假设

证明以下陈述正确的最简单方法是什么？

假设 $Y_1, \dots, Y_n \overset{\text{iid}}{\sim} \text{Exp}(1)$ 。显示 $\sum_{i=1}^{n}(Y_i - Y_{(1)}) \sim \text{Gamma}(n-1, 1)$ 。

注意， $Y_{(1)} = \min\limits_{1 \leq i \leq n}Y_i$ 。

通过 $X \sim \text{Exp}(\beta)$ ，这意味着， $f_{X}(x) = \dfrac{1}{\beta}e^{-x/\beta} \cdot \mathbf{1}_{\{x > 0\}}$ 。

很容易看到 $Y_{(1)} \sim \text{Exponential}(1/n)$ 。此外，我们也有 $\sum_{i=1}^{n}Y_i \sim \text{Gamma}(\alpha = n, \beta = 1)$ 的参数化下

f_{Y} (y) = \frac{1}{Γ (α) β^{α}} x^{α - 1} e^{- x / β} 1_{{x > 0}}, α, β > 0 .

$f_{Y}(y) =\dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta}\mathbf{1}_{\{x > 0\}}\text{, }\qquad \alpha, \beta> 0\text{.}$

西安人给出的解决方案答案：在原始问题中使用符号：由此，我们得到了

\begin{aligned} \sum_{i = 1}^{n} [Y_{i} - Y_{(1)}] & = \sum_{i = 1}^{n} [Y_{(i)} - Y_{(1)}] \\ = \sum_{i = 1}^{n} Y_{(i)} - n Y_{(1)} \\ = \sum_{i = 1}^{n} {Y_{(i)} - Y_{(i - 1)} + Y_{(i - 1)} - \dots - Y_{(1)} + Y_{(1)}} - n Y_{(1)} \\ = \sum_{i = 1}^{n} \sum_{j = 1}^{i} {Y_{(j)} - Y_{(j - 1)}} - n Y_{(1)} where Y_{(0)} = 0 \\ = \sum_{j = 1}^{n} \sum_{i = j}^{n} {Y_{(j)} - Y_{(j - 1)}} - n Y_{(1)} \\ = \sum_{j = 1}^{n} (n - j + 1) [Y_{(j)} - Y_{(j - 1)}] - n Y_{(1)} \\ = \sum_{i = 1}^{n} (n - i + 1) [Y_{(i)} - Y_{(i - 1)}] - n Y_{(1)} \\ = \sum_{i = 2}^{n} (n - i + 1) [Y_{(i)} - Y_{(i - 1)}] + n Y_{(1)} - n Y_{(1)} \\ = \sum_{i = 2}^{n} (n - i + 1) [Y_{(i)} - Y_{(i - 1)}] . \end{aligned}

$\begin{align} \sum_{i=1}^{n}[Y_i - Y_{(1)}] &= \sum_{i=1}^{n}[Y_{(i)}-Y_{(1)}] \\ &= \sum_{i=1}^{n}Y_{(i)}-nY_{(1)}\\ &= \sum_{i=1}^{n}\{Y_{(i)}-Y_{(i-1)}+Y_{(i-1)}-\cdots-Y_{(1)}+Y_{(1)}\}-nY_{(1)}\\ &= \sum_{i=1}^n\sum_{j=1}^{i}\{Y_{(j)}-Y_{(j-1)}\}-nY_{(1)}\text{ where } Y_{(0)} = 0 \\ &= \sum_{j=1}^n\sum_{i=j}^{n}\{Y_{(j)}-Y_{(j-1)}\}-nY_{(1)}\\ &= \sum_{j=1}^{n}(n-j+1)[Y_{(j)}-Y_{(j-1)}]-nY_{(1)}\\ &= \sum_{i=1}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}]-nY_{(1)}\\ &= \sum_{i=2}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}]+nY_{(1)}-nY_{(1)} \\ &= \sum_{i=2}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}]\text{.} \end{align}$

。

\sum_{i = 2}^{n} (n - i + 1) [Y_{(i)} - Y_{(i - 1)}] \sim Gamma (n - 1, 1)

$\sum_{i=2}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}] \sim \text{Gamma}(n-1, 1)$

— 单簧管
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@MichaelChernick 1）我不确定我的独立性证明是否正确，以及2）我不确定我上面猜测的关于Gamma分布差异的结果是否正确。这似乎与此处给出的内容相矛盾，但是也许这种情况有所不同，因为这种差异包括订单统计之一？我不确定。

— 单簧管手

@Clarinetist, I'm not certain. Maybe try to work with

\sum_{i = 2}^{n} (Y_{(i)} - Y_{(1)})

$\sum_{i = 2}^n (Y_{(i)} - Y_{(1)})$ , which clearly equals the sum you're working with. The answer here might be helpful: math.stackexchange.com/questions/80475/…

— gammer

Have you tried to prove that each

(Y_{i} - Y_{(1)}) \sim Expon (1)

$(Y_i - Y_{(1)}) \sim \text{Expon}(1)$ -- except for one

i

$i$ , for which

Y_{i} - Y_{(1)} = 0

$Y_i - Y_{(1)} = 0$ and, then, using the fact that the sum of

(n - 1)

$(n-1)$ iid Exponential variates will be Gamma distributed?

— Marcelo Ventura

@jbowman We have

f_{Z_{i}} (z_{i}) = f_{Y_{i}} (z_{i} + a) = e^{- (z_{i} + a)}

$f_{Z_i}(z_i) = f_{Y_i}(z_i + a) = e^{-(z_i + a)}$ and conditioned on

Y_{i} \geq a

$Y_i \geq a$ , we divide this by

e^{- a}

$e^{-a}$ , giving

e^{- z_{i}}

$e^{-z_i}$ , hence we have

(Z_{i} ∣ Y_{i} \geq A) \sim Exp (1)

$(Z_i \mid Y_i \geq A) \sim \text{Exp}(1)$ . Now here's what's bugged me about this proof: I regarded

A

$A$ as a constant. But

Y_{(1)}

$Y_{(1)}$ isn't a constant. Why would this work?

— Clarinetist

The point is that it doesn't matter what

a

$a$ is! The distribution is always

Exp (1)

$\text{Exp}(1)$ ! Remarkable, isn't it? And from this you can conclude that the distribution of

Y_{i} - Y_{[1]}

$Y_i - Y_{[1]}$ is always

Exp (1)

$\text{Exp}(1)$ for

i > 1

$i > 1$ , regardless of the actual value of

Y_{[1]}

$Y_{[1]}$ .

— jbowman

Answers:

该证明在第211页上的所有随机数生成之书，Devroye的非均匀随机变量生成中给出（这是一个非常优雅的方法！）：

$E_{(0)}=0$
$(n - i + 1) (E_{(i)} - E_{(i - 1)})$ $(n-i+1)(E_{(i)}-E_{(i-1)})$ derived from the order statistics $E_{(1)}\le\ldots\le E_{(n)}$ of an i.i.d. exponential sample of size $n$ are themselves i.i.d. exponential variables

Proof. Since

\begin{aligned} \sum_{i = 1}^{n} e_{i} & = \sum_{i = 1}^{n} e_{(i)} = \sum_{i = 1}^{n} \sum_{j = 1}^{i} (e_{(j)} - e_{(j - 1)}) \\ = \sum_{j = 1}^{n} \sum_{i = j}^{n} (e_{(j)} - e_{(j - 1)}) = \sum_{j = 1}^{n} (n - j + 1) (e_{(j)} - e_{(j - 1)}) \end{aligned}

$\begin{align*} \sum_{i=1}^n e_i &= \sum_{i=1}^n e_{(i)} =\sum_{i=1}^n \sum_{j=1}^i(e_{(j)}-e_{(j-1)})\\ &=\sum_{j=1}^n \sum_{i=j}^n(e_{(j)}-e_{(j-1)}) =\sum_{j=1}^n (n-j+1)(e_{(j)}-e_{(j-1)}) \end{align*}$ the joint density of the order statistic

(E_{(1)}, \dots, E_{(n)})

$(E_{(1)},\ldots,E_{(n)})$ writes as

f (e) = n! \exp {- \sum_{i = 1}^{n} e_{(i)}} = n! \exp {- \sum_{i = 1}^{n} (n - i + 1) (e_{(i)} - e_{(i - 1)})}

$f(\mathbf{e})=n!\,\exp\left\{-\sum_{i=1}^ne_{(i)}\right\}=n!\,\exp\left\{-\sum_{i=1}^n (n-i+1)(e_{(i)}-e_{(i-1)})\right\}$ Setting

Y_{i} = (E_{(i)} - E_{(i - 1)})

$Y_i=(E_{(i)}-E_{(i-1)})$ , the change of variables from

(E_{(1)}, \dots, E_{(n)})

$(E_{(1)},\ldots,E_{(n)})$ to

(Y_{1}, \dots, Y_{n})

$(Y_1,\ldots,Y_n)$ has a constant Jacobian [incidentally equal to

1 / n!

$1/n!$ but this does not need to be computed] and hence the density of

(Y_{1}, \dots, Y_{n})

$(Y_1,\ldots,Y_n)$ is proportional to

\exp {- \sum_{i = 1}^{n} y_{i}}

$\exp\left\{-\sum_{i=1}^n y_i \right\}$ which establishes the result. Q.E.D.

An alternative suggested to me by Gérard Letac is to check that

(E_{(1)}, \dots, E_{(n)})

$(E_{(1)},\ldots,E_{(n)})$ has the same distribution as

(\frac{E_{1}}{n}, \frac{E_{1}}{n} + \frac{E_{2}}{n - 1}, \dots, \frac{E_{1}}{n} + \frac{E_{2}}{n - 1} + \dots + \frac{E_{n}}{1})

$\left(\frac{E_1}{n},\frac{E_1}{n}+\frac{E_2}{n-1},\ldots,\frac{E_1}{n}+\frac{E_2}{n-1}+\ldots+\frac{E_n}{1}\right)$ (by virtue of the memoryless property), which makes the derivation of

\sum_{k = 1}^{n} (E_{k} - E_{(1)}) \sim \sum_{k = 1}^{n - 1} E_{k}

$\sum_{k=1}^n(E_k-E_{(1)})\sim \sum_{k=1}^{n-1}E_k$ straightforward.

— Xi'an
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Thank you for this answer! I'd like to fill in some details for anyone who is reading this in the future:

e_{i}

$e_i$ are observed values of the

E_{i}

$E_i$ , and the easiest way to see that

\sum_{i = 1}^{n} e_{(i)} = \sum_{i = 1}^{n} (n - i + 1) (e_{(i)} - e_{(i - 1)}) = \sum_{i = 1}^{n} e_{(i)}

$\sum_{i=1}^{n}e_{(i)} = \sum_{i=1}^{n}(n-i+1)(e_{(i)} - e_{(i-1)}) = \sum_{i=1}^{n}e_{(i)}$ is to write out

\sum_{i = 1}^{n} (n - i + 1) (e_{(i)} - e_{(i - 1)})

$\sum_{i=1}^{n}(n-i+1)(e_{(i)} - e_{(i-1)})$ term-by-term. Because the density of

(Y_{1}, \dots, Y_{n})

$(Y_1, \dots, Y_n)$ is proportional to

\exp (- \sum_{i = 1}^{n} y_{i})

$\exp\left(-\sum_{i=1}^{n}y_i \right)$ , separate the

y_{i}

$y_i$ to see the density is proportional to

\prod_{i = 1}^{n} e^{- y_{i}}

$\prod_{i=1}^{n}e^{-y_i}$ , hence

Y_{1}, \dots, Y_{n} \overset{iid}{\sim} Exp (1)

$Y_1, \dots, Y_n \overset{\text{iid}}{\sim} \text{Exp}(1)$ .

— Clarinetist

I lay out here what has been suggested in comments by @jbowman.

Let a constant $a\geq 0$ . Let $Y_i$ follow an $\text{Exp(1)}$ and consider $Z_i = Y_i-a$ . Then

Pr (Z_{i} \leq z_{i} ∣ Y_{i} \geq a) = Pr (Y_{i} - a \leq z_{i} ∣ Y_{i} \geq a)

$\Pr(Z_i\leq z_i \mid Y_i \geq a) = \Pr(Y_i-a\leq z_i \mid Y_i \geq a)$

⟹ Pr (Y_{i} \leq z_{i} + a ∣ Y_{i} \geq a) = \frac{Pr (Y_{i} \leq z_{i} + a, Y_{i} \geq a)}{1 - Pr (Y_{i} \leq a)}

$\implies \Pr(Y_i\leq z_i+a \mid Y_i \geq a) = \frac {\Pr(Y_i\leq z_i+a,Y_i \geq a)}{1-\Pr(Y_i\leq a)}$

⟹ \frac{Pr (a \leq Y_{i} \leq z_{i} + a)}{1 - Pr (Y_{i} \leq a)} = \frac{1 - e^{- z_{i} - a} - 1 + e^{- a}}{e^{- a}} = 1 - e^{- z_{i}}

$\implies \frac {\Pr(a\leq Y_i\leq z_i+a)}{1-\Pr(Y_i\leq a)} = \frac {1-e^{-z_i-a}-1+e^{-a}}{e^{-a}}=1-e^{-z_i}$

which is the distribution function of $\text{Exp(1)}$ .

Let's describe this: the probability that an $\text{Exp(1)}$ r.v. will fall in a specific interval (the numerator in the last line), given that it will exceed the interval's lower bound (the denominator), depends only on the length of the interval and not on where this interval is placed on the real line. This is an incarnation of the "memorylessness" property of the Exponential distribution, here in a more general setting, free of time-interpretations (and it holds for the Exponential distribution in general)

Now, by conditioning on $\{Y_i \geq a\}$ we force $Z_i$ to be non-negative, and crucially, the obtained result holds $\forall a\in \mathbb R^+$ . So we can state the following:

If $Y_i\sim \text{Exp(1)}$ , then $\forall Q\geq 0 : Z_i = Y_i-Q \geq 0$ $\implies$ $Z_i\sim \text{Exp(1)}$ .

Can we find a $Q\geq 0$ that is free to take all non-negative real values and for which the required inequality always holds (almost surely)? If we can, then we can dispense with the conditioning argument.

And indeed we can. It is the minimum-order statistic, $Q=Y_{(1)}$ , $\Pr(Y_i \geq Y_{(1)})=1$ . So we have obtained

Y_{i} \sim Exp(1) ⟹ Y_{i} - Y_{(1)} \sim Exp(1)

$Y_i\sim \text{Exp(1)} \implies Y_i-Y_{(1)} \sim \text{Exp(1)}$

This means that

Pr (Y_{i} - Y_{(1)} \leq y_{i} - y_{(1)}) = Pr (Y_{i} \leq y_{i})

$\Pr(Y_i-Y_{(1)} \leq y_i-y_{(1)}) = \Pr(Y_i \leq y_i)$

So if the probabilistic structure of $Y_i$ remains unchanged if we subtract the minimum order statistic, it follows that the random variables $Z_i=Y_i-Y_{(1)}$ and $Z_j=Y_j-Y_{(1)}$ where $Y_i, Y_j$ independent, are also independent since the possible link between them, $Y_{(1)}$ does not have an effect on the probabilistic structure.

Then the sum $\sum_{i=1}^{n}(Y_i - Y_{(1)})$ contains $n-1$ $\text{Exp(1)}$ i.i.d. random variables (and a zero), and so

\sum_{i = 1}^{n} (Y_{i} - Y_{(1)}) \sim Gamma (n - 1, 1)

$\sum_{i=1}^{n}(Y_i - Y_{(1)}) \sim \text{Gamma}(n-1, 1)$

— Alecos Papadopoulos
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