# Asymptotic distribution of censored samples from $\exp(\lambda)$

Let $X_{(1)}, \ldots, X_{(n)}$ be the order statistic of an i.i.d. sample of size $n$ from $\exp(\lambda)$. Suppose the data is censored so we see only the top $(1-p) \times 100%$ percent of the data, that is $$X_{(\lfloor p n \rfloor )}, X_{(\lfloor p n\rfloor + 1)}, \ldots, X_{(n)}\,.$$ Put $m = \lfloor p n \rfloor$, what is the asymptotic distribution of $$\left(X_{(m)}, \frac{\sum_{i= m+1}^n X_{(i)}}{(n-m)} \right)?$$

This is somewhat related to this question and this and also marginally to this question.

Any help would be appreciated. I tried different approaches but was not able to progress much.

• One can show that conditioned on $X_{(m)}$, vector $(X_{(m+1)}−X_{(m)},…,X_{(n)}−X_{(m)}|X_{(m)})$ is distributed as an order statistic of $\{Y_i\}_{1}^{n−m}$ i.i.d samples from $\exp(1)$ (with $m$ as defined in the question i.e. $m = \lfloor pn \rfloor$), therefor $\frac{1}{m−n}\sum_{i=m+1}^nX_{(i)}−X_{(m)}|X_{(m)} = \frac{1}{m−n} \sum_{i=1}^{n-m} Y_{(i)}$ so at the limit $n \to \infty$, we recover the CLT due to independence of $Y_i$, this seems to be the right track, but I am not able to push this argument further and find asymptotic for $(X_{(m)}, \frac{1}{m−n}\sum_{i=m+1}^nX_{(i)})$... – them Aug 31 '16 at 13:20
• To OP: Why do you refer to your sample as being censored? The term censored would indicate that values below the censoring point are recorded as 0, or recorded at the censoring point, etc. But that is not what you are doing ... you are discarding them, which is not censoring ... it is more like truncating them. And since you are considering the asymptotic distribution, and taking $n$ to be large, why you care about first ordering the sample, and truncating the ordered sample??? Why not simply consider a truncated Exponential distribution, truncated below at p%, and then sum terms of that? – wolfies Aug 31 '16 at 15:42
• @wolfies, I fixed all the typos you have pointed out. I will look into Truncated distribution. Regarding the censoring, I have deleted the note. However some sources I have looked at refer to similar problem as type II censoring top of page 6 here – them Aug 31 '16 at 16:16
• @them that's non-standard terminology as far as I know. You should use a truncated model here. – shadowtalker Sep 6 '16 at 0:18

Since $\lambda$ is just a scale factor, without loss of generality choose units of measurement that make $\lambda=1$, making the underlying distribution function $F(x)=1-\exp(-x)$ with density $f(x)=\exp(-x)$.

From considerations paralleling those at Central limit theorem for sample medians, $X_{(m)}$ is asymptotically Normal with mean $F^{-1}(p)=-\log(1-p)$ and variance

$$\operatorname{Var}(X_{(m)}) = \frac{p(1-p)}{n f(-\log(1-p))^2} = \frac{p}{n(1-p)}.$$

Due to the memoryless property of the exponential distribution, the variables $(X_{(m+1)}, \ldots, X_{(n)})$ act like the order statistics of a random sample of $n-m$ draws from $F$, to which $X_{(m)}$ has been added. Writing

$$Y = \frac{1}{n-m}\sum_{i=m+1}^n X_{(i)}$$

for their mean, it is immediate that the mean of $Y$ is the mean of $F$ (equal to $1$) and the variance of $Y$ is $1/(n-m)$ times the variance of $F$ (also equal to $1$). The Central Limit Theorem implies the standardized $Y$ is asymptotically Standard Normal. Moreover, because $Y$ is conditionally independent of $X_{(m)}$, we simultaneously have the standardized version of $X_{(m)}$ becoming asymptotically Standard Normal and uncorrelated with $Y$. That is,

$$\left(\frac{X_{(m)} + \log(1-p)}{\sqrt{p/(n(1-p))}}, \frac{Y - X_{(m)} - 1}{\sqrt{n-m}}\right)\tag{1}$$

asymptotically has a bivariate Standard Normal distribution.

The graphics report on simulated data for samples of $n=1000$ ($500$ iterations) and $p=0.95$. A trace of positive skewness remains, but the approach to bivariate normality is evident in the lack of relationship between $Y-X_{(m)}$ and $X_{(m)}$ and the closeness of the histograms to the Standard Normal density (shown in red dots).

The covariance matrix of the standardized values (as in formula $(1)$) for this simulation was $$\pmatrix{0.967 & -0.021 \\ -0.021 & 1.010},$$ comfortably close to the unit matrix which it approximates.

The R code that produced these graphics is readily modified to study other values of $n$, $p$, and simulation size.

n <- 1e3
p <- 0.95
n.sim <- 5e3
#
# Perform the simulation.
# X_m will be in the first column and Y in the second.
#
set.seed(17)
m <- floor(p * n)
X <- apply(matrix(rexp(n.sim * n), nrow = n), 2, sort)
X <- cbind(X[m, ], colMeans(X[(m+1):n, , drop=FALSE]))
#
# Display the results.
#
par(mfrow=c(2,2))

plot(X[,1], X[,2], pch=16, col="#00000020",
xlab=expression(X[(m)]), ylab="Y",
main="Y vs X", sub=paste("n =", n, "and p =", signif(p, 2)))

plot(X[,1], X[,2]-X[,1], pch=16, col="#00000020",
xlab=expression(X[(m)]), ylab=expression(Y - X[(m)]),
main="Y-X vs X", sub="Loess smooth shown")
lines(lowess(X[,2]-X[,1] ~ X[,1]), col="Red", lwd=3, lty=1)

x <- (X[,1] + log(1-p))  / sqrt(p/(n*(1-p)))
hist(x, main="Standardized X", freq=FALSE, xlab="Value")