3
$\begingroup$

I went through a derivation of the normal distribution probability density function here but was surprised how few assumptions are made in the derivation. In particular it seemed that any probability density function that was differentiable and integrable, symmetric about the mean, always greater than zero, and with finite variance (and possibly monotonically decreasing with increasing $|x|$) must always have this form:

\begin{equation} f(x;\mu,\sigma^2) = \frac{1}{\sigma\sqrt{2\pi}} e^{ -\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2 } \end{equation}

What am I missing here? Does the mere existence of a pdf with the properties I mentioned above imply that it describes a normal distribution?

$\endgroup$

1 Answer 1

3
$\begingroup$

This is clearly not the case, soncider e.g. the density that is proportional to $\exp(-x^4)$.

The important part is that the two-dimensional Gaussian distribution has the property that it has independent coordinates and the distribution is rotationally invariant. This does in fact characterise the Gaussian distribution uniquely (if we say that a constant random variable has distribution $\mathcal{N}(\mu, 0)$).

I couldn't find a simple proof for this general characterisation, but under the further assumption of finite variance it is relatively easily proven.

If $X$ has a finite variance and for every independent copy $Y$ of $X$, the distribution of $\frac{X + Y}{\sqrt{2}}$ is the same distibution as the distribution of $X$, then $X$ is normally distributed.

From our assumption we can conclude $E[X] = \frac{E[X] + E[Y]}{\sqrt{2}} = \sqrt{2} E[X]$, which means that $E[X] = 0$.

Let $X_1, X_2, \ldots$ be i.i.d. copies of $X$. Then by assumption we know that $\frac{X_1 + X_2}{\sqrt{2}}$ and $\frac{X_3 + X_4}{\sqrt{2}}$ have the same distribution as $X$. Since they are also independent, we can conclude that $X$ has the same distribution as $$\frac{\frac{X_1 + X_2}{\sqrt{2}} + \frac{X_3 + X_4}{\sqrt{2}}}{\sqrt{2}} = \frac{X_1 + X_2 + X_3 + X_4}{2}.$$

Now an inductive argument shows that $X$ has the same distribution as $2^{-n/2} \sum \limits_{i = 1}^{2^n} X_i$. But by the central limit theorem, this sequence converges in distribution to a normal distribution, which means that $X$ has a normal distribution.

Remark: $\frac{X + Y}{\sqrt{2}}$ is the first coordinate of the rotation of $(X, Y)$ by $45^\circ$, which connects this theorem to the above mentioned characterisation.

$\endgroup$
1
  • $\begingroup$ Brilliant, thank you very much! $\endgroup$
    – Dave Moten
    Commented Dec 7, 2016 at 9:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .