Can we characterize the family of normal (Gaussian) distributions by closedness of linear combinations?

Question

It is well known that, let's focus on $\mathbb{R}^1$ for now, the normal distribution $N(\mu, \sigma^2)$ has the following property:

If $X\sim \mathcal N(\mu_1, \sigma_1^2)$, $Y \sim \mathcal N (\mu_2, \sigma_2^2)$ are two independent normal variables, then $$ aX+bY \sim \mathcal N(a\mu_1 + b\mu_2, a^2\sigma_1^2 + b^2\sigma_2^2). $$ which is also a normal distribution.

So my question is: Is the inverse true? If a family probability distribution on $\mathbb{R}^1$ is

• parameterized by its expectation $\mu$ and variance $\sigma^2$, i.e. for each $\mu \in \mathbb{R}$ and $\sigma^2 \in \mathbb{R}^{+}$, then there is exactly one distribution in this family with expectation $\mu$ and variance $\sigma^2$
• closed under independent addition, i.e. if both $X$, $Y$'s distributions are in this family, and $X, Y$ are independent, then $X+Y$ is in this family
• closed under scalar multiplication, i.e. if $X$ is in this family, then for $a \in \mathbb{R}\backslash\{0\}$, then $a X$ is also in this family

then can we claim that this family of distributions is exactly the family of all normal distributions on $\mathbb{R}^1$?

(Maybe we will have to generalize the normal distribution to include the case when $\sigma^2 = 0$, i.e. a distribution with probability 1 to be $\mu$, or equivalently, we can say it is a "delta distribution at $\mu$")

Answer 1

The central limit theorem does the job. (At least, if we include the variance-$0$ distributions in there, allowing us to add a constant.)

Let $X_1, X_2, \dots, X_n$ be $n$ independent copies of the distribution from this family with mean $0$ and variance $1$. Then the expression $$Z_n = \frac{X_1 + X_2 + \dots + X_n}{\sqrt n}$$ also has mean $0$ and variance $1$, so by assumption, it has the same distribution (for all $n$). By the central limit theorem, however, it converges to the standard normal distribution as $n \to \infty$.

The only way that the sequence $Z_1, Z_2, Z_3, \dots$ of identically distributed variables can converge to the standard normal distribution is if their distribution is already the standard normal. From there, scaling by $\sigma$ and shifting by $\mu$ proves that any other distribution in the family is also normal.

It's inelegant to have to include the $\sigma=0$ case just to be able to shift over the $Z$'s at the end. Maybe we can argue as follows:

• Let $X$ be a random variable from this family with mean $\mu$ and variance $\epsilon$.
• Let $Y$ be a random variable from this family with mean $0$ and variance $\sigma-\epsilon$; by the argument above, $Y$ is normally distributed.

Then if $X$ and $Y$ are independent, $X+Y$ has the distribution from this family with mean $\mu$ and variance $\sigma$. However, as $\epsilon \to 0$, the CDF of $X+Y$ should approach the CDF of $\mu+Y$ (using, for example, Chebyshev's inequality to say that $X$ cannot be too far from $\mu$). This means that $X+Y$ should also be normal.