I need to mathematically define the probability space $(\Omega, \mathcal F, \mathbb P)$ of continuous random variable $X$. I also need to define the continuous random variable $X$ itself. Problem is… I don't really know how.
It is known that $X$ has the following probability density function $f_X: \mathbb{R} \longrightarrow \left[0, \frac{4}{9} \right]$:
$$f_X(x) = \begin{cases} \begin{align*} &\frac{1}{9}\big(3 + 2x – x^2 \big) \; &: 0
\leq x \leq 3 \\ &0 \; \; &: x < 0 \; \lor \; x > 3 \end{align*}\end{cases}$$
and its plot:
Also, the cumulative distribution function of $X$ is $F_X: \; \mathbb{R} \longrightarrow \left[0,1\right]$ and is defined as:
$$F_X(x) = \begin{cases} \begin{align*} &0 \; \; &: x < 0 \\ &\frac{1}{9} \Big(3x + x^2 – \frac{1}{3}x^3 \Big) \; \; &: x \geq 0 \; \land \; x \leq 3 \\ &1 \; \; &: x > 3 \end{align*}\end{cases}$$
and its plot:
(please see this thread where I calculated CDF for reference)
I suppose:
$$X: \Omega \longrightarrow \mathbb{R}$$
and sample space:
$$\Omega = \mathbb{R}$$
How can I define $\mathcal F$ and $\mathbb{P}$, that are the quantities of probability space $(\Omega, \mathcal F, \mathbb{P})$? I was thinking:
$$\mathbb{P} : \mathcal F \longrightarrow \left[0, 1\right] \; \land \; \mathbb{P}(\Omega) = 1$$
I am jumping into statistics/probability and I am lacking the theoretical knowledge. Truth be speaking, the wikipedia definition of probability space for continuous random variable is too difficult to grasp for me.
Thanks!
Best Answer
The usual way to define a probability space on which a single random variable (continuous or not) is defined is to take $\Omega=[0,1]$, $\mathcal{F}$ to be either the Borel or Lebesgue $\sigma$-algebra on $[0,1]$, and $\mathbb{P}$ to be the Lebesgue measure restricted to $\mathcal{F}$. Then given a CDF $F$, one can define the so-called quantile function $Q(y)=\inf \{ x : F(x) \geq y \}$, and then $X(\omega)=Q(\omega)$ has CDF $F$. This technique is sometimes called the "probability integral transformation".
Note that one can also use $(0,1)$ which has the advantage that $X$ will always be finite valued, rather than merely be almost surely finite-valued.