Probability distribution such that the cumulative function is constant: CDF$=c$

Question

I am trying to find a probability distribution such that $\forall x\in\mathbb R$, $CDF(x)=\frac{1}{2}$.

One obviously answer is the improper uniform distribution over $\mathbb R$.

My question is: are there any classical probability distributions that satisfy the requirement? Are there any other well-defined improper uniform distributions, possibly defined by generalized functions, that satisfy this requirement?

Intuitively, since the CDF is a constant, then the PDF must be something like constantly infinitesimal. Therefore, the uniform distribution seems to be the only option.

Answer 1

The cdfs $F$ of all probability distributions on the reals have the property that $\lim_{x\to-\infty} F(x) = 0$ and $\lim_{x\to\infty}F(x)=1$. This reflects the idea that real random variables $X$ are actually finite, that is, $P(|X|<\infty)=1$, which is the same as $\lim_{n\to\infty}P(|X|>n) = 0$. One could image an extended real line $\overline{\mathbb R}$ obtained by adding the two points $\pm\infty$ to $\mathbb R$; then your $F(x)=1/2$ cdf would correspond to an $\overline{\mathbb R}$-valued random variable taking the values $-\infty$ and $+\infty$ with probabilities $c$ and $1-c$, respectively.

Your discussion of uniform distributions makes me think you might be intermittently confusing distribution functions and density functions, a bit. The graph of the cdf of a uniform distribution has a constant slope.

Answer 2

The CDF you have provided isn't a valid distribution function. It violates the condition that

$$\lim_{x \to \infty} F(x) = 1$$

It also violates the condition that:

$$\lim_{x \to -\infty} F(x) = 0$$