1} points
#' \deqn{(0,0), (\frac{1}{n},\frac{R_1(E,d)}{E}),\dots,(\frac{n-1}{n},\frac{\sum_{i=1}^{n-1}R_i(E,d)}{E}),(1,1)}{%
#' (0,0) , (1/n,R1(E,d)/E) , (2/n , (R1(E,d)+R2(E,d))/E ,\dots , (1,1)}
#' Basically, it represents the cumulative percentage of the endowment assigned by the rule to each cumulative percentage of claimants.
#'
#' @seealso \link{giniindex}, \link{cumulativecurve}, \link{proportionalityindex}, \link{indexpath}, \link{lorenzdominance}.
#' @examples
#' E=10
#' d=c(2,4,7,8)
#' Rules=c(AA,RA,Talmud,CEA,CEL)
#' col=c("red","blue","green","yellow","pink")
#' lorenzcurve(E,d,Rules,col)
#' @references  Lorenz, M. O. (1905). Methods of measuring the concentration of wealth. Publications of the American statistical association, 9(70), 209-219.
#' @references Mirás Calvo, M.A., Núñez Lugilde, I., Quinteiro Sandomingo, C., and Sánchez-Rodríguez, E. (2021). Deviation from proportionality and Lorenz-dominance between
#' the average of awards and the standard rules for claims problems. Working paper 2021-01, ECOBAS.
#' @importFrom graphics lines
#' @importFrom graphics legend
#' @importFrom graphics grid
#' @importFrom graphics mtext
#' @export

lorenzcurve = function(E, d, Rules, col = NULL, legend = TRUE) {

# Default colors
  if (is.null(col)) {
  col=c("red","blue","green","yellow","pink","coral4","darkgray","burlywood3","black","darkorange","darkviolet")
  }

  ########################################
  # Required: (E,d) must be a claims problem, i.e., E >=0, d >=0, E <= sum(d)
  ########################################
  n = length(d)
  D = sum(d) #The number of claims and the total claim
  if (E < 0 || sum((d < 0)) > 0 || E > D)
    stop('(E,d) is not a claims problem.',call.=F)

  if (E ==0)
    stop('The endowment must be strictly positive, E>0.',call.=F)

  ###################
  do = sort(d)
  if (sum(do == d) < n){
    message('The result is computed for the rearranged vector of claims.\n')
  }
  ### THE IDENTITY LINE
  plot(
    c(0, 1), c(0, 1),
    type = "l",
    xlim = c(0, 1),
    ylim = c(0, 1),
    xlab = c("Percentage of claimants"),
    ylab = c("Percentage of awards"),
    main = c("Lorenz curve")
  )
  subtitle=paste("E=",toString(E),"; d=(",toString(d),")",sep="")
  mtext(subtitle,side=3,line=0.5,cex=0.7)
  grid()
  ###### claims CURVE (DATA) ######
  numberrules = length(Rules)
  percentAwards = matrix(0, numberrules, n)
  percentClaimants = matrix(0, numberrules, n)
  # THE CURVES
  for (ii in 1:numberrules) {
    # THE CUMULATIVE PERCENTAGES
    rule = Rules[[ii]](E, do)
    percentAwards[ii, ] = cumsum(rule)/E
    percentClaimants[ii, ] =c(1:n)/n
    # THE POLIGONAL CURVE
    lines(
      c(0,percentClaimants[ii,]),
      c(0, percentAwards[ii,]),
      lwd = 2,
      type = "o",
      col = col[ii]
    )
  }
  name = rep(0, numberrules)
  for (kk in 1:numberrules) {
    name[kk] = Rules[[kk]](0,0,TRUE )
  }
  if (legend==TRUE){
  legend(
    x = "topleft",
    legend = c(name),
    col=col[1:numberrules],
    lty=1,
    lwd=2,
    seg.len=0.5,
    cex=0.8,
    y.intersp=0.6,
    bty ="n"
  )
}
}

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ClaimsProblems documentation built on April 7, 2021, 9:07 a.m.