稳定分布证明问询：由分布卷积性质推导稳定分布

Alright, let's break this down step by step to link the given convolution closure property to the formal definition of a stable distribution.

Quick Definitions to Align On #

First, let's recap the key terms we're working with:

• Stable Distribution: A distribution $F$ is stable if for every positive integer $n$, there exist constants $a_n > 0$ and $b_n \in \mathbb{R}$ such that for independent random variables $X_1, \dots, X_n \sim F$, the transformed sum $\frac{1}{a_n}(X_1 + \dots + X_n) + b_n$ also follows $F$.
• Given Condition: For any positive $a, a'$ and real $b, b'$, there exist $a'' > 0$ and $b'' \in \mathbb{R}$ where the convolution of $F(ax + b)$ and $F(a'x + b')$ equals $F(a''x + b'')$. In probability terms, this means if $Y \sim F(ax + b)$ (i.e., $Y = aX + b$ for $X \sim F$) and $Z \sim F(a'x + b')$, then $Y + Z \sim F(a''x + b'')$ (so $Y + Z = a''W + b''$ for $W \sim F$).

Step 1: Base Case (n=2) #

Let's start with $n=2$, the smallest non-trivial case. Take two independent variables $X_1, X_2 \sim F$. Their sum $X_1 + X_2$ corresponds to the convolution of $F(x)$ with itself (since $a=a'=1$, $b=b'=0$ in the given condition).

By the given condition, there exist $a_2 > 0$ and $b'' \in \mathbb{R}$ such that:
$$F(x) * F(x) = F(a_2 x + b'')$$
This convolution describes the distribution of $X_1 + X_2$. Rearranging the right-hand side, we can write:
$$X_1 + X_2 = a_2 W - b''$$
where $W \sim F$. Divide both sides by $a_2$ and rearrange terms:
$$\frac{1}{a_2}(X_1 + X_2) + \frac{b''}{a_2} = W \sim F$$
If we set $b_2 = \frac{b''}{a_2}$, this exactly matches the stable distribution requirement for $n=2$.

Step 2: Inductive Step #

Now let's use mathematical induction to extend this to all positive integers $n$:

• Inductive Hypothesis: Assume for some $k \geq 2$, there exist $a_k > 0$ and $b_k \in \mathbb{R}$ such that:
  $$\frac{1}{a_k}(X_1 + \dots + X_k) + b_k \sim F$$
  Rearranging this, the sum $X_1 + \dots + X_k$ can be written as $a_k(Y - b_k)$ where $Y \sim F$.

• Prove for n=k+1: Consider the sum $X_1 + \dots + X_{k+1} = (X_1 + \dots + X_k) + X_{k+1}$. Substitute the inductive hypothesis:
  $$X_1 + \dots + X_{k+1} = a_k(Y - b_k) + X_{k+1} = a_k Y + X_{k+1} - a_k b_k$$
  Now, $a_k Y$ has distribution $F\left(\frac{x}{a_k}\right)$ (since $Y \sim F$, scaling by $a_k$ shifts the distribution function), and $X_{k+1} - a_k b_k$ has distribution $F(x + a_k b_k)$ (shifting $X_{k+1}$ by $-a_k b_k$ shifts the distribution function right by $a_k b_k$).

  Applying the given condition to these two distributions ($a = \frac{1}{a_k}$, $b=0$ for the first; $a'=1$, $b'=-a_k b_k$ for the second), there exist $a_{k+1} > 0$ and $b''' \in \mathbb{R}$ such that their convolution equals $F(a_{k+1} x + b''')$. This convolution describes the distribution of $a_k Y + X_{k+1} - a_k b_k$, so:
  $$X_1 + \dots + X_{k+1} = a_{k+1} W - b'''$$
  where $W \sim F$. Rearranging again:
  $$\frac{1}{a_{k+1}}(X_1 + \dots + X_{k+1}) + \frac{b'''}{a_{k+1}} = W \sim F$$
  Setting $b_{k+1} = \frac{b'''}{a_{k+1}}$, we satisfy the stable distribution requirement for $n=k+1$.

Step 3: Conclusion #

Since we've verified the base case ($n=2$) and shown that if the property holds for $n=k$, it holds for $n=k+1$, we can conclude that for every positive integer $n$, there exist $a_n > 0$ and $b_n \in \mathbb{R}$ such that $\frac{1}{a_n}(X_1 + \dots + X_n) + b_n \sim F$. This is exactly the definition of a stable distribution.

内容的提问来源于stack exchange，提问作者Squird37