Continuous Random Variable

Definition We say a real function $F$ is absolutely continuous if there exists real function $f$ such that $\forall_a \forall_x \ \ F(x) = f(a) + \int_a^x f(t) \mathrm{d}t$

Definition We say a random variable $\mathbb{X} \sim F$ is continuous if its c.d.f $F$ is absolutely continuous

Definition The function $f$ such that $\forall_a \forall_x \ \ F(x) = f(a) + \int_a^x f(t) \mathrm{d}t$ is called the probability density function (p.d.f) of $\mathbb{X}$.

Now let's take a look at the probability mass function of $\mathbb{\mathbb{X}}$.

For any number $a\in\mathbb{R}$, let $A_n$ be the half open interval $(a-\frac{1}{n}, a]$ $\mathbb{P}(\mathbb{X}=a) = \mathbb{P}( \mathbb{X} = \cap_{n=1}^\infty A_n ) \text{ (because } a = \cap_{n=1}^\infty A_n )$ $= \mathbb{P}(\cap_{n=1}^\infty\{\mathbb{X} = A_n \}) = \cap_{n=1}^\infty\mathbb{P}(\mathbb{X} = A_n )$ where the last equality comes from the continuity of probability function on monotonic sequence of sets. Now $\cap_{n=1}^\infty\mathbb{P}(\mathbb{X} = A_n ) = \cap_{n=1}^\infty F(a) - F(a-\frac{1}{n}) = \lim_{n\to\infty} F(a) - F(a-\frac{1}{n})$

If $\mathbb{X}$ is continuous r.v, then $\mathbb{P}(\mathbb{X}=a) = \lim_{n\to\infty} F(a) - F(a-\frac{1}{n}) = 0$ because $F$ is continuous.

Therefore, p.m.f of a continuous r.v is meaningless, because it's p.m.f is 0 every where. So p.d.f is what we need to capture the spirit of $\mathbb{P}\{\mathbb{X}=x\}$, even though $\mathbb{P}\{\mathbb{X}=x\}$ is $0$. \textbf{Remark}

1. When something has a probability 0, it doesn't mean it won't happen! As we just see, any p.m.f of a contunuous random variable is 0.

2. Given a c.d.f $F$, its corresponding p.d.f $f$, is not unique. However, it rarely matters because the results we need generally come from imtergrating $f$, and even if there are differen $f$'s, the result after intergrating is always the same. Why? To answer this you will have to take some courses in measure theory.