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I have recently started studying probability. Kindly explain me following

Random variable means a function from $\Omega \to R$(set of real numbers). This is what i understood from my school books. But when i started reading Ross, it is given as in addition to above, any Borel set in R must have preimage in $\Omega$ i.e, preimage of Borel set in R must be an event.

Why this 2nd part is needed? Does it mean Random function(variable) X is "onto"

Another doubt, I thought random variable is some variable. But i understood now that it is infact a function. Then why it is called random "variable"?

If X, Y are 2 random variables, i know that XY is also random variable

My proof:

$X: \Omega \to R \\ Y: \Omega \to R $

then

$XY(\alpha) = X(Y(\alpha)), \alpha$ is some event in $\Omega$ sample space. Since composition of functions is a function from same domain $\Omega$ to R and so XY is a function and so it is a random variable. Please correct me if i am wrong.

StubbornAtom
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Nascimento de Cos
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    A random variable is defined as a measurable function etc. That means it needs the extra condition (a measurable function is one in which the preimage of every measurable set is itself measurable). It is known as a random variable for historical reasons. – Brian Tung Jul 06 '20 at 16:40
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    You can't compose two functions unless the range of one is the domain of the other. – Brian Tung Jul 06 '20 at 16:41
  • For completeness: a random variable, i.e. a measurable function $\Omega\to\mathbb R$ is not necessarily "onto". – drhab Jul 06 '20 at 16:43

2 Answers2

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The measure-theoretic restriction is to avoid pathological situations where $X \ge c$ can't be assigned a probability because the set $\{\alpha: X(\alpha) \ge c\}$ is not measurable.

Although in principle the random variable $X$ is a function on the sample space $\Omega$, in practice probabilists rarely think of it that way, in fact they often don't bother spelling out what the sample space is.

Your proof makes no sense because $X Y$ is not a composition, it is just the ordinary product of real functions: $(XY)(\alpha) = X(\alpha) Y(\alpha)$.

Robert Israel
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  • Sir if we are making $(c, +\infinity)$ as not measurable then why Borel set $(-\infinity , c)$ is measurable(we are assigning $(-\infintiy, c)$ to some event in sample space. ? then why can not we do the same thing by pre-mapping $(c, +\infinity)$ to some event. Pls correct my ignorance – Nascimento de Cos Jul 06 '20 at 17:19
  • I think you misunderstand me. In order for $X$ to be a random variable, all the sets $X^{-1}(A)$ where $A$ is a Borel set must be measurable. It suffices to consider the cases $A = [c, \infty)$ because the others can be built from these using complements, countable unions and countable intersections. For example, $(c,\infty) = \bigcup_{n=1}^\infty [c+1/n, \infty)$. – Robert Israel Jul 06 '20 at 19:52
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There is a common saying: "A random variable is neither random nor a variable." The terminology might have come from some intuition at a non-measure-theoretic level, but in the formal definition it is simply a [measurable] function.

This "measurability" (preimage of Borel set must be measurable in $\Omega$) is important because you would like to assign a probability to events of the form $\{X \in B\}$ for Borel sets $B$. This probability is assigned from a probability measure on $\Omega$. If the preimage $X^{-1}(B)$ is not measurable in $\Omega$, then you cannot assign a probability to that event.

$XY$ is defined as the product $X(\alpha) Y(\alpha)$, not the composition.

angryavian
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    I saw once in a book the quite funny sentence that "a random variable is neither random nor a variable" just like "the Holy Roman Empire was neither holy, nor Roman, nor an Empire" – Maximilian Janisch Jul 06 '20 at 16:49
  • Sir but product of 2 functions need not be a function right? say $f(x)=x, g(y)=y^2, xy \to f(x)g(x) , (x,y)=(2,3), (3,2), 23 \to 23^2=18, 32 \to 32^2 = 12, So f(x)*g(x) $ is not a function right? Pls correct me if wrong – Nascimento de Cos Jul 06 '20 at 16:50
  • Sir i do not know what is meaning of measurability? can you please explain me in layman terms? Further if every borel set in R, say $(-\inf , x)$ preimage is an event, Does it mean it is onto function? Am i missing something. Kindly elaborate – Nascimento de Cos Jul 06 '20 at 16:52
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    @NascimentodeCos The product of two functions from any set to a group $(G,\cdot)$ is always a function. In particular if $f,g:\Omega\to\mathbb R$, then $f\cdot g$ is also a function for all sets $\Omega$. In your case, if $f(x)=x$ and $g(y)=y^2$, then $(f\cdot g)(x)\overset{\text{Def.}}=f(x)\cdot g(x)=x^3$ for $x\in\mathbb R$. – Maximilian Janisch Jul 06 '20 at 16:53
  • @MaximilianJanisch understood now sir. Can you please have a look at above doubt?? – Nascimento de Cos Jul 06 '20 at 16:54
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    @NascimentodeCos The preimage of a Borel set $B$ under a measurable function $X:\Omega \to \mathbb{R}$ is $X^{-1}(B) = {\alpha \in \Omega : X(\alpha) \in B}$. This preimage is allowed to be empty, so $X$ need not be onto. e.g., if $B={3}$, then $X^{-1}({3})$ might be empty, in which case no value of $\alpha$ has $X(\alpha) = 3$ so $X$ is not onto. – angryavian Jul 06 '20 at 16:57
  • @angryavian Understood now. Sir can you pls explain what is measurability and measurable function , if possible, in layman terms? – Nascimento de Cos Jul 06 '20 at 17:01
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    @NascimentodeCos See here for measurable sets. A measurable function is a function such that the preimage of any measurable set under that function is also measurable. – Maximilian Janisch Jul 06 '20 at 17:05
  • @angryavian sir from that post i understood that for borel set we assign a number or something which is event in sample space and call this as measure (something like set to its cardinality as measure). If i am speaking correctly, then does it mean the same thing as mapping event to some real number(here random variable)? 1st part of defination of random variable seem same as 2nd part of defination of random variable. In that case cann't we exclude? THere is some intrinsic part i am missing here. Request your patience in understanding it further. – Nascimento de Cos Jul 06 '20 at 17:16
  • @MaximilianJanisch actually you want the preimages of Borel sets to be measurable, not necessarily all (Lebesgue) measurable sets. See e.g. my answer here – Robert Israel Jul 06 '20 at 20:01
  • @RobertIsrael I get you. Maybe my sentence was formulated a bit too short. (I mean your definition of measurable function depends also on the sigma algebra you choose on $\mathbb R$ .) – Maximilian Janisch Jul 06 '20 at 21:55