Random variable

Definition

A Random variable is a measurable map from a probability space to any measurable space

Let $(\Omega,\mathcal{A},\mathbb{P})$ be a probability space and let $X:(\Omega,\mathcal{A})\rightarrow(V,\mathcal{U})$ be a random variable

Then:

$$X^{-1}(U\in\mathcal{U})\in\mathcal{A}$$ , but anything $$\in\mathcal{A}$$ is $\mathbb{P}$-measurable! So we see:

$$\mathbb{P}(X^{-1}(U\in\mathcal{U}))\in[0,1]$$ which we may often write as: $$\mathbb{P}(X=U)$$ for simplicity (see Mathematicians are lazy)

Notation

Often a measurable space that is the domain of the RV will be a probability space, given as $$(\Omega,\mathcal{A},\mathbb{P})$$ , and we may write either:

• $X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(V,\mathcal{U})$
• $X:(\Omega,\mathcal{A})\rightarrow(V,\mathcal{U})$

With the understanding we write $\mathbb{P}$ in the top one only because it is convenient to remind ourselves what probability measure we are using.

Pitfall

Note that it is only guaranteed that $$X^{-1}(U\in\mathcal{U})\in\mathcal{A}$$ but it is not guaranteed that $$X(A\in\mathcal{A})\in\mathcal{U}$$ , it may sometimes be the case.

For example consider the trivial $\sigma$-algebra $$\mathcal{U}=\{\emptyset,V\}$$

Example

Discrete random variable

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Recall the roll two die example from probability spaces, we will consider the RV $X$ = the sum of the scores

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Recall the die example from probability spaces (which is restated less verbosely here), there:

Component	Definition
$\Omega$	$$\Omega=\{(a,b)|\ a,b\in\mathbb{N},\ a,b\in[0,6]\}$$
$\mathcal{A}$	$$\mathcal{A}=\mathcal{P}(\Omega)$$
$\mathbb{P}$	$$\mathbb{P}(A) = \frac{1}{36}|A|$$

Let us define the Random variable that is the sum of the scores on the die, that is $$X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(\{2,\cdots,12\},\mathcal{P}(\{2,\cdots,12\}))$$ .

It should be clear that $$(\{2,\cdots,12\},\mathcal{P}(\{2,\cdots,12\}))$$ is a measurable space however we need not consider a measure on it.

Writing $X$ out explicitly is hard but there are two parts to it:

Warning - the first bullet point is a suspected claim

• We can look at what generates a space, we need only consider the single events really, that is to say:

  $$X(A\in\mathcal{A})\cup X(B\in\mathcal{A})=X(A\cup B\in\mathcal{A})$$ , so we need only look at $X$ of the individual events

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TODO: Prove this

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• We can write it more explicitly as:

  $$X(A\in\mathcal{A})=\{a+b|(a,b)\in A\}$$

Example of pitfall

Take $$X:(\Omega,\mathcal{P}(\Omega),\mathbb{P})\rightarrow(V,\mathcal{U})$$ , if we define $$\mathcal{U}=\{\emptyset,V\}$$ then clearly:

$$X(\{(1,2)\})=\{3\}\notin\mathcal{U}$$ . Yet it is still measurable.

So an example! $$\mathbb{P}(X^{-1}(\{5\}))=\mathbb{P}(X=5)=\mathbb{P}(\{(1,4),(4,1),(2,3),(3,2)\})=\frac{4}{36}=\frac{1}{9}$$