Chapter 03 - Discrete Random Variables

Random variables

Theory

Theory 1

Random variable

A random variable (RV) $X$ on a probability space $(S,\mathcal{F},P)$ is a function $X:S\to \mathbb{R}$.

So $X$ assigns to each outcome a number.

Note: The word ‘variable’ indicates that an RV outputs numbers.

Random variables can be formed from other random variables using mathematical operations on the output numbers.

Given random variables $X$ and $Y$, we can form these new ones:

$$\frac{1}{2}(X+Y),X\cdot Y,\cos X,X^2,\text{etc.}$$

Suppose $s\in S$ is some particular outcome. Then, for example, $(X+Y)(s)$ is by definition $X(s)+Y(s)$.

---

Random variables determine events.

• Given $a\in \mathbb{R}$, the event “$X=a$” is equal to the set $X^{-1}(a)$
• That is: the set of outcomes mapped to $a$ by $X$
• That is: the event “$X$ took the value $a$”

Such events have probabilities. We write them like this:

$$P[X=a]\gg \gg P[X^{-1}(a)]$$

This generalized to events where $X$ lies in some range or set, for example:

$$P[a\le X<b],P[X\in \{2,4,5,6,9\}]$$

---

The axioms of probability translate into rules for these events.

For example, additivity leads to:

$$P[X<0]+P[X=0]+P[0<X\le 3]+P[3<X]=1$$

---

A discrete random variable has probability concentrated at a discrete set of real numbers.

• A ‘discrete set’ means finite or countably infinite.
• The distribution of probability is recorded using a probability mass function (PMF) that assigns probabilities to each of the discrete real numbers.
• Numbers with nonzero probability are called possible values.

PMF

The PMF function $P_X(k)$, for $X$ a discrete RV, is defined by:

$$P_X(k)=P[X=k]$$

A continuous random variable has probability spread out over the space of real numbers.

• The distribution of probability is recorded using a probability density function (PDF) which is integrated over intervals to determine probabilities.

PDF

The PDF function for $Y$ (a CRV) is written $f_Y(x)$ for $x\in \mathbb{R}$, and probabilities are calculated like this:

$$P[a\le Y\le b]={\int }_a^b{f}_Y(x)dx$$

---

For any RV, whether discrete or continuous, the distribution of probability is encoded by a function:

CDF

The cumulative distribution function (CDF) for a random variable $X$ is defined for all $x\in \mathbb{R}$ by: $$F_X(x)=P[X\le x]$$

Notes:

• Sometimes the relation to $X$ is omitted and one sees just “$F(x)$.”
• Sometimes the CDF is called, simply, “the distribution function” because:

The CDF works for a discrete, continuous, or mixed RV

• PMF is for discrete only
• PDF is for continuous only
• CDF covers both and covers mixed RVs

---

The CDF of a discrete RV is always a stepwise increasing function. At each step up, the jump size matches the PMF value there.

From this graph of $F_X(x)$:

we can infer the PMF values based on the jump sizes:

$P_X(-1)$	$P_X(0)$	$P_X(1)$	$P_X(2)$	$P_X(3)$	$P_X(4)$
$0$	$1/8$	$3/8$	$3/8$	$1/8$	$0$

---

For a discrete RV, the CDF and the PMF can be calculated from each other using formulas.

PMF from CDF

Given a PMF $P_X(x)$, the CDF is determined by:

$$F_X(x)=\sum _{k_i\le x}{P}_X(k_i)$$

where $\{k_1,k_2,\dots \}$ is the set of possible values of $X$.

Link to original

Illustration

Example - PDF and CDF: Roll 2 dice

PDF and CDF: Roll 2 dice

Roll two dice colored red and green. Let $X_R$ record the number of dots showing on the red die, $X_G$ the number on the green die, and let $S$ be a random variable giving the total number of dots showing after the roll, namely $S=X_R+X_G$.

• Find the PMFs of $X_R$ and of $X_G$ and of $S$.
• Find the CDF of $S$.
• Find $P[S=8]$.

Solution

(1) Construct sample space:

Denote outcomes with ordered pairs of numbers $(i,j)$, where $i$ is the number showing on the red die and $j$ is the number on the green one.

Therefore $i,j\in \mathbb{N}$ are integers satisfying $1\le i,j\le 6$.

---

(2) Create chart of outcomes:

---

(3) Define random variables:

We have $X_R(i,j)=i$ and $X_G(i,j)=j$.

Therefore $S(i,j)=i+j$.

---

(4) Find PMF of $X_R$:

Use variable $k$ for each possible value of $X_R$, so $k=1,2,\dots ,6$. Find $P_{X_R}(k)$:

$$\begin{array}{c}{P}_{X_R}(k)=P[X_R=k]\\ \\ \gg \gg \frac{|\text{outcomes with k on red}|}{|\text{all outcomes}|}\gg \gg \frac{6}{36}=\frac{1}{6}\end{array}$$

Therefore $P_{X_R}(k)=1/6$ for every $k$.

---

(5) Find PMF of $X_G$ similarly:

$$P_{X_G}(k)=\frac{1}{6}\text{for all}k$$

---

(6) Find PMF of $S$:

$$P_S(k)=P[S=k]\gg \gg \frac{|\text{outcomes with sum k}|}{|\text{all outcomes}|}$$

Count outcomes along diagonal lines in the chart.

$k$	2	3	4	5	6	7	8	9	10	11	12
$P_S(k)$	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

Evaluate: $P[S=8]=5/36$.

---

(7) Find CDF of $S$:

CDF definition: $F_S(x)=P[S\le x]$

Apply definition: add new PMF value at each increment:

$$F_S(x)=\{\begin{array}{ll}0& x<2\\ 1/36& 2\le x<3\\ 3/36& 3\le x<4\\ 6/36& 4\le x<5\\ 10/36& 5\le x<6\\ 15/36& 6\le x<7\\ 21/36& 7\le x<8\\ 26/36& 8\le x<9\\ 30/36& 9\le x<10\\ 33/36& 10\le x<11\\ 35/36& 11\le x<12\\ 36/36& 12\le x\end{array}$$ Link to original

Example - Total heads count; binomial expansion of 1

PMF for total heads count; binomial expansion of 1

A fair coin is flipped $n$ times.

Let $X$ be the random variable that counts the total number of heads in each sequence.

The PMF of $X$ is given by:

$$P_X(k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right){\left(\frac{1}{2}\right)}^n$$

Since the total probability must add to 1, we know this formula must hold:

$$\begin{array}{rl}1& =\sum _{\text{possible k}}{P}_X(k)\\ & \\ \gg \gg 1& =\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\left(\frac{1}{2}\right)}^n\end{array}$$

Is this equation really true?

There is another way to view this equation: it is the binomial expansion $(x+y{)}^n$ where $x=\frac{1}{2}$ and $y=\frac{1}{2}$:

$${\left(\frac{1}{2}+\frac{1}{2}\right)}^n=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\left(\frac{1}{2}\right)}^n$$ Link to original

Example - Life insurance payouts

Life insurance payouts

A life insurance company has two clients, $A$ and $B$, each with a policy that pays $100,000 upon death. Consider events $D_1$ that the older client dies next year, and $D_2$ that the younger dies next year. Suppose $P[D_1]=0.10$ and $P[D_2]=0.05$.

Define a random variable $X$ measuring the total money paid out next year in units of $1,000. The possible values for $X$ are 0, 100, 200. Now calculate:

$$\begin{array}{rl}P[X=0]\gg \gg & P[D_1^c]P[D_2^c]\gg \gg 0.95\cdot 0.90\gg \gg 0.86\\ & \\ P[X=100]\gg \gg & 0.05\cdot 0.90+0.95\cdot 0.10\gg \gg 0.14\\ & \\ P[X=200]\gg \gg & 0.05\cdot 0.10\gg \gg 0.005\end{array}$$Link to original

Bernoulli process

Theory - Bernoulli, binomial, geometric, Pascal, uniform

Theory 1 - Bernoulli, binomial, geometric, Pascal, uniform

In a Bernoulli process, an experiment with binary outcomes is repeated; for example flipping a coin repeatedly. Several discrete random variables may be defined in the context of some Bernoulli process.

Notice that the sample space of a Bernoulli process is infinite: an outcome is any sequence of trial outcomes, e.g. $HTHHTTHHHTTTHHHHTTTT\cdots$

Bernoulli variable

A random variable $X_i$ is a Bernoulli indicator, written $X_i\sim \mathrm{Ber}(p)$, when $X_i$ indicates whether a success event, having probability $p$, took place in trial number $i$ of a Bernoulli process.

Bernoulli PMF:

$$P_{X_i}(k)=\{\begin{array}{ll}p& k=1\\ q& k=0\\ 0& \text{else}\end{array}$$

Here $q=1-p$.

An RV that always gives either $0$ or $1$ for every outcome is called an indicator variable.

Binomial variable

A random variable $S$ is binomial, written $S\sim \text{Bin}(n,p)$, when $S$ counts the number of successes in a Bernoulli process, each having probability $p$, over a specified number $n$ of trials.

Binomial PMF:

$$P_S(k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}\text{for }k=0,1,2,\dots ,n$$

• For example, if $S\sim \mathrm{Bin}(10,0.2)$, then $P_S(5)$ gives the odds that success happens exactly 5 times over 10 trials, with probability $0.2$ of success for each trial.
• In terms of the Bernoulli indicators, we have: $S=X_1+X_2+\cdots +X_n$
• If $A$ is the success event, then $p=P[A]$ is the success probability, and $q=1-p$ is the failure probability.

---

Geometric variable

A random variable $N$ is geometric, written $N\sim \mathrm{Geom}(p)$, when $N$ counts the discrete wait time in a Bernoulli process until the first success takes place, given that success has probability $p$ in each trial.

Geometric PMF:

$$P_N(k)=q^{k-1}p\text{for }k=1,2,3,\dots$$

Here $q=1-p$.

• For example, if $N\sim \mathrm{Geom}(30\%)$, then $P_N(7)$ gives the probability of getting: failure on the first $6$ trials AND success on the $7^{\text{th}}$ trial.

---

Pascal variable

A random variable $L$ is Pascal, written $L\sim \mathrm{Pasc}(\ell ,p)$, when $L$ counts the discrete wait time in a Bernoulli process until success happens $\ell$ times, given that success has probability $p$ in each trial.

Pascal PMF:

$$P_L(k)=\left(\genfrac{}{}{0ex}{}{k-1}{\ell -1}\right)(1-p{)}^{k-\ell }{p}^{\ell }\text{for }k=\ell ,\ell +1,\ell +2,\dots$$

• For example, if $L\sim \mathrm{Pasc}(3,0.25)$, then $P_L(8)$ gives the probability of getting: the $3^{\text{rd}}$ success on (precisely) the $8^{\text{th}}$ trial.
• Interpret the formula: $\#$ ways to arrange $2$ successes among $7$ ‘prior’ trials, times the probability of exactly $3$ successes and $5$ failures in one specific sequence.
• The Pascal distribution is also called the negative binomial distribution, e.g. $L\sim \mathrm{Negbin}(\ell ,p)$.

---

Uniform variable

A discrete random variable $X$ is uniform on a finite set $A\subset S$, written $X\sim \mathrm{Unif}(\mathrm{A})$, when the probability is a fixed constant for outcomes in $A$ and zero for outcomes outside $A$.

Discrete uniform PMF:

$$P_X(k)=\{\begin{array}{ll}\frac{1}{|A|}& \text{when }k\in A\\ 0& \text{when }k\notin A\end{array}$$

Continuous uniform PDF:

$$f_X(x)=\{\begin{array}{ll}\frac{1}{P[A]}& \text{when }x\in A\\ 0& \text{when }x\notin A\end{array}$$

Link to original

Illustration

Example - Roll die until

Roll die until

Roll a fair die repeatedly. Find the probabilities that:

(a) At most 2 threes occur in the first 5 rolls.

(b) There is no three in the first 4 rolls, using a geometric variable.

Solution

(a)

(1) Label variables and events:

Use a variable $S\sim \mathrm{Bin}(5,1/6)$ to count the number of threes among the first six rolls.

Seek $P[S\le 2]$ as the answer.

---

(2) Calculations:

Divide into exclusive events:

$$\begin{array}{c}P[S\le 2]\gg \gg P_S(0)+P_S(1)+P_S(2)\\ \\ \gg \gg \left(\genfrac{}{}{0ex}{}{5}{0}\right){\left(\frac{1}{6}\right)}^0{\left(\frac{5}{6}\right)}^5+\left(\genfrac{}{}{0ex}{}{5}{1}\right){\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^4+\left(\genfrac{}{}{0ex}{}{5}{2}\right){\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^3\\ \\ \gg \gg \frac{625}{648}\gg \gg \approx 0.965\end{array}$$

---

(b)

(1) Label variables and events:

Use a variable $N\sim \mathrm{Geom}(1/6)$ to give the roll number of the first time a three is rolled.

Seek $P[N>4]$ as the answer.

---

(2) Compute:

Sum the PMF formula for $\mathrm{Geom}(1/6)$:

$$P[N>4]\gg \gg \sum _{k=5}^{\infty }{\left(\frac{5}{6}\right)}^{k-1}\left(\frac{1}{6}\right)$$

---

(3) Recall geometric series formula:

For any geometric series:

$$a+ar+a{r}^2+a{r}^3+\cdots =\frac{a}{1-r}$$

Therefore:

$$P[N>4]=\sum _{k=5}^{\infty }{\left(\frac{5}{6}\right)}^{k-1}\left(\frac{1}{6}\right)\gg \gg {\left(\frac{5}{6}\right)}^4$$ Link to original

Example - Cubs winning the World Series

Cubs winning the World Series

Suppose the Cubs are playing the Yankees for the World Series. The first team to 4 wins in 7 games wins the series. What is the probability that the Cubs win the series?

Assume that for any given game the probability of the Cubs winning is $p=45\%$ and losing is $q=55\%$.

Solution

Method (a): We solve the problem using a binomial distribution.

(1) Label variables and events:

Use a variable $X\sim \mathrm{Bin}(7,p)$. This $X$ counts the number of wins over 7 games. Thus, for example, $P_X(4)$ is the probability that the Cubs win exactly 4 games over 7 played.

Seek $P_X(4)+P_X(5)+P_X(6)+P_X(7)$ as the answer.

---

(2) Calculate using binomial PMF:

$$P_X(k)=\left(\genfrac{}{}{0ex}{}{7}{k}\right)p^k{q}^{7-k}$$

Insert data:

$$\begin{array}{c}{P}_X(4)+\cdots +P_X(7)\\ \\ \gg \gg \left(\genfrac{}{}{0ex}{}{7}{4}\right)p^4{q}^3+\left(\genfrac{}{}{0ex}{}{7}{5}\right)p^5{q}^2+\left(\genfrac{}{}{0ex}{}{7}{6}\right)p^6{q}^1+\left(\genfrac{}{}{0ex}{}{7}{7}\right)p^7{q}^0\end{array}$$

Compute:

$$\begin{array}{c}\gg \gg \frac{7\cdot 6\cdot 5}{3\cdot 2}{p}^4{q}^3+\frac{7\cdot 6}{2}{p}^5{q}^2+\frac{7}{1}{p}^6{q}^1+1\cdot p^7{q}^0\\ \\ \gg \gg p^4(35q^3+21p^1{q}^2+7p^{2}q+p^3)\end{array}$$

Convert $q\gg (1-p)$:

$$\begin{array}{c}\gg \gg p^4(35(1-p{)}^3+21p(1-p{)}^2+7p^2(1-p)+p^3)\\ \\ \gg \gg 35p^4-84p^5+70p^6-20p^7\gg \gg \approx 0.39\end{array}$$

---

Method (b): We solve the problem using a Pascal distribution instead.

(1) Label variables and events:

Use a variable $Y\sim \mathrm{Pasc}(4,p)$. This $Y$ measures the discrete wait time until the $4^{\text{th}}$ win. Thus, for example, $P_Y(k)$ is the probability that the Cubs win their $4^{\text{th}}$ game on game number $k$.

Seek $P_Y(4)+P_Y(5)+P_Y(6)+P_Y(7)$ as the answer.

---

(2) Calculate using Pascal PMF:

$$P_Y(k)=\left(\genfrac{}{}{0ex}{}{k-1}{3}\right)p^4{q}^{k-4}$$

Insert data:

$$\begin{array}{c}{P}_Y(4)+\cdots +P_Y(7)\\ \\ \gg \gg \left(\genfrac{}{}{0ex}{}{3}{3}\right)p^4{q}^0+\left(\genfrac{}{}{0ex}{}{4}{3}\right)p^4{q}^1+\left(\genfrac{}{}{0ex}{}{5}{3}\right)p^4{q}^2+\left(\genfrac{}{}{0ex}{}{6}{3}\right)p^4{q}^3\end{array}$$

Compute:

$$\begin{array}{c}\gg \gg 1\cdot p^4+\frac{4}{1}\cdot p^4{q}^1+\frac{5\cdot 4}{2}{p}^4{q}^2+\frac{6\cdot 5\cdot 4}{3\cdot 2}{p}^4{q}^3\\ \\ \gg \gg p^4(1+4q+10q^2+20q^3)\end{array}$$

Convert $q\gg (1-p)$:

$$\begin{array}{c}\gg \gg p^4(1+4(1-p)+10(1-p{)}^2+20(1-p{)}^3)\\ \\ \gg \gg 35p^4-84p^5+70p^6-20p^7\gg \gg \approx 0.39\end{array}$$

---

Notice: The calculation seems very different than method (a), right up to the end!

Link to original

Expectation and variance

Theory - Expectation and variance

Theory 1

Expected value

The expected value $E[X]$ of random variable $X$ is the weighted average of the values of $X$, weighted by the probability of those values.

Discrete formula using PMF:

$$E[X]=\sum _{k}k\cdot P_X(k)$$

Continuous formula using PDF:

$$E[X]={\int }_{-\infty }^{+\infty }x\cdot f_X(x)dx$$

Notes:

• Expected value is sometimes called expectation, or even just mean, although the latter is best reserved for statistics.
• The Greek letter $\mu$ is also used in contexts where ‘mean’ is used.

---

Let $X$ be a random variable, and write $E[X]=\mu$.

Variance

The variance $\mathrm{Var}[X]$ measures the average squared deviation of $X$ from $\mu$. It estimates how concentrated $X$ is around $\mu$.

• Defining formula:

$$\mathrm{Var}[X]=E[(X-\mu {)}^2]$$

• Shorter formula:

$$\mathrm{Var}[X]=E[X^2]-E[X{]}^2$$

Calculating variance

• Discrete formula using PMF:

$$\mathrm{Var}[X]=\sum _k(k-\mu {)}^2{P}_X(k)$$

• Continuous formula using PDF:

$$\mathrm{Var}[X]={\int }_{-\infty }^{+\infty }(x-\mu {)}^2{f}_X(x)dx$$

Standard deviation

The quantity ${\sigma }_X=\sqrt{\mathrm{Var}[X]}$ is called the standard deviation of $X$.

Link to original

Illustration

Example - Tokens in bins

Gambling game - tokens in bins

Consider a game like this: a coin is flipped; if $H$ then draw a token from Bin 1, if $T$ then from Bin 2.

• Bin 1 contents: 1 token $1,000, and 9 tokens $1
• Bin 2 contents: 5 tokens $50, and 5 tokens $1

It costs $50 to enter the game. Should you play it? (A lot of times?) How much would you pay to play?

Solution

(1) Setup:

Let $X$ be a random variable measuring your winnings in the game.

The possible values of $X$ are 1, 50, and 1000.

---

(2) Find the PDF of $X$:

For $k=1$ have $P_X(1)\gg \gg \frac{1}{2}\cdot \frac{9}{10}+\frac{1}{2}\cdot \frac{5}{10}\gg \gg \frac{7}{10}$

For $k=50$ have $P_X(50)\gg \gg \frac{1}{2}\cdot \frac{5}{10}\gg \gg \frac{1}{4}$

For $k=1000$ have $P_X(1000)\gg \gg \frac{1}{2}\cdot \frac{1}{10}\gg \gg \frac{1}{20}$

These add to 1, and $P_X(x)=0$ for all other $x$.

---

(3) Find $E[X]$ using the discrete formula:

$$\begin{array}{c}E[X]=\sum _{k}k\cdot P_X(k)\gg \gg 1\cdot P_X(1)+50\cdot P_X(50)+1000\cdot P_X(1000)\\ \\ \gg \gg 1\cdot \frac{7}{10}+50\cdot \frac{1}{4}+1000\cdot \frac{1}{20}\gg \gg \approx 63.2\end{array}$$

Since $63.2>50$, if you play it a lot at $50 you will generally make money.

---

Challenge Q:
If you start with $200 and keep playing to infinity, how likely is it that you go broke?

Link to original

Example - Expected value: rolling dice

Expected value: rolling dice

Let $X$ be a random variable counting the number of dots given by rolling a single die.

Then:

$$E[X]\gg \gg 1\cdot \frac{1}{6}+2\cdot \frac{1}{6}+\cdots +6\cdot \frac{1}{6}\gg \gg \frac{7}{2}$$

Let $S$ be an RV that counts the dots on a roll of two dice.

The PMF of $S$:

Then:

$$E[S]\gg \gg 2\cdot \frac{1}{36}+3\cdot \frac{2}{36}+4\cdot \frac{3}{36}+\cdots +12\cdot \frac{1}{36}\gg \gg 7$$

Notice that $\frac{7}{2}+\frac{7}{2}=7$.

In general, $E[X+Y]=E[X]+E[Y]$.

Let $X$ be a green die and $Y$ a red die.

From the earlier calculation, $E[X]=\frac{7}{2}$ and $E[Y]=\frac{7}{2}$.

Since $S=X+Y$, we derive $E[S]=7$ by simple addition!

Link to original

Example - Expected value by finding new PMF

Expected value by finding new PMF

Let $X$ have distribution given by this PMF:

Find $E[|X-2|]$.

Solution

(1) Compute the PMF of $|X-2|$:

PMF arranged by possible value:

$$\begin{array}{rl}P[|X-2|=0]& \gg \gg P[X=2]=\frac{1}{14}\\ P[|X-2|=1]& \gg \gg P[X=1]+P[X=3]=\frac{1}{7}+\frac{3}{14}=\frac{5}{14}\\ P[|X-2|=2]& \gg \gg P[X=4]=\frac{2}{7}\\ P[|X-2|=3]& \gg \gg P[X=5]=\frac{2}{7}\\ P[|X-2|=k]& \gg \gg 0\text{for}k\ne 0,1,2,3.\end{array}$$

---

(2) Calculate the expectation:

Using formula for discrete PMF:

$$E[|X-2|]\gg \gg 0\cdot \frac{1}{14}+1\cdot \frac{5}{14}+2\cdot \frac{2}{7}+3\cdot \frac{2}{7}\gg \gg \frac{25}{14}$$ Link to original

Exercise - Variance using simplified formula

Variance for composite using PMF and simpler formula

Suppose $X$ has this PMF:

$k:$	1	2	3
$P_X(k):$	$1/7$	$2/7$	$4/7$

Find $\mathrm{Var}\left[\frac{1}{1+X}\right]$ using the formula $\mathrm{Var}[Y]=E[Y^2]-E[Y{]}^2$ with $Y=\frac{1}{1+X}$.

(Hint: you should find $E[Y]=\frac{13}{42}$ and $E[Y^2]=\frac{13}{126}$ along the way.)

Link to original

Discrete families summary

Theory

Theory 1

Bernoulli: $X\sim \mathrm{Ber}(p)$

• Indicates a win.
• $P_X(1)=p,P_X(0)=q$
• $E[X]=p$
• $\mathrm{Var}[X]=pq$

Binomial: $X\sim \mathrm{Bin}(n,p)$

• Counts number of wins.
• $P_X(k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}$
• $E[X]=np$
• $\mathrm{Var}[X]=npq$

Geometric: $X\sim \mathrm{Geom}(p)$

• Counts discrete wait time until first win.
• $P_X(k)=q^{k-1}p$
• $E[X]=\frac{1}{p}$
• $\mathrm{Var}[X]=\frac{q}{p^2}$

Pascal: $X\sim \mathrm{Pasc}(\ell ,p)$

• Counts discrete wait time until ${\ell }^{\text{th}}$ win.
• $P_X(k)=\left(\genfrac{}{}{0ex}{}{k-1}{\ell -1}\right)q^{k-\ell }{p}^{\ell }$
• $E[X]=\frac{\ell }{p}$
• $\mathrm{Var}[X]=\frac{\ell q}{p^2}$

Poisson: $X\sim \mathrm{Pois}(\lambda )$

• Counts “arrivals” during time interval.
• $P_X(k)=e^{-\lambda }\frac{{\lambda }^k}{k!}$
• $E[X]=\lambda$
• $\mathrm{Var}[X]=\lambda$

Link to original