What does determine if a distribution is Gaussian?

Question

If the probability density function is unknown, having a mean $\mu$ and a variance $\sigma^2$ doesn't mean a distribution must be Gaussian.

Answer 1

If you have a random variable $X$ that has a mean and variance, this doesn’t uniquely identify it’s distribution. Only certain properties. For example, if you know that the mean $\mu < \infty$ and that $\sigma^2 < \infty$ then this would imply that you don’t have a Cauchy random variable, but lots of distributions have a finite mean and variance.

If you introduce charactaristic functions into the situtation,

$$E[e^{itX}]$$

Then you can use The Uniqueness Theorem of charactaristic functions which says that if you have a charactaristic function that comes out to be

$$e^{it\mu - \frac{1}{2}t^2\sigma^2}$$

For $\mu \in \mathbb{R}$ and $\sigma \geq 0$, Then $X$ has the pdf of a $N(\mu, \sigma^2)$.

The more general way to say this in measure theory is if you have two Fourier transforms of probability measures (charactaristic functions) $\hat{\mu}$ and $\hat{\nu}$ and you find that $\hat{\mu}=\hat{\nu}$ then $\mu = \nu$.

Where $\mu$ and $\nu$ are two finite measures on $\mathbb{R}$, othewise known as pdfs in probability language.

So to summarize, charactaristic functions equal $\to$ pdfs equal.

NOTE: this doesn’t translate to moment generating functions, not all random variables have moment generating functions but they all have charactaristic functions.