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Let $X_{1},...,X_{n}$ be independent random variables with exponential distribution of parameter $\lambda$.

The task is to find the maximum likelihood estimator and the first moment estimator of the percentile 0.85.

How can I do that? I know how to calculate those estimations of $\lambda$.

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Note: the negative exponential law can be parametrized in several ways, so it would be better to specify the density. Let's suppose that the density is the following

$$f_X(x)=\lambda e^{-\lambda x}\mathbb{1}_{[0;+\infty)}(x)$$

  1. The first moment is

$$\mathbb{E}[X]=\frac{1}{\lambda}$$

  1. The 85° percentile is the following

$$x=F^{-1}_X(0.85)=g(\lambda)$$

If you can estimate $\lambda$ then you can use the Invariance Property of ML Estimator

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  • $\begingroup$ The percentile 0.85 is x such $F_{X}(x)=0.85$ not $F_{X}(0.85)$ $\endgroup$ – lmglm Jun 30 at 6:08
  • $\begingroup$ @Lucy...ops, thank you, amended. It's very early in the morning here in Italy...I'm still sick :( $\endgroup$ – tommik Jun 30 at 6:22
  • $\begingroup$ That's okay. Thank you! So you said that ''the estimation by maximum likelihood of the percentile'' is just calculating the one for lambda as we know and replacing in $g(\lambda)$? $\endgroup$ – lmglm Jun 30 at 6:28
  • $\begingroup$ @Lucy : yes, by invairance property if $\hat{\lambda}$ is MLE for $\lambda$ then $g(\hat{\lambda})$ is MLE for $g(\lambda)$ $\endgroup$ – tommik Jun 30 at 6:37
  • $\begingroup$ @tommik. Direct and to the point. (+1). Get better soon. $\endgroup$ – BruceET Jun 30 at 9:09
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Suppose $\lambda = 0.2$ so that $X \sim \mathsf{Exp}(\mathrm{rate}=0.2),$ has $E(X) = 1/\lambda = 5.$

Then a sample of size $n = 10$ from this distribution might be the values in the vector x below, as sampled in R:

set.seed(629)  # for reproducibility
x = rexp(10, .2)
summary(x)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
 0.09785  1.69027  3.03245  4.81721  4.45529 18.73775 

Then the MME $\hat\mu$ of $\mu = 5$ is $\hat \mu = \bar X = 4.8172.$

The quantile function (inverse CDF) in R, is qexp, so then MLE of the 85th percentile is $F_X^{-1}(\bar X) = 0.3938.$

qexp(.85, 1/mean(x))
[1] 9.13883

This estimates the true 85th percentile of $X \sim \mathsf{Exp}(\mathrm{rate}=0.2),$ which is 9.4856,

qexp(.85, .2)
[1] 9.4856

Because $\bar X/\mu = \lambda \bar X \sim \mathsf{Gamma}(\mathrm{shape}=n,\mathrm{rate}=n),$ we have

$$0.95 = P\left(L < \frac{\bar X}{\mu} < U\right) = P\left(\frac{\bar X}{U} < \mu < \frac{\bar X}{L}\right),$$ where $L$ and $U$ cut probability 0.025 from the lower and upper tails, respectively of $\mathsf{Gamma}(n,n).$ Thus a 95% CI for $\mu$ is of the form $\left(\frac{\bar X}{U},\frac{\bar X}{L}\right).$ For our sample with $\bar X = 4.8172,$ this computes to $(2.82, 10.05).$

ci.mu = mean(x)/qgamma(c(.975,.025), 10, 10); ci.mu
[1]  2.819589 10.045510

Then a 95% CI for the 85th percentile based on the MLE $\hat\mu$ is $(5.35, 19.06).$ Of course, a larger sample size $n$ would give shorter confidence intervals.

qexp(.85, 1/ci.mu)
[1]  5.349099 19.057537

Note: I experimented with a 95% parametric bootstrap CI $(4.38, 15.61)$ based on a crude quantile method. The re-sampling is shown below.

set.seed(2020)
q.85 = replicate(10^6, qexp(.85, 1/mean( 
                 rexp(10,1/mean(x)))))
quantile(q.85, c(.025,.975))
     2.5%     97.5% 
 4.381346 15.615057 
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  • $\begingroup$ the MME for $\mu$ is $1/ \bar{X}$ isn't it? I was looking for a theoric approach of the estimation of the percentile 0.85 $\endgroup$ – lmglm Jun 30 at 16:00
  • $\begingroup$ No. MME for population mean $\mu$ is sample mean $\bar X,$ No matter which parameterization of the exponential dist'n your're using. Easy to get confused between parameterization of exponential using rate and using scale. Even more so if you're looking at various sources. Usual practice is to use $\theta$ for scale and $\lambda$ for rate, As @tommik says, best to specify the density function at the start and stick with that. R uses rate parameterization. // Except for the bootstrap in the note, my development is totally theoretical, please don't be distracted by use of R for numerical exmp. $\endgroup$ – BruceET Jun 30 at 17:39
  • $\begingroup$ Yes, my bad that one is for $\lambda$. I get then the maximum likelihood version but which is "the first moment estimator of the percentile 0.85"? There I don't have the invariance property. Thanks! $\endgroup$ – lmglm Jun 30 at 17:44
  • $\begingroup$ Reading through all the comments here, it seems to me you may be confused on one or more fundamental principles of estimation. I hope you can discuss some problems about which you are confused in person with an instructor or TA. Pretending to understand and 'discussing' with people who try to help here doesn't seem to be getting at your fundamental difficulties. $\endgroup$ – BruceET Jun 30 at 18:13

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