更新： 现在问题更具体了，我添加了左截断的对数似然和 R 代码。

右截断。 为了$Y_i$分布为带参数的截断泊松$\lambda$和截断参数$k$, 的对数似然$n$随机样本由下式给出

$$\log(\lambda) \sum_{i=1}^n y_i -n \lambda -n \log \left({\sum_{j=0}^k {{e^{-\lambda}\lambda^j}\over{j!}}}\right)$$

这是最大化的任何值$\lambda$什么时候$\hat{k}=\max(y_1, y_2,\ldots, y_n)$可以取的最小值）。的最大似然估计需要一个迭代过程。Moore (1952, Biometrika) 提供了一个很好且易于计算的初始估计：$k$$\lambda$$\hat{\lambda}$

$$\hat{\lambda}_0 ={{\sum_{i=1}^n y_i}\over{\sum_{i=1}^n I(y_i<\hat{k})}}$$

其中是指示函数，如果取 1 ，否则取 0。（所以这是一个向上调整的平均值。）$I(\cdot)$$y_i<\hat{k}$

一些不那么优雅的 R 代码如下：

# MLE estimation of a truncated Poisson with unknown truncation level

# Objective is to find estimates of lambda (underlying Poisson mean) and
# k (unknown trunction value).

# Generate some samples from a known truncated Poisson distribution
  lambda <- 10  # Poisson mean
  k <- 8        # Truncation level
  n <- 100      # Sample size
  y <- rpois(n*4,lambda) # Oversample to be more certain of getting n samples
  # Keep just the first n legitimate observations
  y <- y[y <= k][1:n]

# MLE for k and initial estimate for lambda
# From Moore (1952)  Biometrika
# "The estimation of the Poisson parameter from a truncated distribution"
  khat <- max(y)
  lambda0 <- sum(y)/length(y[y < khat])

# Define log of the likelihood function
  logL <- function(lambda,ysum,n,k) {
    log(lambda)*ysum - n*lambda - n*ppois(k,lambda,log.p=TRUE)
  }

# Find maximum likelihood estimate of lambda  
  trPoisson <- optim(lambda0, logL, ysum=sum(y), n=length(y), k=khat, 
    method="BFGS", control=list(fnscale=-1))

# Show results
  cat("MLE of truncation value =",khat,"\n")
  # MLE of truncation value = 8 
  cat("MLE of lambda =",trPoisson$par,"\n")
  # MLE of lambda = 9.765054 

左截断。 对于左截断，对数似然如下：

$$\log(\lambda) \sum_{i=1}^n y_i -n \lambda -n \log \left(1-{\sum_{j=0}^{k-1} {{e^{-\lambda}\lambda^j}\over{j!}}}\right)$$

时，最右边的总和为零。下面是一些R代码：$k<0$

# MLE estimation of a left-truncated Poisson with unknown truncation level

# Objective is to find estimates of lambda (underlying Poisson mean) and
# k (unknown trunction value).

# Generate some samples from a known truncated Poisson distribution
  lambda <- 10   # Poisson mean
  k <- 7         # Truncation level
  n <- 100       # Sample size
  y <- rpois(n*30,lambda) # Oversample to be more certain of getting n samples
  y <- y[y >= k][1:n]

# MLE for k and initial estimate for lambda
  khat <- min(y)
  lambda0 <- mean(y)  # Probably any starting value greater than zero will work

# Define log of the likelihood function
  logL <- function(lambda,ysum,n,k) {
       log(lambda)*ysum - n*lambda - n*log(1-ppois(k-1,lambda))
  }

# Find maximum likelihood estimate of lambda  
  trPoisson <- optim(lambda0, logL, ysum=sum(y), n=length(y), k=khat, 
    method="L-BFGS-B", lower=0, upper=Inf, control=list(fnscale=-1))

# Show results
  cat("MLE of truncation value =",khat,"\n")
  cat("MLE of lambda =",trPoisson$par,"\n")