Consider a fair coin and define an experiment of throwing the coin twice.

Define the random variable $X$ to be the total number of heads in an experiment. (i.e if you throw 1 head and 1 tail the total number of heads in this experiment is 1).

What is the probability of getting 1 head in an experiment, i.e. $\mathbb{P}[X=1]$?

Solution

We define the sample space $\mathrm{\Omega }$ of this experiment to be $\{(HH),(HT),(TH),(TT)\}$.

We enumerate each outcome ${\xi }_i$ in the sample space as

• ${\xi }_1=HH$

• ${\xi }_2=HT$

• ${\xi }_3=TH$

• ${\xi }_4=TT$

$\mathbf{\text{First}}$, recall that $X$ is a function that map an outcome $\xi$ from the sample space $\mathrm{\Omega }$ to a number $x$ in the real space $\mathbb{R}$. In this context it means that $X$ maps one of the four outcomes ${\xi }_i$ to the total number of heads in the experiment (i.e $X(\cdot )=\mathbf{\text{number of heads}}$).

It is important to note that the codomain of $X$ is not any arbitrary number. We can only map our 4 outcomes ${\xi }_i$ in the domain to 3 distinct numbers $0$, $1$ or $2$, which we will see by manually enumerating each case below.

With that, this random variable $X$ is completely determined.

$\mathbf{\text{Secondly}}$, we need to examine carefully what is meant by $\mathbb{P}[X(\xi )=1]$ since this will answer the question on what is the probability of getting 1 head.

However, $X(\xi )=1$ is an expression and not an event that the probability measure $\mathbb{P}$ expects. Here we should recall that the probability law $\mathbb{P}(\cdot )$ is always applied to an event $E\in \mathcal{F}$ where $E$ is a set.

So we need to map this expression to an event $E\in \mathcal{F}$. So you can ask yourself how to establish this “mapping” of $X(\xi )=1$ to an event in our event space $\mathcal{F}$. This seems pretty easy since we already know that $X(\xi )=1$ has two cases matched in (174), namely $X({\xi }_2)=1$ and $X({\xi }_3)=1$. So we can simply define the event $E$ to be $\{{\xi }_2,{\xi }_3\}=\{(HT),(TH)\}$.

We verify that $E=\{{\xi }_2,{\xi }_3\}$ is indeed an event in $\mathcal{F}$:

More concretely, given an expression $X(\xi )=x$, we construct the event set $E$ by enumerating all the outcomes ${\xi }_i$ in the sample space $\mathrm{\Omega }$ that satisfy $X(\xi )=x$.

and this coincides with the pre-image of $x$ in the random variable $X$ as defined in Definition 76.

$\mathbf{\text{Consequently}}$, we have

Consider a fair dice and define an experiment of rolling the dice once.

The sample space is then

If we roll two fair dice, then the sample space is

So the probability of rolling say $(1,2)$ is,

Professor Stanley Chan gave a good example of the difference between a variable and a random variable.

The main difference is that a variable is deterministic while a random variable is non-deterministic.

Consider solving the following equation:

Then, if $x$ is a fixed constant, then $X={\displaystyle \frac{x}{2}}$ is a variable.

However, if $x$ is not fixed, meaning that it can have multiple states, then $X$ is a random variable since it is not deterministic.

Tie back to the example in Example 22, we note that $X$ is a random variable since the total number of heads $x$ in an experiment is not fixed. It can be 0, 1 or 2 depending on your toss.