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We all know that if we have a variable x, then there is a meaning to - for example - $$y=e^x$$.

And we all know how to manipulate that algebraically and to do calculus. For example, if

$$y_1=e^{x_1}$$

and

$$y_2=e^{x_2}$$

then

$$y_1y_2 = e^{x_1+x_2}$$

But random variables are a very different beast. Ultimately we all know intuitively what we mean when we draw a PDF or, more accurately, the CDF which is itself an algebraic function mapping $[-\infty, \infty]$ to $[0,1]$ as part of that defines a random variable.

But what is the actual math that makes that definition of an RV actually act algebraically, so that, as above, we can happily write things like

$$Y_1=e^{X_1}$$

and

$$Y_2=e^{X_2}$$

then

$$Y_1Y_2 = e^{X_1+X_2}$$

where $X_1$ and $X_2$ are random variables - and get away with it???

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  • $\begingroup$ If you are worried about measurability issues, there are basic theorems that say the product of two measurable functions is measurable, and so on. Otherwise, the fact $Y_1(\omega)Y_2(\omega)=e^{X_1(\omega)+X_2(\omega)}$ for all $\omega$ follows by substituting the definitions of $Y_i(\omega)$ (as the answers below indicate) and the PDF/CDF has nothing to do with it...this is a more basic fact about substitution. $\endgroup$ – Michael May 22 at 2:22
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Formally $X_i$ are real valued (measurable) functions i.e $X_i\colon \Omega\to \mathbb{R}$ for $i=1,2$ where $\Omega$ is the sample space. So $$ Y_i=e^{X_i} $$ for $i=1,2$ as an equality of functions means that $$ Y_i(\omega)=e^{X_i(\omega)} $$ for all $\omega\in \Omega$. Hence since $X_i(\omega)$ is a real number it follows using the first part of your question that $$ Y_1(\omega)Y_2(\omega)=e^{X_1(\omega)}e^{X_2(\omega)}=e^{X_1(\omega)+X_2(\omega)} $$ for all $\omega$ whence as functions $$ Y_1Y_2=e^{X_1+X_2}. $$

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  • $\begingroup$ It’s probably worth pointing out that one can also consider equality in distribution, i.e. $Y$ is a random variable with distribution equal to the distribution of $e^X$, as well as almost sure equality. $\endgroup$ – icurays1 May 22 at 0:26
  • $\begingroup$ Good answer. but why do you say Omega maps to R, when, it seems, the mappimg is always to [0,1]? $\endgroup$ – eSurfsnake May 22 at 4:27
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    $\begingroup$ @eSurfsnake: Random variables need not take values in the interval $[0,1]$, the values can be anything. You might be confusing it with probabilites, which must of course be in the interval $[0,1]$. An event is a (measurable) subset of $\Omega$, and the probability of an event is its measure, which is at most $1$, since a probability space $\Omega$ by definition has measure $1$. $\endgroup$ – Hans Lundmark May 22 at 5:57
  • $\begingroup$ yes, to me an RV is tightly bound with a probability - otherwise, what is the point of probability theory? But, am I right to assume that the abstract concept of a measurable function means precisely that if I have and RV, X, with its CDF, then a "measurable function" g(X) maintains certain logical relations; i.e., that a domain of measure $\mu$ on X maps to a range of meaningful measure such that no contradictions occur between X and g(x)? $\endgroup$ – eSurfsnake May 23 at 6:04
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Formally a random variable is simply a measurable function $X:\Omega\to \mathbb R$, where $\Omega$ is a sample space equipped with some sigma algebra, $\Sigma$, containing the events we are interested in, and a probability measure $\mathbb P$. We call the triple $(\Omega, \Sigma, \mathbb P)$ a probability space. Thus if $g:\mathbb R\to \mathbb R$ is some "regular" function, $g\circ X:\Omega\to \mathbb R$ is simply the function mapping $\Omega\ni\omega\mapsto g(X(\omega))$. Thus if $X_1$ and $X_2$ are both random variables on the same probability space it is easy to talk about things like $g\circ(X_1+X_2)$. This is simply the map $\omega\mapsto g(X_1(\omega)+X_2(\omega))$.

For an example we can consider the most simple of experiments: flipping a fair coin twice. Then our sample space is $\Omega=\{HH,HT,TH,TT\}$, and our $\sigma$-algebra is simply the power set. The most natural probability measure is the one assigning a value of $1/4$ to each singleton in $\Omega$. Suppose we win a bet if in the experiment two heads occur. We can then capture this by the random variable $X:\Omega\to \mathbb R$ given by $X(HH)=1$ and $X$ is $0$ on all other elements of $\Omega$. Thus we see $X$ is simply a function, just not from $\mathbb R$ to itself.

Often in statistics we actually don't see $\Omega$ written out explicitly, because it often consists of weird things, like the population of the earth, but it is always exists underneath when we're doing probability calculations. In fact for a r.v. $X$ the CDF, $F_X$ is defined in terms of the probability space as follows $$F_X(x)=\mathbb P\{\omega\in \Omega:X(\omega)\leq x\}.$$

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