Carmér‘s Theorem

我们知道,两个独立的正态随机变量加起来还是一个正态的随机变量。但是反过来呢?是不是两个独立的随机变量加起来是个正态,它们就都是正态呢?Carmér定理说的就是这样一件事情:假设X_1, \cdots, X_n是一组独立的随机变量,并且X_1+\cdots+X_n是服从正态分布的,那么X_1,\cdots,X_n都是服从正态分布的。

这是这学期概率论课上老师给的bonus问题,这个定理原始论文是“Über eine Eigenschaft der normalen Verteilungsfunktion”,一篇用德文写的。然后我看着它的公式大概猜了一下这篇文章在干啥…… 证明见下面正文,应该是对的吧。感觉这个证明把这学期实分析、复分析学的东西都用上了一些。

主要的思路是,先证明特征函数\mathbb E[e^{itX}]可以定义在整个复平面上,并且是一个解析函数。同时,证明这个函数的增长阶是2并且没有零点,之后用Hadamard因子分解定理说明它的形式。

还有一个叫做Raikov定理,和这个定理非常相似,把定理中的“正态分布”换成“Poisson分布”就是Raikov定理的内容。证明和这个定理也很像,不过需要先说明两个随机变量都是离散的。

Proof. We may assume that n = 2 and X_1 + X_2 \sim \mathcal N(0, 1).

Let F_1(x), F_2(x) and \Phi(x) be the distribution functions of X_1, X_2 and X_1 + X_2, respectively. We know that

\begin{equation} \Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt \label{eqn:df-normal} \end{equation}

Let \mu_1, \mu_2 be the laws of X_1, X_2, respectively. Since X_1 and X_2 are independent, we have

\begin{equation} \Phi(x) = \int_{-\infty}^\infty F_1(x - t) \mu_2[dt] = \int_{-\infty}^\infty F_2(x - t) \mu_1[dt] \label{eqn:conv-x1x2} \end{equation}

Since \{X_1 + X_2 \leq -4\} \subseteq \{X_1 \leq -2\} \cup \{X_2 \leq -2\} and \mathbb P[X_1 + X_2 \leq -4] = \Phi(-4) > 0, we have either F_1(-2) = \mathbb P[X_1 \leq -2] > 0 or F_2(-2) = \mathbb P[X_2 \leq -2] > 0.

We may assume F_2(-2) > 0, then for every x \in \mathbb R

\begin{equation*} \Phi(x - 2) \geq \int_{-\infty}^{-2}F_1(x - 2 - t)\mu_2[dt] \geq F_1(x)F_2(-2) \end{equation*}

because F_1 is increasing and non-negative. From this equation, we know that for x \leq 0

\begin{equation} \begin{aligned} F_1(x) &\leq \frac{\Phi(x-2)}{F_2(-2)} = A_0 \int_{-\infty}^{x-2} \frac{t}{t}e^{-t^2/2}dt\\&\leq A_0 \int_{-\infty}^{x-2} -t e^{-t^2/2}dt = A e^{-x^2/2 + 2x} \end{aligned} \label{eqn:F1-neg-x} \end{equation}

By equation (\ref{eqn:conv-x1x2}) and \int_\mathbb R \mu_2[dt] = 1, we know

\begin{equation*} 1 - \Phi(x) = \int_{\infty}^\infty [1 - F_1(x - t)]\mu_2[dt] \end{equation*}

Similarly, we can show that for x > 0

\begin{equation} 1 - F_1(x) \leq Be^{-x^2/2-2x} \label{eqn:F1-pos-x} \end{equation}

Next, we will show that \int_\mathbb R e^{x^2/2} \mu_1[dx] < \infty.

\begin{equation*} \begin{aligned} \int_\mathbb R e^{x^2/2} \mu_1[dx] &\leq \sum_{k=1}^\infty \int_{-k}^{-k + 1}e^{k^2/2} \mu_1[dx] + \sum_{k=0}^\infty \int_{k}^{k + 1}e^{(k + 1)^2/2} \mu_1[dx] \\ &= \sum_{k=1}^\infty e^{k^2/2} [F_1(-k + 1) - F_1(-k)] + \sum_{k=0}^\infty e^{(k+1)^2/2} [F_1(k + 1) - F_1(k)] \\ &\leq \sum_{k=1}^\infty e^{k^2/2} F_1(-k + 1) + \sum_{k=0}^\infty e^{(k+1)^2/2} [1 - F_1(k)] \\ &\leq C\left [\sum_{k=1}^\infty e^{-k} + \sum_{k=0}^\infty e^{-k}\right ] \leq \frac{2C}{1 - e^{-1}} < \infty \end{aligned} \end{equation*}

Therefore, the following integral exists for every z \in \mathbb C

\begin{equation} \left |\int_{-\infty}^\infty e^{izx} \mu_1[dx] \right| \leq e^{|z|^2/2} \underbrace{\int_{-\infty}^\infty e^{x^2/2} \mu_1[dx]}_{\text{denoted by } M} = Me^{|z|^2/2} \label{eqn:chf-bound} \end{equation}

because |e^z| \leq e^{|z|} and x^2 + |z|^2 \geq 2|xz|. Then, the characteristic function f_1 of X_1 can be defined on the complex plane \mathbb C.

By a similar proof for e^{x^2/2} \in L^1(\mu_1), we can show that |x|e^{x^2/2} \in L^1(\mu_1),

\begin{equation*} \begin{aligned} \int_\mathbb R |x|e^{x^2/2} \mu_1[dx] &\leq \sum_{k=1}^\infty \int_{-k}^{-k + 1}ke^{k^2/2} \mu_1[dx] + \sum_{k=0}^\infty \int_{k}^{k + 1} (k + 1)e^{(k + 1)^2/2} \mu_1[dx] \\ &\leq \sum_{k=1}^\infty ke^{k^2/2} F_1(-k + 1) + \sum_{k=0}^\infty (k + 1)e^{(k+1)^2/2} [1 - F_1(k)] \\ &\leq C\left [\sum_{k=1}^\infty ke^{-k} + \sum_{k=0}^\infty (k + 1)e^{-k}\right ] < \infty \end{aligned} \end{equation*}

Then, given R > 0, for every |z| < R

\begin{equation*} \left | \frac{\partial e^{izx}}{\partial z} \right | \leq |x| e^{x^2/2} e^{R^2/2} \in L^1(\mu_1) \end{equation*}

By dominated convergence theorem, f_1 is differentiable at z, and hence is an entire function.

Therefore, \mathbb E[|X_1|] < \infty and then \mathbb E[|X_2|] \leq \mathbb E[|X_1 + X_2|] + \mathbb E[|X_1|] < \infty. Let m = \mathbb E[X_1] and Y_1 = X_1 - m - 2, Y_2 = X_2 + m + 2. Then, Y_1 and Y_2 satisfy the condition of this theorem. Moreover, we know \mathbb P[Y_1 \leq -2] > 0 since \mathbb E[Y_1] = -2, and therefore equations (\ref{eqn:F1-neg-x}) and (\ref{eqn:F1-pos-x}) hold for F_{Y_2}. By a similar argument, we know the characteristic function of Y_2 is an entire function and then f_{X_2} = e^{iz(m + 2)}f_{Y_2} is also an entire function.

Let f_2 and f be the characteristic functions of X_2 and X_1 + X_2, respectively. Since X_1 + X_2 is normally distributed, f(z) \neq 0 for z \in \mathbb C. Then it follows from f_1(z)f_2(z) = f(z) that f_1 has no zeros on \mathbb C.

From equation (\ref{eqn:chf-bound}), f_1 has an order of growth \leq 2. By Hadamard's factorization theorem, f_1 takes the form f_1(z) = e^{P(z)}, where P(z) is a polynomial with degree \leq 2.

Since f_1(0) = 1, we know that P(0) = 0 and then P(z) = c_2z^2 + c_1z for some c_2, c_1 \in \mathbb C. And it follows from \overline{f_1(t)} = f_1(-t), \forall t \in \mathbb R that \overline{c_2}t^2 + \overline{c_1}t = c_2t^2 - c_1t, \forall t\in\mathbb R. Then, c_1 \in i\mathbb R and c_2 \in \mathbb R. Moreover, we know c_2 < 0 because f_1 is bounded in \mathbb R.

Then we can rewrite P(z) into P(z) = im_1z - \sigma_1^2z^2/2 for some m_1, \sigma_1 \in \mathbb R. And hence X_1 \sim \mathcal N(m_1, \sigma_1^2). Then, we have X_2 \sim \mathcal N(-m_1, 1 - \sigma_1^2).

References

Cramér, H. (1936). Über eine eigenschaft der normalen verteilungsfunktion. Mathematische Zeitschrift, 41(1):405–414.

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顺利从福州一中毕业!感觉大学周围都是聚聚十分可怕QAQ 想要联系的话欢迎发邮件:miskcoo [at] gmail [dot] com

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