# Is a collection of random variables always a random vector?

Let $X_1, \ldots, X_n$ be a collection of $n$ random variables with the same sample space $\Omega$, the same $\sigma$-algebra $\mathcal{F}$ but not identically distributed, i.e., $P(X_1 = \omega)$ is not necessarily the same as say $P(X_2 = \omega)$.

Wikipedia defines a random vector as

[...] a column vector $\mathbf{X} = (X_1, \ldots, X_n)^T$ [...] whose components are scalar-valued random variables on the same probability space $(\Omega, \mathcal{F}, P)$ [...].

My question is whether the collection of $n$ random variables described in the first paragraph is a random vector or not. If not, how would you call it? An indexed family of random variables perhaps?

## 1 Answer

Let us first recall the definition of random variable $X$ defined on probability space $(\Omega,\mathcal F,P)$ and with values from some measurable space $(S,\mathcal S)$; it is $(\mathcal F, \mathcal S)$-measurable function $$X: \Omega \to S$$

For definition of measurable function, see, for example, this wikipedia article.

The wikipedia page you quote say that random vector $X=(X_1,\ldots,X_n)$ is function $$X: \Omega \to S^n$$ where $X_1,\ldots,X_n$ are random random variables defined on the same probability space $(\Omega,\mathcal F,P)$ with values from measurable space $(S,\mathcal S)$, defined as $$X(\omega)=(X_1(\omega),\ldots,X_n(\omega)).$$ This definition doesn't even mention the distributions of $X_1,\ldots,X_n$; in particular, it does not say that they have to be the same. Therefore, the function above is a random vector for any random variables $X_1,\ldots,X_n$.

Actually, random vector $X$ is simply any random variable taking values from measurable space $(S^n,\mathcal S^n)$, where $\mathcal S^n$ is product $\sigma$-algebra, since $(\mathcal F,\mathcal S)$-measurability of it's components $X_1,\ldots,X_n$ is necessary and sufficient condition for $(\mathcal F,\mathcal S^n)$-measurability of $(X_1,\ldots,X_n)$.

• Alright, so the fact that $P(X_1 = \omega) \neq P(X_2 = \omega)$ doesn't imply that the $P$ in $(\Omega, \mathcal{F}, P)$ is different for $X_1$ and $X_2$? – Cromack Apr 15 '16 at 13:43
• $P$ in $(\Omega,\mathcal F,P)$ is measure on $\mathcal F$. Distributions of random variable $X_1$ and $X_2$ are measures $P \circ X_1^{-1}$ and $P \circ X_2^{-1}$ on $\mathcal S$. The probabilites $P(X_1 = \omega)=P(X_1^{-1}(\{\omega\})$ and $P(X_2=\omega)=P(X_2^{-1}(\{\omega\})$ can be different (if $X_1 \neq X_2$) even though measure $P$ in both equations is the same. – Zoran Loncarevic Apr 15 '16 at 13:53
• You need to understand the concept of random variable as measurable function. – Zoran Loncarevic Apr 15 '16 at 13:57
• I am sorry I dragged measure theory into this, but either you take for granted that $(X_1,\ldots,X_n)$ is random vector for any random variables $X_1,\ldots,X_n$, or you need some measure theory to justify that. – Zoran Loncarevic Apr 15 '16 at 14:01
• No need to be sorry :) In fact, that makes me want to learn more about measure theory. For now, I will take for granted that $(X_1, \ldots, X_n)$ is a random vector for any collection of random variables, but I will try to understand the reason why. Thanks, Zoran! – Cromack Apr 15 '16 at 16:48