PSTAT 5A: Lecture 08

Discrete Random Variables

Ethan P. Marzban

2023-04-18

Recap of Probability

Recall the basic ingredients of probability we have discussed so far:
- Experiment: any procedure we can repeat an infinite number of times, where on each reptition there is a fixed set of things (called outcomes) that can happen.
- Outcome space ($\boldsymbol{\Omega}$): the set of all outcomes associated with a particular experiment
- Event: a subset of the outcome space
- Probability: a function that maps events to a number; specifically, one that quantifies our beliefs about a particular event

An Experiment

Let’s actually conduct an experiment together!
Specifically, suppose we toss a coin 3 times and record the outcomes.

viewof toss = Inputs.button("Toss")

dummy = toss + 1
coin = ["H", "T"]
s1 = coin[Math.floor(Math.random()*coin.length*dummy/dummy)];
s2 = coin[Math.floor(Math.random()*coin.length*dummy/dummy)];
s3 = coin[Math.floor(Math.random()*coin.length*dummy/dummy)];
s4 = coin[Math.floor(Math.random()*coin.length*dummy/dummy)];
[s1, s2, s3];

Additionally, let’s keep track of the number of heads we observe each time we run this experiment.
- In fact, let’s do this on the chalkboard.

An Experiment

Alright, let’s make note of a few things.

Note that each time we run this experiment, we (sure enough) get an element of the outcome space, which is

An Experiment

But, also note that each time we run the experiment, the number of heads doesn’t necessarily remain fixed.

In fact, each outcome in the outcome space corresponds to a different number of heads:

Outcome	Number of Heads
(`H`, `H`, `H`)	3
(`H`, `H`, `T`)	2
(`H`, `T`, `H`)	2
(`T`, `H`, `H`)	2
(`H`, `T`, `T`)	1
(`T`, `H`, `T`)	1
(`T`, `T`, `H`)	1
(`T`, `T`, `T`)	0

Random Variables

This leads us to the notion of random variables.
Loosely speaking, a random variable is a variable or process with a numerical outcome.
We denote random variables using capital letters; e.g. $X$, $Y$, $Z$, $W$, etc.
So, for example, $X =$ “the number of heads in 3 tosses of a coin” is a random variable because (a) it is a numerical outcome of an experiment and (b) it is random (i.e. its value changes depending on the outcome of the experiment).

State Space

A key part of the definition of random variables is that they must be numerical.
What this means is we can always look at the set of values a random variable could take: this is what we call the state space of a random variable.
For example: if $X =$ “number of heads in 3 tosses of a coin”, we see that $X$ will only ever be $0$, $1$, $2$, or $3$.
- This is because it is not possible to toss 3 coins and get, say, 5 heads, or a negative number of heads!
We often denote the state space of a random variable using the notation $S_{\verb|<variable>|}$; e.g. $S_X$ to mean the state space of $X$, $S_Y$ to mean the state space of $Y$, etc.

Classifying Random Variables

Because random variables are numerical, their state spaces will always be numerical sets of values.
This means we can classify state spaces using our Variable Classification scheme from Week 1!
- Specifically: the state space $S_X$ of a random variable will either have “jumps”, or not.
We extend the same classification language to random variables:

Definition

Given a random variable $X$, we say that:

$X$ is a discrete random variable (or just “$X$ is discrete) if $S_X$ is has jumps
$X$ is a continuous random variable (or just “$X$ is continuous) if $S_X$ has no jumps

Leadup

Let’s return to our coin tossing example.
What is the probability that we observe zero heads?
Well, in the language of our random variable $X$ (which counts the number of heads in these three tosses of our fair coin), we can translate “zero heads” to the event “$\{X = 0\}$’’, meaning we want to find $\mathbb{P}(X = 0)$.
Observing zero heads is equivalent to observing all tails, meaning the event $\{X = 0\}$ is equivalent to the event { (T, T, T) }.
Now, up to this point I have been careful to avoid explicitly mentioning whether our coin is fair or not.
- For the time being, let’s assume that the probability our coin lands ‘heads’ on any given toss is some fixed value $p$.

Now, we have assumed that the coin tosses are performed independently of each other.

What this means is the probability of the outcome (T, T, T) is simply $(1-p)^3$
- If it’s not clear why, think of it this way: (T, T, T) means “T first and T second and T third’’. By independence, the probability of this string of occurrences is simply \[ \mathbb{P}(\text{\texttt{T} first}) \times \mathbb{P}(\text{\texttt{T} second}) \times \mathbb{P}(\text{\texttt{T} third}) \] which is just $(1 - p) \times (1 - p) \times (1 - p) = (1 - p)^3$.
Hence, $\mathbb{P}(X = 0) = (1 - p)^3$.
- For example, if the coin were fair (i.e. $p = 1/2$), then $\mathbb{P}(X = 0) = (1/2)^3 = (1/8)$.

What about the probability of observing exactly 1 head?

I.e., how would we go about computing $\mathbb{P}(X = 1)$?
Well, let’s use the same logic we used before: that is, let’s see what outcomes comprise the event $\{X = 1\}$.
Upon inspection, we see that the event $\{X = 1\}$ is equivalent to the event {(H, T, T), (T, H, T), (T, T, H) }.
- The probability of the outcome (H, T, T) is $p \times (1 - p) \times (1 - p) = p \times (1 - p)^2$
- The probability of the outcome (T, H, T) is $(1 - p) \times p \times (1 - p) = p \times (1 - p)^2$
- The probability of the outcome (T, T, H) is $(1 - p)^2 \times p = p \times (1 - p)^2$
Therefore, putting these facts together, we find \[ \mathbb{P}(X = 1) = 3 \times p \times (1 - p)^2 \]

We could play the same game to find $\mathbb{P}(X = 2)$ and $\mathbb{P}(X = 3)$:

Probability	Value
$\mathbb{P}(X = 0)$	$(1 - p)^3$
$\mathbb{P}(X = 1)$	$3 \times p \times (1 - p)^2$
$\mathbb{P}(X = 2)$	$3 \times p^2 \times (1 - p)$
$\mathbb{P}(X = 3)$	$p^3$

Another way to display these results is in the following format:

\[\begin{array}{r|cccc} \boldsymbol{k} & 0 & 1 & 2 & 3 \\ \hline \boldsymbol{\mathbb{P}(X = k)} & (1 - p)^3 & 3 p (1 - p)^2 & 3 p^2 (1 - p) & p^3 \end{array}\]

Probability Mass Function

The table on the previous slide is called a probability mass function, and is often abbreviated as p.m.f..
In general, the p.m.f. of an arbitrary random variable $X$ is a table or formula that specifies all the possible values a random variable can take (i.e. the state space), along with the probability with which the random variable attains those values.
We use the term “function” to describe this because, in abstraction, we can notate the p.m.f. as \[ p_X(k) := \mathbb{P}(X = k) \] where $k$ can be any value in the state space of $X$.

Example

Worked-Out Example 1

Suppose we toss three fair coins independently, and let $X$ denote the number of heads observed. Construct the p.m.f. (probability mass function) of $X$.

By our work from above, the p.m.f. of $X$ is given by \[\begin{array}{r|cccc} \boldsymbol{k} & 0 & 1 & 2 & 3 \\ \hline \boldsymbol{\mathbb{P}(X = k)} & 1/8 & 3/8 & 3/8 & 1/8 \end{array}\]
By the way, notice that the probabilities in the p.m.f. sum up to 1.
- This is not a coincidence! Because the probabilities represent the probabilities of all values $X$ can take, they must sum up to 1.

Properties of PMF’s

This leads us to posit the following two properties of probability mass functions:

Properties of a PMF

The values in a PMF must sum to 1
The values in a PMF must always be nonnegative

Also: we implicitly set probabilities not contained in the p.m.f. to be zero.
- For instance: in our coin tossing example, $\mathbb{P}(X = 1.5) = 0$.
- This makes sense! If $k \notin S_X$, then by definition of the state space it is impossible for $X$ to attain the value $k$, and so $\mathbb{P}(X = k) = 0$.

Worked-Out Example

Worked-Out Example 2

A random variable $X$ has the following p.m.f.: \[\begin{array}{r|cccc} \boldsymbol{k} & -1.4 & 0 & 3 & 4.15 \\ \hline \boldsymbol{\mathbb{P}(X = k)} & 0.1 & 0.2 & \boldsymbol{a} & 0.6 \end{array}\] What must be the value of $\boldsymbol{a}$?

Because the values in a p.m.f. must sum to 1, we must have \[ 0.1 + 0.2 + a + 0.6 = 1 \] which means \[ a = 1 - (0.1 + 0.2 + 0.6) = \boxed{0.1} \]

Worked-Out Example

Worked-Out Example 3

A random variable $X$ has the following p.m.f.: \[\begin{array}{r|cccc} \boldsymbol{k} & -1.4 & 0 & 3 & 4.15 \\ \hline \boldsymbol{\mathbb{P}(X = k)} & 0.1 & 0.2 & 0.1 & 0.6 \end{array}\] Compute both $\mathbb{P}(X = 0)$ and $\mathbb{P}(X \leq 0)$.

For $\mathbb{P}(X = 0)$, we can simply read off the corresponding element from the p.m.f.:

Now, $\mathbb{P}(X \leq 0)$ means “what’s the probability that $X$ is less than or equal to $0$”?

There are only two values in $S_X$ [by the way, as an aside, can anyone tell me what $S_X$ is in this problem?] that are less than or equal to $0$: $0$ itself, and $-1.4$.
Hence, saying that $X$ was less than or equal to $0$ is equivalent to saying $X$ was either $-1.4$ or $0$.
Therefore, \[ \mathbb{P}(X \leq 0) = \mathbb{P}(X = -1.4) + \mathbb{P}(X = 0) = 0.1 + 0.2 = \boxed{0.3} \]

Computing $\mathbb{P}(X \leq k)$

To compute $\mathbb{P}(X \leq k)$, we sum up the values $\mathbb{P}(X = x)$ for all values of $x$ in the state space that are less than or equal to $k$.

Expected Value

Definition

The expected value (or just expectation) of a discrete random variable $X$ is \[ \mathbb{E}[X] = \sum_{\text{all $k$}} k \cdot \mathbb{P}(X = k) \] where the sum ranges over all values of $k$ in the state space.

In words: multiply each value in the state space by the corresponding probability, and then sum.
The expected value is a sort of ‘center’ of a random variable.

Worked-Out Example

Worked-Out Example 4

We compute \[\begin{align*} \mathbb{E}[X] & = \sum_{\text{all $k$}} k \cdot \mathbb{P}(X = k) \\[5mm] & = (-1.4) \cdot \mathbb{P}(X = -1.4) + (0) \cdot \mathbb{P}(X = 0) + (3) \cdot \mathbb{P}(X = 3) \\ & \hspace{10mm} + (4.15) \cdot \mathbb{P}(X = 4.15) \\[5mm] & = (-1.4) \cdot (0.1) + (0) \cdot (0.2) + (3) \cdot (0.1) + (4.15) \cdot (0.6) = \boxed{2.65} \end{align*}\]

Variance and SD

Definition

The variance of a discrete random variable $X$ is \[ \mathrm{Var}(X) = \sum_{\text{all $k$}} (k - \mathbb{E}[X])^2 \cdot \mathbb{P}(X = k) \] where the sum ranges over all values of $k$ in the state space. The standard deviation is the square root of the variance: \[ \mathrm{SD}(X) = \sqrt{\mathrm{Var}(X)} \]

Second Formula for Variance

\[ \mathrm{Var}(X) = \left( \sum_{\text{all $k$}} k^2 \cdot \mathbb{P}(X = k) \right) - \left( \mathbb{E}[X] \right)^2 \]

Worked-Out Example

Worked-Out Example 5

We previously found that $\mathbb{E}[X] = 2.65$.
Hence, we need only to find $\sum_{k} k^2 \cdot \mathbb{P}(X = x)$: \[\begin{align*} \sum_{\text{all $k$}} k^2 \cdot \mathbb{P}(X = k) & = (-1.4)^2 \cdot \mathbb{P}(X = -1.4) + (0)^2 \cdot \mathbb{P}(X = 0) + (3)^2 \cdot \mathbb{P}(X = 3) \\ & \hspace{10mm} + (4.15)^2 \cdot \mathbb{P}(X = 4.15) \\[5mm] & = (-1.4)^2 \cdot (0.1) + (0)^2 \cdot (0.2) + (3)^2 \cdot (0.1) + (4.15)^2 \cdot (0.6) = 11.4295 \end{align*}\]

Therefore, by the second formula for variance, \[ \mathrm{Var}(X) = 11.4295 - (2.65)^2 = \boxed{4.407} \] and so $\mathrm{SD}(X) = \sqrt{4.407} \approx \boxed{2.099}$

We could have used the definition of variance as well: \[\begin{align*} \mathrm{Var}(X) & = \sum_{\text{all $k$}} (k - \mathbb{E}[X])^2 \cdot \mathbb{P}(X = k) \\[3mm] & = (-1.4 - 2.65)^2 \cdot \mathbb{P}(X = -1.4) + (0 - 2.65)^2 \cdot \mathbb{P}(X = 0) \\ & \hspace{10mm} + (3 - 2.65)^2 \cdot \mathbb{P}(X = 3) + (4.15 - 2.65)^2 \cdot \mathbb{P}(X = 4.15) \\[5mm] & = (-1.4 - 2.65)^2 \cdot (0.1) + (0 - 2.65)^2 \cdot (0.2) + (3 - 2.65)^2 \cdot (0.1) \\ & \hspace{10mm} + (4.15 - 2.65)^2 \cdot (0.6) = 4.407 \end{align*}\]

Your Turn!

Exercise 1

Suppose $X$ is a random variable with p.m.f. (probability mass function) given by \[\begin{array}{r|cccc} \boldsymbol{k} & -1 & 0 & 1 & 2 \\ \hline \boldsymbol{\mathbb{P}(X = k)} & 0.3 & 0.2 & 0.1 & \boldsymbol{a} \end{array}\]

Find the state space $S_X$ of $X$.
Find the value of $\boldsymbol{a}$
Find $\mathbb{P}(X = 0.5)$
Find $\mathbb{P}(X \leq 1)$
Find $\mathbb{P}(X > 1)$
Find $\mathbb{E}[X]$
Find $\mathrm{Var}(X)$ and $\mathrm{SD}(X)$

Your Turn!

Exercise 2

Consider the following game: a fair six-sided die is rolled. If the number showing is 1 or 2, you win a dollar; if the number showing is 3, 4, or 5 you win 2 dollars; if the number showing is 6, you lose 1 dollar. Let $W$ denote your net winnings after playing this game once.

Write down the state space $S_W$ of $W$.
Find the p.m.f. of $W$.
What are your expected winnings after one round of the game?

Back to Coins

Alright, let’s close out this lecture by returning to our coin tossing example.
As a reminder: if we let $X$ denote the number of heads in 3 tosses of a $p-$coin (i.e. a coin that lands ‘heads’ with probability $p$), the p.m.f. of $X$ is given by

\[\begin{array}{r|cccc} \boldsymbol{k} & 0 & 1 & 2 & 3 \\ \hline \boldsymbol{\mathbb{P}(X = k)} & (1 - p)^3 & 3 p (1 - p)^2 & 3 p^2 (1 - p) & p^3 \end{array}\]

What if instead of tossing 3 coins, we had tossed 4? Or 5? Or 10?
We could go through the same steps we did before, when deriving the p.m.f. for three tosses, but let’s be a little smarter about this; let’s answer the following more general question:

If $X$ denotes the number of heads in $n$ tosses of a $p-$coin, what is the p.m.f. of $X$?

It’s always a good idea to start with the support. If we are tossing $n$ coins, we cannot observe any more than $n$ heads, nor can we observe any less than $0$ heads. As such, \[ S_X = \{0, 1, 2, \cdots, n\} \]
Now, let’s consider an arbitrary $k \in S_X$, and examine the event $\{X = k \}$.
In words, $\{X = k \}$ means “we observe exactly $k$ heads”, which is equivalent to “we observe exactly $k$ heads and $(n - k)$ tails.”
Suppose, for the moment, that these $k$ heads occurred consecutively, and at the beginning of our tosses. I.e., suppose we have the outcome \[ (\underbrace{H, \ H, \ \cdots, \ H}_{\text{$k$ heads}}, \ \underbrace{T, \ T, \ \cdots, \ T}_{\text{$n - k$ tails}}) \]
- The probability of this outcome is $p^k \cdot (1 - p)^{n - k}$.

However, the event $\{X = k\}$ does not mean “$k$ heads at the beginning followed by $(n - k)$ tails”; it just means “$k$ heads anywhere among our $n$ tosses and $(n - k)$ tails”.

As such, there are $\binom{n}{k}$ other outcomes in the event $\{X = k\}$, corresponding to the $\binom{n}{k}$ different ways to place the $k$ heads among the $n$ tosses.
So, what we have is \[ \mathbb{P}(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k} \]
- As an aside, note that this agrees with our three-coin case from above: plugging in $n = 3$ yields exactly the same p.m.f. we found before.

Binomial Distribution

The Binomial Distribution

Suppose the probability of a single trial resulting in a ‘success’ is $p$. Letting $X$ denote the number of successes in $n$ independent trials, then we say that $X$ follows the Binomial Distribution with parameters $n$ and $p$. We use the notation $X \sim \mathrm{Bin}(n, p)$ to denote this.

Facts about the Binomial Distribution

If $X \sim \mathrm{Bin}(n, p)$, then

$\displaystyle \mathbb{P}(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}$
$\mathbb{E}[X] = np$ and $\mathrm{Var}(X) = np(1 - p)$

Binomial Conditions

Four Conditions to Check

If $X$ counts the number of successes in $n$ trials, there are four conditions that need to be satisfied in order for $X$ to follow the Binomial Distribution:

The trials must be independent
The number of trials, $n$, must be fixed
There should be a well-defined notion of “success” and “failure” on each trial
The probability of “success” must remain constant across trials.

So, remember: $X \sim \mathrm{Bin}(n, p)$ just means “$X$ counts the number of successes in $n$ trials, where success occurs with probability $p$ on any given trial, subject to the four conditions above being satisfied.

Worked-Out Example

Worked-Out Example 6

If we roll a fair $6-$sided die $13$ times (assume rolls are independent of each other) and let $X$ denote the number of times we observe an even number, is $X$ binomially distributed?
In a large population of $100$ students, of which $70$ own Android phones, we draw a random sample of 10 without replacement and let $Y$ denote the number of students in this sample that have Android phones. Is $Y$ binomially distributed?
Consider the same setup as in part (b) above, except this time suppose students are selected with replacement. Is $Y$ binomially distributed?

Part (a)

We check the Binomial Conditions.
1. Independent trials? Yup!
2. Fixed number of trials? Yup! ($n = 13$)
3. Well-defined notion of success? Yup! (“success” = “rolling an even number” and “failure” = “rolling an odd number”)
4. Fixed probability of success? Yup! ($p = 1/2$).
Since all 4 conditions are satisfied, $X$ binomially distributed: specifically, \[ X \sim \mathrm{Bin}(13, \ 1/2) \]

Part (b)

We check the Binomial Conditions.
1. Independent trials? ; because sampling is done without replacement, trials are no longer independent (i.e. the result of our second trial is very much dependent on the result of our first).
Since at least one condition is violated, $Y$ does follow the binomial distribution.

Part (c)

We check the Binomial Conditions.
1. Independent trials? Yup!
2. Fixed number of trials? Yup! ($n = 10$)
3. Well-defined notion of success? Yup! (“success” = “owning an Android phone” and “failure” = “not owning an Android phone”)
4. Fixed probability of success? Yup! ($p = 7/10$).
Since all 4 conditions are satisfied, $Y$ binomially distributed: specifically, \[ Y \sim \mathrm{Bin}(10, \ 7/10) \]

Your Turn!

Exercise 3

Suppose Jana toss $65$ different $12-$sided dice, independently of each other; let $Z$ denote the number of times a multiple of three results.

Verify that $Z$ follows the Binomial Distribution, and identify its parameters.
What is the probability that Jana observes exactly 23 multiples of three?
What is the expected number of multiples of three Jana will observe?
What is the standard deviation of the number of multiples of three Jana will observe?

Probability	Value
\(\mathbb{P}(X = 0)\)	\((1 - p)^3\)
\(\mathbb{P}(X = 1)\)	\(3 \times p \times (1 - p)^2\)
\(\mathbb{P}(X = 2)\)	\(3 \times p^2 \times (1 - p)\)
\(\mathbb{P}(X = 3)\)	\(p^3\)