$$X_{1}\sim \mathcal{Uniform}([0,2])$$ $$X_{2}\sim \mathcal{Uniform}([1,2])$$

$X_{1}$ and $X_{2}$ are independent random variables.

My question is whether joint distribution of those variables is: $$f(x_{1},x_{2})=f_{1}(x_{1})\,f_2(x_{2})=1/2$$

... for all $(x_1, x_2) \in\ [0,2]{\times}[1,2]$?

If answer is incorrect, then how to approach this issue?

If answer is correct, then I am thinking wheter exist more rigorous and formal approach to derive this?


Short answer: Yes, you are correct.

But you asked for rigor... I'll try to make the rationale rigorous for your edification :P Hopefully I don't end up confusing you, but rather send you on an adventure to learn more formal probability theory.

Let $(\Omega, \mathscr{F}, P)$ be our underlying probability space (meaning all random variables we discuss here are assumed to be $\mathscr{F}$-measurable functions of $\omega \in \Omega$).

Consider the following random variable $X: \Omega \to \mathbb{R}^2$, $$X = \begin{bmatrix}X_1 \\ X_2\end{bmatrix}$$

Notice that the components of $X$ are also random variables, $X_1: \Omega \to \mathbb{R}$ and $X_2: \Omega \to \mathbb{R}$.

Let the probability density function (PDF) of $X$ be called $f(x) = f(x_1,x_2)$, and let the PDFs of its components $X_1$ and $X_2$ be called $f_1(x_1)$ and $f_2(x_2)$ respectively. We define a conditional PDF as, $$f_1(x_1 | x_2) := \frac{f(x_1, x_2)}{f_2(x_2)}$$ When we say "$X_1$ and $X_2$ are independent" we strictly mean, $$f_1(x_1 | x_2) \equiv f_1(x_1)$$

So like you said, $$f(x_1, x_2) = f_1(x_1)f_2(x_2)$$

If $f_1(x_1)$ is uniform on $[0,2]$ then, $$f_1(x_1) := \begin{cases} \frac{1}{2}, & x_1 \in [0,2] \\ 0, & \text{else} \end{cases}$$ and similarly, $f_2(x_2)$ uniform on $[1,2]$ means, $$f_2(x_2) := \begin{cases} 1, & x_2 \in [1,2] \\ 0, & \text{else} \end{cases}$$

So we must have, $$f(x_1,x_2) = f_1(x_1)f_2(x_2) = \begin{cases} (\frac{1}{2})(1), & x_1 \in [0,2]\ \cap\ x_2 \in [1,2] \\ (\frac{1}{2})(0), & x_1 \in [0,2]\ \cap\ x_2 \not\in [1,2] \\ (0)(1), & x_1 \not\in [0,2]\ \cap\ x_2 \in [1,2] \\ (0)(0), & x_1 \not\in [0,2]\ \cap\ x_2 \not\in [1,2] \\ \end{cases}$$

which can be expressed more simply as, $$f(x_1,x_2) = \begin{cases} \frac{1}{2}, & x_1, x_2 \in [0,2]\times[1,2] \\ 0, & \text{else} \end{cases}$$

It is good to verify that your result is indeed a valid PDF, $$\iint_{\mathbb{R}^2}f(x)dx = \int_1^2 \int_0^2\frac{1}{2}dx_1dx_2 = 1$$

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Jyrki Lahtonen Apr 7 '17 at 6:09

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