Probability – Is a Uniformly Random Number Over the Real Line a Valid Distribution?

distributionsprobabilityuniform distribution

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.

x <- runif(1e6, -1, 3)
y <- -log(4/(x+1) - 1)
hist(y, breaks=100)

As you can see, the result looks nothing like a uniform distribution.

Histogram showing a distribution with a peak at zero and exponentially decaying tails on both sides.

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