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In Davison A.C., Hinkley D.V. - Bootstrap methods and their application It say's $\bar{X}/\lambda$ "has the Gamma distribution with index n and unit mean". I don't understand the sentence. Isn't gamma distribution a generell distribution, including the exponential distribution, Erlang distribution, and chi-squared distribution? And which one would it be here with which parameters?

Thank you.

StubbornAtom
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Joshua
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2 Answers2

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Let's $X\sim Exp(\lambda)$, say, for $x>0$

$$f_X(x)=\frac{1}{\lambda} e^{-\frac{x}{\lambda}}$$

That is equivalent to a Gamma distribution $\Gamma(1;\lambda)$

Let's have $X_1,...,X_n$ iid rv's with the same denisity,

$$\frac{\sum_i X_i}{n \lambda}\sim \Gamma(n;n)$$

That is a Gamma distribution with index $n$ and mean $\frac{n}{n}=1$

Due to the fact that $n$ is integer, this distribution is a.k.a. Erlang

The proof is trivial using Moment Generating Function and its properties

BruceET
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tommik
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  • Thank you, and these parameters are called shape and rate? So shape is n and rate is 1? – Joshua Jun 29 '20 at 20:22
  • $n$ is shape, 1 can be eihter scale or rate; in this case $\lambda$ is scale while rate is $\frac{1}{\lambda}$. Read here https://en.wikipedia.org/wiki/Gamma_distribution – tommik Jun 29 '20 at 20:29
  • ok, but if $\lambda$ is unknown, then You don't exactly know how $\bar{X}/\lambda$ is distributed right? Because one parameter is missing, do I understand that correctly? – Joshua Jun 29 '20 at 20:37
  • As you can see reading my answer, the law of $\frac{\overline{X}}{\lambda}$ is independent by $\lambda$. This is an important result: $\frac{\overline{X}}{\lambda}$ is a rv depending on the parameter but with a law that is independent by the parameter: it is a pivotal quantity so you can use it to estimate the unknown parameter :) – tommik Jun 29 '20 at 20:40
  • Sorry i don't get it. How is $\bar{X}/\lambda$ independent of $\lambda$ if it follows a distribution with one parameter being $1/\lambda$ – Joshua Jun 29 '20 at 20:44
  • If you read well what i wrote, I said that $\frac{\overline{X}}{\lambda}\sim \Gamma(n;n)$, INDEPENDENT BY $\lambda$. If you do not get it do all the calculation to get this result. It is very easy and you can get it with MGF properties or with Fundamental Transformation Theorem – tommik Jun 29 '20 at 20:48
  • Further more: if you use the following quantity: $\frac{2\sum_i X_i}{\lambda}\sim \chi_{(2n)}^2$ so you can use the tables to estimate your parameter – tommik Jun 29 '20 at 20:57
  • ok, so the parameter "rate" of the gamma distribution which is $1/\lambda$ has absolutely no meaning or what? – Joshua Jun 29 '20 at 20:57
  • come on....initially you have a distribution (neg exp) with an unknown parameter to estimate....with the help of your textbook you have a new rv that depends on the n observations (observable data), on the parameter, whose distribution is independent by the parameter....this is the best scenario you can have to do inference on $\lambda$ – tommik Jun 29 '20 at 20:59
  • Thank you for your help. Im about to flip out. I better go to bed and give it a try again tomorrow. – Joshua Jun 29 '20 at 21:01
  • I am here, if you need help. I suggest you to read this https://en.wikipedia.org/wiki/Pivotal_quantity before try again, to understand what a pivotal quantity is – tommik Jun 29 '20 at 21:04
  • (+1) Just fixed a minor typo. But also, don't you mean gamma with shape $n$ and rate = $n$ in displayed eqn? – BruceET Jun 29 '20 at 23:28
  • If you're estimating the mean μ of a normal population with mean μ and SD σ, then you might take the mean $\bar X$ of n obs. to estimate μ. Then the dist'n of $\bar X$ is normal with mean μ and SD $\sigma/\sqrt{n}.$ Also, $\bar X−μ$ has a normal dist'n with mean 0 and SD $\sigma/\sqrt{n}.$. (No mention of μ.) // Similarly, if you take n obs. from an exponential population with mean λ, then you might use $\bar X$ to estimate λ. But here, $\bar X/λ$ is gamma dist'd with parameters shape=n and rate n. (No mention of λ). – – BruceET Jun 30 '20 at 00:14
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Comment continued: Let $n=10; \lambda =\mathrm{scale} = 2.$ Vector a has 100,000 sample averages. Simulation in R.

set.seed(2020);  n = 10;  scale=2
a = replicate(10^5, mean(rexp(n,1/scale)))  
 # In R 2nd param = rate
mean(a); var(a)
[1] 1.998913    # aprx E(A) = 2
[1] 0.3998701   # aorx Var(A) = 4/10
mean(a/scale)
[1] 0.9994563   # aprx 1
var(a/scale)
[1] 0.09996752  # aprx 1/10

hist(a/scale, prob=T, col="skyblue2")
 curve(dgamma(x, n, n), add=T, col="red")


density of gamma with shape=n, rate=n

enter image description here

BruceET
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