Let $X = (X_1, \ldots, X_n)$ be a random vector with i.i.d. components. I am trying to understand the following expression from my probability textbook:

$\displaystyle P(\mathbf{X} \in B) = P(X_1 \in B_1, \, X_2 \in B_2, \ldots, X_n \in B_n) =$ $\displaystyle = \prod_{i=1}^n P(X_i \in B_i), \qquad \forall B = B_1 \times \ldots \times B_n \in \mathcal{B}(\mathbb{R}^n), \qquad \qquad \qquad \qquad \qquad \qquad \quad ~~~(1)$
where $B_i \in \mathcal{B}(\mathbb{R}), ~ \forall i$.

I have two questions about this expression.

I think that we should use the same symbol $P$ in all places of this expression because random vector $\mathbf{X}$ and random variable $X_i$ have the same domain probability space $(\Omega, \mathcal{F}, P)$ (by definition of a random vector).

1) Does this imply that event $\{\omega: X_i(\omega) \in B_i\}$ is a shorthand notation for event $\{ \omega: X_1(\omega) \in \mathbb{R} , \ldots , X_{i-1}(\omega) \in \mathbb{R} , X_i(\omega) \in B_i , X_{i+1}(\omega) \in \mathbb{R}, \ldots, X_n(\omega) \in \mathbb{R} \}$ ?

If no, I don't understand which measure $P$ is used in expression $\prod_{i=1}^n P(X_i \in B_i)$.

2) Random vector $\mathbf{X}$ induces probability space $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n), \mathcal{P}_{\mathbf{X}})$ and r.v. $X_i$ induces probability space $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mathrm{P}_{X})$.
How to rewrite expression (1) in terms of induced measures (i.e. joint distribution $\mathcal{P}_{\mathbf{X}}$ and marginal distribution $\mathrm{P}_X$)?

I got the following result but I'm not sure: $$\mathcal{P}_{\mathbf{X}}(B) = \prod_{i=1}^n \mathrm{P}_X(B_i), \qquad \forall B = B_1 \times \ldots \times B_n \in \mathcal{B}(\mathbb{R}^n)$$


1 Answer 1

  1. No, the product will range over $i$, and the expression $(X_i\in B_i)$ is a shorthand for $\{\omega:X_i(\omega)\in B_i\}$, and $(X_1\in B_1,\dots,X_n\in B_n)$ is a shorthand for the set you wrote up.
    As you observed, each $X_i$ and $\bf X$ has the same domain $\Omega$, so the above set with fixed $i$ makes perfect sense.

    And by independency, we also have $$P({\bf X}\in{\bf B}) =\prod_iP(X_i\in B_i) $$

  2. Yes, it seems correct.
  • 1
    $\begingroup$ Thanks, I think I understand now. Events $\{\omega: \mathbf{X}(\omega) \in B\}$, $\{ \omega: X_i(\omega) \in B_i\}$, $\cap_{i=1}^n \{\omega: X_i(\omega) \in B_i \}$ belong to the same sigma-algebra $\mathcal{F}$ so we can take the same probability measure $P$ on them. $\endgroup$
    – Rodvi
    May 27, 2018 at 10:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .