Misalkan

Apa cara termudah untuk melihat bahwa pernyataan berikut ini benar?

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

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

Dengan , ini berarti bahwa . $X \sim \text{Exp}(\beta)$ $f_{X}(x) = \dfrac{1}{\beta}e^{-x/\beta} \cdot \mathbf{1}_{\{x > 0\}}$

Mudah untuk melihat . Selain itu, kami juga memiliki bawah parametrization $Y_{(1)} \sim \text{Exponential}(1/n)$ $\sum_{i=1}^{n}Y_i \sim \text{Gamma}(\alpha = n, \beta = 1)$

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

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

Solusi yang diberikan Jawaban Xi'an : Menggunakan notasi dalam pertanyaan asli: Dari ini, kita dapat1) .

\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)

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

self-study distributions exponential order-statistics jacobian Klarinetis
sumber

@MichaelChernick 1) Saya tidak yakin apakah bukti kemandirian saya benar, dan 2) Saya tidak yakin apakah hasil yang saya duga di atas berkaitan dengan perbedaan distribusi Gamma bahkan benar. Ini tampaknya bertentangan dengan apa yang diberikan di sini , tetapi mungkin situasi ini berbeda karena perbedaan ini termasuk salah satu statistik urutan? Saya tidak yakin.

Klarinetis

@Clarinetist, saya tidak yakin. Mungkin mencoba untuk bekerja dengan

, yang jelas sama dengan jumlah Anda bekerja dengan. Jawabannya di sini mungkin bisa membantu: math.stackexchange.com/questions/80475/...

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

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

gammer

Sudahkah Anda mencoba membuktikan bahwa masing-masing

- kecuali untuk satu

, untuk yang

dan, kemudian, menggunakan fakta bahwa jumlah dari

variasi eksponensial akan didistribusikan oleh Gamma?

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

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

i

$i$

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

$Y_i - Y_{(1)} = 0$

(n - 1)

$(n-1)$

Marcelo Ventura

@jbowman Kami memiliki

dan dikondisikan pada

, kami membaginya dengan

, memberikan

, maka kita memiliki

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

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

Y_{i} \geq a

$Y_i \geq a$

e^{- a}

$e^{-a}$

e^{- z_{i}}

$e^{-z_i}$

. Sekarang inilah yang menyadap saya tentang bukti ini: Saya menganggap

sebagai konstanta. Tapi

tidak konstan. Mengapa ini bisa berhasil?

(Z_{i} ∣ Y_{i} \geq A) \sim Exp (1)

$(Z_i \mid Y_i \geq A) \sim \text{Exp}(1)$

A

$A$

Y_{(1)}

$Y_{(1)}$

Klarinetis

Intinya adalah bahwa tidak peduli apa

adalah! Distribusi selalu

! Luar biasa, bukan? Dan dari sini Anda dapat menyimpulkan bahwa distribusi

selalu

untuk

, terlepas dari nilai aktual

a

$a$

Exp (1)

$\text{Exp}(1)$

Y_{i} - Y_{[1]}

$Y_i - Y_{[1]}$

Exp (1)

$\text{Exp}(1)$

i > 1

$i > 1$

Y_{[1]}

$Y_{[1]}$

jbowman

Jawaban:

Buktinya diberikan di Bunda Semua Buku Generasi Acak, Generasi Acak Non-seragam Non-seragam Devroye , di hal.211 (dan ini sangat elegan!):

Teorema 2.3 (Sukhatme, 1937) Jika kita mendefinisikan maka jarak eksponensial yang dinormalisasi berasal dari statistik urutan dari sampel eksponensial ukuran i sendiri variabel i eksponensial $E_{(0)}=0$
$(n - i + 1) (E_{(i)} - E_{(i - 1)})$ $(n-i+1)(E_{(i)}-E_{(i-1)})$ $E_{(1)}\le\ldots\le E_{(n)}$ $n$

Bukti. Karena kepadatan gabungan dari statistik urutan

\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}

$\begin{align*} \sum_{i=1}^n e_i &= \sum_{i=1}^n e_{(i)} =\sum_{i=1}^n \sum_{j=1}^i(e_{(j)}-e_{(j-1)})\\ &=\sum_{j=1}^n \sum_{i=j}^n(e_{(j)}-e_{(j-1)}) =\sum_{j=1}^n (n-j+1)(e_{(j)}-e_{(j-1)}) \end{align*}$

menulis sebagai

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

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

Pengaturan

, perubahan dari variabel dari

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

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

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

$Y_i=(E_{(i)}-E_{(i-1)})$

(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
sumber

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
sumber