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$X=(X_1,X_2,\ldots, X_P)$ is a $p$-dimensional random variable on $(\Omega, S, P) $ iff $X_i$'s are univariate random variables on the same probability space $(\Omega, S, P)$ ."

We all know that. it is written and proved in several books.

But I'm confused with the 'same probability space' thing.

Let me explain my problem with an example : Suppose $X_1$= the height of a boy selected at random from a class of 50 boys. $X_2$ =the weight of the same boy selected at random from a class of 50 boys.

Now in both the cases the sample space $\Omega = (-\infty,+\infty)$ . but the $P$ in the probability space $(\Omega, S, P)$ for both the random variable must be different . let's consider $X_1 \sim \mathrm{N}(5, 1)$ and $X_2\sim \mathrm{N}(45, 4)$. Then $X_1$ and $X_2$ are not defined on same probability space. (or , are they ? I'm kinda confused)

So , $X = ( X_1, X_2)$ , is it not a bi-variate random variable?

[Our professor was talking about constructing multivariate random variable taking different bio-metric measures of an individual . But, if, all the univariate random variables (components of a multivariate random variable) to be defined on the same probability space be a necessary and sufficient condition, then constructing a multivariate random variable taking different bio-metric measures must not be possible.]

This is my first question here, hoping for good responses.Although I don't know whether I'm able to put my query in a compact form. If you find any ambiguity kindly ask.

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    $\begingroup$ Please refrain from adding personal comments such as "Although i don't know whether I'm able to put my query is a compact form. if you find any ambiguity , kindly ask me.", as it has nothing to do with the question. Mentioning "Thanks!" etc. are discouraged here; if you find something helpful, then upvote and/ or accept it. $\endgroup$
    – hola
    Aug 10, 2014 at 7:56
  • $\begingroup$ okay..!! I got it. $\endgroup$
    – saudade
    Aug 10, 2014 at 7:58
  • $\begingroup$ actually, the query is a little complicated , and people may find it ambiguous . that's why , i mentioned it. $\endgroup$
    – saudade
    Aug 10, 2014 at 7:59
  • $\begingroup$ Just an indomitable curiosity: Which college/ university do you read in? I guess you are a B.Sc. student from Kolkata. $\endgroup$
    – hola
    Aug 10, 2014 at 8:03
  • $\begingroup$ "We all know that. It is written and proved in several books." However, if you were familiar with the said construction of appropriate probability space, there wouldn't be any confusion. If you want, I can outline how it's done in an answer $\endgroup$
    – Alen
    Aug 10, 2014 at 8:10

2 Answers 2

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First, the text below requires that you know definitions of a real random variable (or vector), measurable function, Borel $\sigma $-algebra and cumulative distribution function.

Let $X$ be an $n$-dimensional random vector with the CDF ${F}$ (this is a fundamental assumption which begs the question does the appropriate probability space and the random vector exist, we proceed to show that they do).

Then the relation ${F}\left( x \right) = {\mathbb{P}}\left( {X \leqslant x} \right)$ (here $X \leqslant x \Leftrightarrow {X_i} \leqslant {x_i},\forall i$) uniquely determines a probability $\mathbb{P}$ on ${\mathbb{R}^n}$ (this is non-trivial result and the proof is cumbersome). Let's first construct the underlying probability space.

Let $\Omega = {\mathbb{R}^n}$, $\mathcal{F} = {\mathcal{B}^n}$ the Borel $\sigma $-algebra on ${\mathbb{R}^n}$ and $\mathbb{P}$ as above. Now we define the random vector $X:{\mathbb{R}^n} \to {\mathbb{R}^n},X\left( x \right) = x$. It follows, using the standard notation ${\mathbb{P}_X}\left( B \right) = \mathbb{P}\left( {{X^{ - 1}}\left( B \right)} \right)$ for the probability ${\mathbb{P}_X}$ induced by $X$ on ${\mathbb{R}^n}$, that ${F_X}\left( x \right) = {\mathbb{P}_X}\left( {\left( { - \infty } \right.,\left. x \right]} \right) = \mathbb{P}\left( {{X^{ - 1}}\left( { - \infty } \right.,\left. x \right]} \right) = \mathbb{P}\left( {\left( { - \infty } \right.,\left. x \right]} \right) = F\left( x \right)$, so $\left( {\Omega ,\mathcal{F},\mathbb{P}} \right)$ and $X$ are indeed such a probability space and a random variable that the given $F$ is the corresponding cumulative distribution function.

Now on to your specific example: ${X_i}$ is a random variable on $\left( {\Omega ,\mathcal{F},\mathbb{P}} \right)$ because $X_i^{ - 1}\left( {{B_i}} \right) = {\left( {{\pi _i}\left( X \right)} \right)^{ - 1}}\left( {{B_i}} \right) = {X^{ - 1}}\left( {\pi _i^{ - 1}\left( {{B_i}} \right)} \right)$, where ${{\pi _i}}$ is $i$-th coordinate projection and ${B_i} \in \mathcal{B}\left( \mathbb{R} \right)$. Since projections are continuous, they are measurable, so $\pi _i^{ - 1}\left( {{B_i}} \right) \in {\mathcal{B}^n}$ (it's helpful to imagine this Borel $\sigma $ algebra as a $\sigma $ algebra in the codomain of $X$) which implies that ${X^{ - 1}}\left( {\pi _i^{ - 1}\left( {{B_i}} \right)} \right) \in \mathcal{F}$, since $X$ is a random vector.

So, if ${X_1}$ is the height and ${X_2}$ the weight of an individual, then $\Omega = {\mathbb{R}^2}$ (and not $\Omega = \mathbb{R}$ as you assumed) and $X_1^{ - 1}\left( {\left[ {a,b} \right]} \right) = {X^{ - 1}}\left( {\left[ {a,b} \right] \times \mathbb{R}} \right) = \left[ {a,b} \right] \times \mathbb{R} \in {\mathcal{B}^2} = \mathcal{F}$. Also, ${X_1},{X_2}:\Omega \to \mathbb{R}$ and $X = \left( {{X_1},{X_2}} \right):\Omega \to {\mathbb{R}^2}$, so ${X_1},{X_2},X$ have the same domain.

The reason why you CAN forget about the underlying probability space is exactly the construction given above which guarantees that such a space always exists. However, I've found that it's sometimes helpful to be aware of it.

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  • $\begingroup$ hey , i have understood it. thank ya. a lot...!! it was lucid and neat. $\endgroup$
    – saudade
    Aug 11, 2014 at 6:27
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The answer to your question is a matter of interpretation. If you formalize the sample spaces abstractly enough, then the sample spaces for $X_1$ and $X_2$ can be the same. After all, both weight and height are continuous quantities, and so can be modeled as real numbers. If the real numbers aren't a general enough setting, you can go into more abstract spaces like "Polish spaces".

The real answer is that you typically don't want to lose the "typing" information (what kind of number does your variable represent). That is certainly a worthy goal. And to do that, probablists and statisticians try to forget about the sample space as quickly as possible. In that sense, the definition in your book is too specific, and only corresponds to the one real people use "in principle".

The point about probability measures is similar enough.

Your primary focus "should" be the random variables and their properties, not the sample spaces they are defined on.

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  • $\begingroup$ i appreciate your response. but , somehow it doesn't answer question properly. $\endgroup$
    – saudade
    Aug 10, 2014 at 8:10
  • $\begingroup$ In what way? The question of whether $(X_1, X_2)$ is an RV is only a problem if you use an artificially restrictive definition of an RV. $\endgroup$
    – nomen
    Aug 10, 2014 at 14:22
  • $\begingroup$ what is 'an artificially restrictive definition' and what is 'not an artificially restrictive definition'? a definition is a definition. we choose to forget about the restriction of the definition , because it's artificial , that can't be any answer. anyway , a sensible answer has been submitted . you can check it above. $\endgroup$
    – saudade
    Aug 11, 2014 at 6:29
  • $\begingroup$ You are confused. "a definition is a definition" is wrong. In particular, the definition you are using for your probability space is unsuitable for defining $(X_1, X_2)$ as an RV. So you must use a definition of random variables that defines, or at least includes, $(X_1, X_2)$ as an RV. Note that the answer you are happy with assumes that you already have the definition of a random vector. $\endgroup$
    – nomen
    Aug 11, 2014 at 16:19
  • $\begingroup$ sure, we already have a definition of a random vector . almost all the popular books have defines it in one same way. so , what's wrong with that? ' a definition is a definition is wrong' ! ok , then define 'definition'. $\endgroup$
    – saudade
    Aug 13, 2014 at 7:32

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