Normalの合計とNormalのキューブの合計の比率

12

次の極限分布（として $n \rightarrow \infty$ ）を見つけるのを手伝ってください： IIDれる。

U_{n} = \frac{X_{1} + X_{2} + \dots + X_{n}}{X_{1}^{3} + X_{2}^{3} + \dots X_{n}^{3}},

$U_n = \frac{X_1 + X_2 + \ldots + X_n}{X_1^3 + X_2^3 + \ldots X_n^3},$

X_{i}

$X_i$

N (0, 1)

$N(0,1)$

distributions normal-distribution asymptotics

— アルナンシュ・ビスワス
ソース

1

ランダム変数の変換を見てみましたか？たとえば、特徴的な機能、ラプラススティールチェス変換などを試すことができます。

— スティン

1

ヒント：分子と分母は漸近的に二変量の正規です。モーメントを直接計算できます。平均は明らかにゼロ、分子の分散は

、分母の分散は

、共分散は

です。（したがって、相関は

n

$n$

15 n

$15n$

3 n

$3n$

。）通常、任意のゼロ平均二変量を発現し、極限分布を見つけるには

の形で

の独立したゼロ平均法線のための

と

と定数

、その注意比率

は、シフトされたスケーリングされたコーシー分布です。

3 / \sqrt{15} \approx 0.775

$3/\sqrt{15} \approx 0.775$

(U, V)

$(U,V)$

(A, β A + B)

$(A, \beta A + B)$

A

$A$

B

$B$

β

$\beta$

V / U = β + B / A

$V/U = \beta + B/A$

— whuber

2

公式がどこと独立している、それだけで古典的な教科書の練習になります。という事実を使用します

U_{n} = \frac{X_{1} + X_{2} + \dots + X_{n}}{Y_{1}^{3} + Y_{2}^{3} + \dots Y_{n}^{3}}

$U_n = \frac{X_1 + X_2 + \ldots + X_n}{Y_1^3 + Y_2^3 + \ldots Y_n^3}$

X_{i} \sim N (0, 1)

$X_i\sim N(0,1)$

Y_{i} \sim N (0, 1)

$Y_i \sim N(0,1)$

F_{n} \overset{d}{\to} F, G_{n} \overset{d}{\to} G \Rightarrow \frac{F_{n}}{G_{n}} \overset{d}{\to} \frac{F}{G}

$F_n \stackrel{d}{\to} F,\quad G_n \stackrel{d}{\to}G \Rightarrow \frac{F_n}{G_n} \stackrel{d}{\to} \frac{F}{G}$ and we can conclude that

U

$U$ asymptotes to scaled Cauchy distribution.

But in your formulation, we can't apply the theorem due to dependence. My Monte-Carlo suggests that the limit distribution of $U_n$ is non-degenerate and it has no first moment and is not symmetric. I'd be interested in whether there is a explicit solution to this problem. I feel like the solution can only be written in terms of Wiener process.

[EDIT] Following whuber's hint, note that

(\frac{1}{\sqrt{n}} \sum X_{i}, \frac{1}{\sqrt{n}} \sum X_{i}^{3}) \overset{d}{\to} (Z_{1}, Z_{2})

$(\frac{1}{\sqrt{n}}\sum{X_i},\frac{1}{\sqrt{n}}\sum{X_i^3})\stackrel{d}{\to}(Z_1,Z_2)$

(Z_{1}, Z_{2}) \sim N (0, (\begin{matrix} 1 & 3 \\ 3 & 15 \end{matrix}))

$(Z_1,Z_2) \sim N(0,\pmatrix{1 &3 \\3 &15})$

E [X_{1}^{4}] = 3

$E[X_1^4]=3$

E [X_{1}^{6}] = 15

$E[X_1^6]=15$

(n - 1)!!

$(n-1)!!$

n

$n$

U_{n} \overset{d}{\to} \frac{Z_{1}}{Z_{2}}

$U_n \stackrel{d}{\to} \frac{Z_1}{Z_2}$

Z_{1} = \frac{1}{5} Z_{2} + \sqrt{\frac{2}{5}} Z_{3}

$Z_1 = \frac{1}{5}Z_2+\sqrt{\frac{2}{5}}Z_3$ where

Z_{3} \sim N (0, 1)

$Z_3\sim N(0,1)$ and independent of

Z_{2}

$Z_2$ , we conclude that

U_{n} \overset{d}{\to} \frac{1}{5} + \sqrt{\frac{2}{5}} \frac{Z_{3}}{Z_{2}} \equiv \frac{1}{5} + \sqrt{\frac{2}{75}} Γ

$U_n \stackrel{d}{\to} \frac{1}{5}+\sqrt{\frac{2}{5}}\frac{Z_3}{Z_2} \equiv \frac{1}{5}+\sqrt{\frac{2}{75}}\Gamma$ where

Γ \sim C a u c h y

$\Gamma \sim Cauchy$

— Julius
ソース

0

Some comments, not a full solution. This is to long for a comment, but really only a comment. Some properties of the solution. Since the $X_i$ are iid standard normal, which is a symmetric (about zero) distribution, $X_i^3$ will also have symmetric distributions, and sums of (independent) symmetric rv's will be symmetric. So this is a ratio with the numerator and denominator both symmetric, so will be symmetric. The denominator will have a continuous density which is positive at zero, so we will expect the ratio to lack expectation (It is a general result that if $Z$ is a random variable with continuous density positive at zero, then the $1/X$ will lack expectation. See I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?). But here, there is dependence between numerator and denominator which complicates the matter ... (Clearly needs more thought here).

The interesting paper https://projecteuclid.org/download/pdf_1/euclid.aop/1176991795 shows that $x_i^3$ above, the cube of standard normal variables, has an indeterminate distribution "in the hamburger sense", that is, it is not determined by its moments! So the comment above about using transforms, might indicate a difficult way to proceed!

— kjetil b halvorsen
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