Is $\text{Unif}(- \infty, \infty ) $ a valid distribution?
I'm trying to capture the idea of a completely random number (where every number has an equal chance of being chosen), but I'm not sure if that idea can be captured using a valid statistical distribution.
I'm thinking there may be some way to do this via a mapping. For example, it looks like we can map numbers from the range $(-1, 3)$ to all of the reals via the function $f(x) = – \log (\frac{4}{x+1} – 1)$ which has a domain of $(-1, 3)$ and a range of $(-\infty, \infty)$.
So, could $\text{Unif}(-\infty, \infty)$ be defined using a mapping somehow from $\text{Unif}(-1, 3)$? If so, how would you define this mapping?
Best Answer
Uniform distribution has a finite range $-\infty < a < b < \infty$. It's probability density function is $p(x) = \frac{1}{b - a}$ for $x \in (a, b)$. If you set the range to infinities, you'd end up with $p(x) = \frac{1}{\infty } \to 0$. It doesn't integrate to unity since it would need to be $p(x) = c$ for some $c > 0$. For a finite support, as the support grows, $c \to 0$. If you think of it differently, for every $c > 0$ there would be an uniform distribution with a finite support with probability density function equal to $\tfrac{1}{b - a} = c$, so there cannot be such distribution with an infinite support. It's not a proper distribution.
As for your idea with mapping, notice that using such transformation would not lead to having uniformly distributed values. Try the following numerical experiment.
As you can see, the result looks nothing like a uniform distribution.