# Carmér‘s Theorem

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

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

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

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

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{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}

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 $x > 0$

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

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 $z \in \mathbb C$

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

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.

Miskcoo's Space，版权所有丨如未注明，均为原创

## 2 thoughts on “Carmér‘s Theorem”

1. Mochizuki_Mk19 says:

原始论文我去sci-hub大致浏览了一下，原文分为123三个部分，1部分中对标准正态分布的情况作出了简要证明，在2部分中因为非标准正态分布的参数经过变换可以变成标准正态分布的参数，因此如果原文(1)式有非一般形式的解，那么可以通过参数变换证明其满足(1a)式。3部分则是借助勒贝格积分将原结论推广到了n维欧氏空间上。

……没上过概率论课看这个还是有点吃力，如果有具体哪里的关键说明需要翻译的话可以找我（

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