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Let $X$ be a random variable on a probability space $(\Omega,\Sigma,\mathbb{P})$.

Let $F_X$ denote the probability distribution function of $X$ given by:

$$F_X(y) = P(X\leq y) \text{ for } y \in \mathbb{R}$$

Then, I came across the following:

Almost everywhere differentiability of $F_X$ does not imply that the probability density function of $X$, denoted by $f_X$, exists. For example, if $X$ is a random variable such that $P(X = \frac{1}{2}) = 1$, then $F_X$ is differentiable almost everywhere but $f_X$ does not exist.

What is the reason that $f_X$ does not exist for $X$ such that $P(X = \frac{1}{2}) = 1$?

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2 Answers 2

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If the density of $X$ is $f$, then $$ \int_{(-\infty,1/2)\cup(1/2,+\infty)} f(x)\,dx = 0 $$ and $$ \int_{\{1/2\}} f(x)\,dx = 1. \tag 1 $$ The problem is the that integral of a function over and interval of length $0$ cannot be anything but $0$, but $(1)$ says it must be $1$.

Even if $F_X$ is continuous everywhere and differentiable almost everywhere, there may be no density. The Cantor distribution is a standard example of that. One way to describe the Cantor distribution is that it's the distribution whose base-$3$ expansion is $0.d_1 d_2 d_3 \ldots$ where $d_i$, $d_i=1,2,3,\ldots$ is $0$ or $2$, each with probability $1/2$, and these digits are independent.

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  • $\begingroup$ @hermes He uses this to derive a contradiction. $\endgroup$
    – Eric Auld
    Commented Aug 15, 2015 at 6:01
  • $\begingroup$ Actually, probability is separate on discrete and continuous case. On continuous case probability is always $0$ on single point. $\endgroup$ Commented Aug 15, 2015 at 6:13
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It would have to be a function such that $\int_0^a f_X = 0$ for all $a<1$ and $\int_0^a f_X = 1$ for all $a\geq 1$. No such function exists (only the Dirac delta "function")

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