# normal distribution with infinite variance and infinite mean

If we have a normal distribution with mean $$\mu=f(x)$$ and variance $$\sigma^2$$ where $$\sigma=g(x)$$ having the form:

$$$$p(y)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}}$$$$ along with the following conditions $$$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=K \\ \lim_{x\to\infty}f(x)=\infty \\ \lim_{x\to\infty}g(x)=\infty$$$$ where $$K$$ is just a constant real number by taking the following limit $$$$\lim_{x\to\infty}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}}$$$$

what kind of distribution will I end up?

is this still a Normal? is there a way to represent it?

• It's worth pointing that $K=0$ by your definition, which isn't too hard to show. – Pepe Silvia Jun 17 at 2:35

## 1 Answer

You're talking about limits, so to re-frame your question, as $$x\rightarrow\infty$$ does the distribution $$\mathcal{N}(f(x),g(x)^2)$$ converge to another distribution on $$\mathbb{R}$$? The weakest converge we usually deal with is convergence in distribution; if these distributions converge then their CDFs $$F_x$$ converge to a CDF $$F$$ wherever $$F$$ is continuous. As $$F_x(y)=\Phi\left(\frac{y-f(x)}{g(x)}\right)\rightarrow\Phi(-K)$$ as $$x\rightarrow\infty$$. So $$F$$ needs to be constant and equal to $$\Phi(-K)$$ everywhere. As CDFs can't be constant, these distributions don't converge.

• So what if I really have some data that has extremely large variance and mean and has the shape of a bell just like Gaussian distribution then what should I do? I mean there should be a distribution that fits those right? – Jason Jun 17 at 8:43
• and one more thing when you take the limit of CDF you only look at x but we have another variable which is y and could also be infinite shouldn't you take it into consideration when you take the limit? – Jason Jun 19 at 15:21
• If your data has very large variance and mean but is still actually Gaussian, then it's just Gaussian and there's no real issue. If you want examples of distributions that are "bell shaped" but not Gaussian and that have infinite mean or variance look at the Cauchy distribution, or more generally distributions with pdf $\frac{c}{1+x^{2n}}$ for $n>0$ and some constant $c$ dependent on $n$. You can see that this distribution has finite $m$-th moment for $m<2n-1$. To answer your second question, no we look at fixed $y$, simply because that's what the definition of pointwise convergence is. – Pepe Silvia Jun 22 at 0:38