Discrete random variables

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.title[
# Discrete random variables
]
.author[
### <a href="https://jmclip.github.io/MACSS_math_camp/">MACS 33000</a> <br /> University of Chicago
]

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`$$\newcommand{\E}{\mathrm{E}} \newcommand{\Var}{\mathrm{Var}} \newcommand{\Cov}{\mathrm{Cov}}$$`
# Misc

* CALCULATORS CLAIMED (need your own if you want it!)
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  * Please BE SPECIFIC so I can address concerns!
    * instead of 'more examples', try 'lecture X was challenging. more examples on Y'
* **Upcoming dates:**
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  * MACSS Econ orientation (9/20)
  * orientation (9/23) + rest of week
  * pub night (tbd)

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# Learning objectives

* Define random variables
* Distinguish between discrete and interval variables
* Identify discrete random variable distributions relevant to social science
* Review measures of central tendency and dispersion
* Define expected value and variance
* Define cumulative mass functions (CMFs) for discrete random variables

---

# Random variable

* A random process or variable with a numerical outcome
* Random variable `$X$` that is a function of the sample space
    * Number of incumbents who win
    * An indicator whether a country defaults on a loan (1 if a default, 0 otherwise)
    * Number of casualties in a war (rather than all possible outcomes)
    
    `$$X:\text{Sample Space} \rightarrow \Re$$`

---

# Treatment assignment

* `$3$` units, flipping fair coin `$\frac{1}{2}$` to assign each unit
* Assign to `$T=$`Treatment or `$C=$`control
* `$X$` = Number of units received treatment
* Defining the function

`$$X  = \left \{ \begin{array} {ll}0  \text{  if  } (C, C, C)  \\1 \text{  if  } (T, C, C) \text{ or } (C, T, C) \text{ or } (C, C, T) \\2 \text{ if  }  (T, T, C) \text{ or } (T, C, T) \text{ or } (C, T, T) \\3 \text{ if } (T, T, T) \end{array} \right.$$`
    
    `$$\begin{aligned}X( (C, C, C) )  & = 0 \\X( (T, C, C)) & =  1 \\X((T, C, T)) & = 2 \\X((T, T, T)) & = 3 \end{aligned}$$`

---

# Examples

* `$X$` = Number of Calls into congressional office in some period `$p$`
    * `$X(c) = c$`
* Outcome of Election
    * Define `$v$` as the proportion of vote the candidate receives
    * Define `$X = 1$` if `$v>0.50$`
    * Define `$X = 0$` if `$v<0.50$`
    * For example, if `$v = 0.48$`, then `$X(v) = 0$`
* An indicator whether a country defaults on a loan (1 if a default, 0 otherwise)
* Not all are experimental - most are observational

---

# Discrete random variables

* A random variable with a finite or countably infinite range is **discrete**
* A random variable with an uncountably infinite number of values is **continuous**

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# Probability mass functions

* Probability of the values that our value can take. 
* FOR DISCRETE VALUES ONLY!!
* Operator is the `$\sum$` (sum) because we are ADDING PIECES
* Need a corresponding probability for each value (remember what we said re: probability ( `$0 \leq p \leq 1$` ))

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# Probability mass function (PMF)
Define these for our domain (x). NOTE: little 'x' is the value that is being taken on while big 'X' is the random variable itself.

So, `$P(X=x)$` refers to the probability that our random variable takes on the value of x. You can also see `$P_X(X=x)$` and similar variants.

---

### Intuition

* `$\Pr(C, T, C) = \Pr(C)\Pr(T)\Pr(C) = \frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{8}$`

`$$\begin{aligned}p(X = 0) & = \Pr(C, C, C) = \frac{1}{8}\\p(X = 1) & = \Pr(T, C, C) + \Pr(C, T, C) + \Pr(C, C, T) = \frac{3}{8} \\p(X = 2) & = \Pr(T, T, C)  + \Pr(T, C, T) + \Pr(C, T, T) = \frac{3}{8} \\p(X = 3) & = \Pr(T, T, T) = \frac{1}{8}\end{aligned}$$`
    
* `$p(X = a) = 0$`, for all `$a \notin (0, 1, 2, 3)$`

---

# Probability mass function

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# Probability mass function

* For a discrete random variable `$X$`

`$$p_X(x) = \Pr(X = x)$$`
    
* Note that

`$$\sum_x p_{X}(x) = 1$$`

--
  
    
* Can also add probabilities for smaller sets `$S$` of possible values of `$X \in S$`

`$$\Pr(X \in S) = \sum_{x \in S} p_X (x)$$`
    `$$\Pr (X > 0) = \sum_{x=1}^2 p_X (x) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$$`

*remind you of anything* **::cough::** *integrals*
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# Calculating our PMF!

To calculate the PMF of a random variable `$X$`

* For each possible value `$x$` of `$X$`, collect all possible outcomes that give rise to the event `$\{ X=x \}$`
* Add their probabilities to obtain `$p_X (x)$`

---

# Topic model example

* Topics: distinct concepts (war in Ukraine, national debt, fire department grants)
* Mathematically: PMF on words
* Set of words `$\{ \text{ukraine, fire, department, soldier, troop, war, grant}\}$`

### Topic 1

`$\Pr(\text{ukraine}) = 0.3$`; `$\Pr(\text{fire}) = 0.0001$`; `$\Pr(\text{department}) = 0.0001$`; `$\Pr(\text{soldier}) = 0.2$`; `$\Pr(\text{troop}) = 0.2$`; `$\Pr(\text{war})=0.2997$`; `$\Pr(\text{grant})=0.0001$`

### Topic 2

`$\Pr(\text{ukraine}) = 0.0001$`; `$\Pr(\text{fire}) = 0.3$`; `$\Pr(\text{department}) = 0.2$`; `$\Pr(\text{soldier}) = 0.0001$`; `$\Pr(\text{troop}) = 0.0001$`; `$\Pr(\text{war})=0.0001$`; `$\Pr(\text{grant})=0.2997$`

* Topic models: take a set of documents and estimate topics

---

# Cumulative mass function

* For a random variable `$X$`

`$$F(x) = \Pr(X \leq x)$$`
  --
* Characterizes how probability **CUMULATES** as `$X$` gets larger
* `$F(x) \in [0,1]$`
* `$F(x)$` is non-decreasing

---

# Three person experiment

Consider the three person experiment: `$\Pr(T) = \Pr(C) =  1/2$`

What is `$F(2)$`?

`$$\begin{aligned}F(2) & = \Pr(X = 0) + \Pr(X = 1) + \Pr(X = 2) \\& = \frac{1}{8} + \frac{3}{8} + \frac{3}{8} \\& = \frac{7}{8} \end{aligned}$$`

What is `$F(2) - F(1)$`?

`$$\begin{aligned} F(2)  - F(1) & = [\Pr(X = 0) + \Pr(X = 1) + \Pr(X = 2)]  \nonumber \\& \quad -[\Pr(X = 0) + \Pr(X = 1)] \\F(2) - F(1) & = \Pr(X = 2)\end{aligned}$$`

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# PMFs and CMFs

PMF is to our original function as CMF is to an integral
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# PMFs and CMFs

* Similar to integration, but over a discrete set of values

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# Distributions and the functions we love

---
# Function & Distribution overview

So far, we've looked at how to add up values. However, the values we get depend upon the underlying function. There are a few well-known distributions that we'll cover as these will be increasingly useful as you move through your coursework.

* Bernoulli
* Binomial
* Geometric
* Poisson

This is a selection of distributions -- each has certain properties that can be useful for predicting outcomes and events but there are many, many more that may be useful to you as well.

---
# Bernoulli

Suppose `$X$` is a random variable, with `$X \in \{0, 1\}$` and `$\Pr(X = 1) = \pi$`. Then we will say that `$X$` is **Bernoulli** random variable,

`$$p_X(k)= \pi^{k} (1- \pi)^{1 - k}$$`

for `$k \in \{0,1\}$` and `$p_X(k) = 0$` otherwise

`$$Y \sim \text{Bernoulli}(\pi)$$`

---

# Bernoulli

Suppose we flip a fair coin and `$Y = 1$` if the outcome is Heads

$$
`\begin{aligned}
Y & \sim \text{Bernoulli}(1/2) \nonumber \\
p_X(1) & = (1/2)^{1} (1- 1/2)^{ 1- 1} = 1/2 \nonumber \\
p_X(0) & = (1/2)^{0} (1- 1/2)^{1 - 0} = (1- 1/2) \nonumber 
\end{aligned}`
$$

---

# Bernoulli PMF

<div class="figure" style="text-align: center">
<img src="10-discrete-random-vars_files/figure-html/bernoulli-1.png" alt="Example Bernoulli probability mass functions" width="864" />
<p class="caption">Example Bernoulli probability mass functions</p>
</div>

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# Binomial

A model to count the number of successes across `$N$` trials.

Suppose `$X$` is a random variable that counts the number of successes in `$N$` independent and identically distributed Bernoulli trials.  Then `$X$` is a **Binomial** random variable,

`$$p_X(k) = {N\choose{k}}\pi^{k} (1- \pi)^{n-k}$$`
    
for `$k \in \{0, 1, 2, \ldots, N\}$` and 0 otherwise

`$$\binom N{k} = \frac{N!}{(N-k)! k!}$$`

--
To say that a variable, `$Y$`, has or follows a binomial distribution we write: 
`$$Y \sim \text{Binomial}(N, \pi)$$`

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# Binomial PMF: trials
--
<img src="10-discrete-random-vars_files/figure-html/binomial-1.png" width="864" style="display: block; margin: auto;" />

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# Binomial

Recall our experiment example

* `$\Pr(T) = \Pr(C) = 1/2$`
* `$Z =$` number of units assigned to treatment

$$
`\begin{aligned}
Z & \sim  \text{Binomial}(1/2)\\
p_Z(0) & = {3\choose0} (1/2)^0 (1- 1/2)^{3-0} = 1 * \frac{1}{8}\\
p_Z(1) & = {3\choose1} (1/2)^1 (1 - 1/2)^{2} = 3 * \frac{1}{8} \\
p_Z(2) & = {3\choose2} (1/2)^2 (1- 1/2)^1 = 3 * \frac{1}{8} \\
p_Z(3) & = {3\choose3} (1/2)^3 (1 - 1/2)^{0} = 1 * \frac{1}{8}
\end{aligned}`
$$

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# Geometric

* A model to **count the number of trials** of a Bernoulli outcome **before success occurs the FIRST time**
--

* Suppose `$X$` is a random variable that counts the number of tosses needed for a head to come up the first time. Its PMF is

`$$p_X(k) = (1 - p)^{k-1}p, \quad k = 1, 2, \ldots$$`
    
--
We can then add up all these possibilities to ensure it satisfies our criteria for probability: 
$$
`\begin{aligned}
\sum_{k=1}^{\infty} p_X(k) &= \sum_{k=1}^{\infty} (1 - p)^{k-1}p \\
&= p \sum_{k=1}^{\infty} (1 - p)^{k-1} \\&
= p * \frac{1}{1 - (1-p)} \\
&= 1
\end{aligned}`
$$

---

# Geometric PMF

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# Poisson

* Often interested in counting number of events that occur
    * Number of wars started
    * Number of speeches made
    * Number of bribes offered
    * Number of people waiting for licenses
* Generally referred to as **event counts**

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# Poisson

Suppose `$X$` is a random variable that takes on values `$X \in \{0, 1, 2, \ldots, \}$` and that `$\Pr(X = k) = p_X(k)$` is,

`$$p_X(k) = e^{-\lambda} \frac{\lambda^{k}}{k!}, \quad k = 0,1,2,\ldots$$`
    
for `$k \in \{0, 1, \ldots, \}$` and `$0$` otherwise. (Here, `$\lambda$` is the mean number of events per year that we're interested in.)

`$$X \sim \text{Poisson}(\lambda)$$`

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# Poisson PMF

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# Example

* Suppose the number of threats a president makes in a term is given by `$X \sim \text{Poisson}(5)$`
* What is the probability the president will make ten threats?

`$$p_X(10) = e^{-5} \frac{5^{10}}{10!}$$`

---
# Example: working backwards

* Suppose we have an event which we believe follows a Poisson distribution. Perhaps this is the number of students who forget their calculator for the exam. If we observed that last year, about `$4.979\%$` of the time all students remembered their calculator. How could we figure out the likelihood that exactly one student forgets theirs in the upcoming exam?

`$$p_X(0) = e^{-\lambda} \frac{\lambda^{x}}{x!}$$`
--
`$$p_X(0) = e^{-\lambda} \frac{\lambda^{0}}{0!}=0.04979$$`
--
`$$0.04979 = e^{-\lambda}$$`
--
`$$log(0.0479) = - \lambda \, \text{ therefore } \lambda \approx 3.0$$`
---
# Poisson plot

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# Approximating a binomial RV

The Poisson PMF with parameter `$\lambda$` is a good approximation for a binomial PMF with parameters `$n$` and `$p$`
    
`$$e^{-\lambda} \frac{\lambda^{k}}{k!} \approx {N\choose{k}}\pi^{k} (1- \pi)^{n-k}, \quad \text{if } k \ll n$$`

* Assumes
    * `$\lambda = np$`
    * `$n$` is very large
    * `$p$` is very small

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# Approximating a binomial RV

* `$n = 100$` and `$p = 0.01$`

#### Using the binomial PMF

`$$\frac{100!}{95! 5!} * 0.01^5 (1 - 0.01)^{95} = 0.00290$$`

#### Using the Poisson PMF

* `$\lambda = np = 100 * 0.01 = 1$`

`$$e^{-1} \frac{1}{5!} = 0.00306$$`
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# Function recap:

* We're dealing here ONLY with **DISCRETE** functions
* Random variable takes on a limited number of values
* Functions offer a form for calculating probability
* Different assumptions for different types of functions

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# More fun with random variables
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# Functions of random variables

Given a random variable `$X$`, you may wish to create a new random variable `$Y$` using transformations of `$X$`

`$$Y = g(X) = aX + b$$`

`$$g(X) = \log(X)$$`

* If `$Y = g(X)$` is a function of a random variable `$X$`, then `$Y$` is also a random variable
* All outcomes in the sample space defined with a numerical value `$x$` for `$X$` also have a numerical value `$y = g(x)$` for `$Y$`

---

# Expectation, mean, and variance

* PMF of a random variable `$X$` provides probabilities of all possible values of `$X$`
* Often desirable to summarize this information into a single representative number
* **Expectation** of `$X$`

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# Motivation

* Consider spinning a wheel of fortune many times
* At each spin, one of the numbers `$m_1, m_2, \ldots, m_n$` comes up with corresponding probability `$p_1, p_2, \ldots, p_n$`, and this is your monetary reward from that spin
* What is the amount of money that you "expect" to get "per spin"?

* Spin `$k$` times
* `$k_i$` and `$m_i$`
* `$m_1 k_1 + m_2 k_2 + \ldots + m_n k_n$`

`$$M = \frac{m_1 k_1 + m_2 k_2 + \ldots + m_n k_n}{k}$$`

`$$\frac{k_i}{k} \approx p_i, i = 1, \ldots,n$$`

`$$M = \frac{m_1 k_1 + m_2 k_2 + \ldots + m_n k_n}{k} \approx m_1p_1 + m_2p_2 + \ldots + m_np_n$$`

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# Expectation

Random variable `$X$`, with PMF `$p_X$`

`$$\E[X] = \sum_{x:p(x)>0} x p(x)$$`

---

# Example of expected value

Suppose `$X$` is number of units assigned to treatment

What is `$\E[X]$`?

`$$\begin{aligned}\E[X]  & = 0* \frac{1}{8} + 1 * \frac{3}{8} + 2 * \frac{3}{8} + 3 * \frac{1}{8} \\& = 1.5 \end{aligned}$$`

* Weighted average
* Measure of central tendency

---

# A single person poll

Suppose that there is a group of `$N$` people

* Suppose `$M< N$` people approve of the president's performance
* `$N - M$` disapprove of their performance

Draw one person `$i$`, with `$\Pr(\text{Draw } i ) = \frac{1}{N}$`

`$$X  = \left \{ \begin{array} {ll}	1  \text{  if  person Approves}  \\    0 \text{  if  Disapproves}   \\\end{array} \right.$$`

What is `$\E[X]$`?

`$$\begin{aligned}\E[X] & = 1 * \Pr(\text{Approve})  + 0 * \Pr(\text{Disapprove}) \\ & = 1 * \frac{M}{N} \\ & = \frac{M}{N}  \end{aligned}$$`

---

# Indicator variables and probabilities

Suppose `$A$` is an event. Define random variable `$I$` such that `$I= 1$` if an outcome in `$A$` occurs and `$I =0$` if an outcome in `$A^{c}$` occurs.

Then,

`$$\E[I] = \Pr(A)$$`

`$$\begin{aligned}\E[I]  & =  1 * \Pr(A) + 0 * \Pr(A^{c}) \\ & = \Pr(A) \end{aligned}$$`

---

# Moments

* 1st moment: `$\E[X^1] = \E[X]$`
* 2nd moment: `$\E[X^2]$`
* `$n$`th moment: `$\E[X^n]$`

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# Variance

* Expected value is a measure of **central tendency**
* What about **spread** or **dispersion**?

For each value, measure distance from the center. Recall that `$E[X]$` is a SET VALUE.

`$$d(x, E[X])^{2} = (x - E[X])^2$$`

Weighted average of these distances
    
`$$\begin{aligned} \E[(X - \E[X])^2] & = \sum_{x:p_X(x)>0} (x  - \E[X])^2p_X(x)\end{aligned}$$`
--

`$$\sum_{x:p_X(x)>0} \left(x^2 p_X(x)\right)- 2 \E[X]\sum_{x:p_X(x)>0} \left(x p_X(x)\right) + \E[X]^2\sum_{x:p_X(x)>0} p_X(x)$$`
--

`$$=  \E[X^2] - 2\E[X]^2 + \E[X]^2 \\ = \E[X^2] - \E[X]^2 \\= \text{Var}(X)$$`

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# Variance

Expected value of the random variable `$(X - \E[X])^2$`

`$$\begin{aligned}\Var(X) &= \E[(X - \E[X])^2] \\&= \E[X^2] - \E[X]^2\end{aligned}$$`
    
* Measure of **dispersion** of `$X$` around its mean
* Standard deviation of `$X$` - `$\sigma_X = \sqrt{\Var(X)}$`

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# Calculating variance of a random variable

* Generate the PMF of the random variable `$(X - \E[X])^2$`
* Calculate the expectation of this function

### Expected value rule for functions of random variables

Let `$X$` be a random variable with PMF `$p_X$`, and let `$g(X)$` be a function of `$X$`. Then, the expected value of the random variable `$g(X)$` is given by

`$$\E[g(X)] = \sum_{x} g(x) p_X(x)$$`

`$$\begin{align}\Var(X) &= \E[(X - \E[X])^2] \\\Var(X) &= \E[X^2] - \E[X]^2\end{align}$$`
    
---

# Bernoulli variable

* Suppose `$Y \sim \text{Bernoulli}(\pi)$`

`$$\begin{aligned} \E[Y] & = 1 * \Pr(Y = 1) + 0 * \Pr(Y = 0) \nonumber \\& = \pi + 0 (1 - \pi) \nonumber  = \pi \\\Var(Y) & = \E[Y^2] - \E[Y]^2 \nonumber  \\\E[Y^2] & = 1^{2} \Pr(Y = 1) + 0^{2} \Pr(Y = 0) \nonumber \\& = \pi \nonumber \\ \Var(Y) & = \pi - \pi^{2} \nonumber \\& = \pi(1 - \pi ) \nonumber\end{aligned}$$`

* `$\E[Y] = \pi$`
* `$\Var(Y) = \pi(1- \pi)$`
* What is the maximum variance?

`$$\begin{aligned} \Var(Y) & = \pi(1- \pi) \nonumber \\& = 0.5(1 - 0.5 ) \\& = 0.25\end{aligned}$$`

---

# Binomial

`$$Z = \sum_{i=1}^N Y_{i} \text{ where } Y_{i} \sim \text{Bernoulli}(\pi)$$`

`$$\begin{aligned}\E[Z] & = \E[Y_{1} + Y_{2} + Y_3 + \ldots + Y_N ] \\& = \sum_{i=1}^N \E[Y_{i} ] \\& = N \pi \\\Var(Z) & = \sum_{i=1}^N \Var(Y_{i}) \\& = N \pi (1-\pi)\end{aligned}$$`

---

# Decision making using expected values

* Optimizes the choice between several candidate decisions that result in random rewards
* View the expected reward of a decision as its average payoff over a large number of trials
* Choose a decision with maximum expected reward

---

# Example: going to war

* Suppose country `$1$` is engaged in a conflict and can either win or lose
* Define `$Y = 1$` if the country wins and `$Y = 0$` otherwise

`$$Y \sim \text{Bernoulli}(\pi)$$`

* Suppose country `$1$` is deciding whether to fight a war
* Engaging in the war will cost the country `$c$`
* If they win, country `$1$` receives `$B$`
* What is `$1$`'s expected utility from fighting a war?

`$$\begin{aligned}\E[U(\text{war})] & = U(\text{war} | \text{win}) * \Pr(\text{win}) + U(\text{war} | \text{lose}) * \Pr(\text{lose}) \\& = (B - c) \Pr(Y = 1) + (- c) \Pr(Y = 0 ) \\& = B * \Pr(Y = 1)  - c(\Pr(Y = 1)  + \Pr(Y = 0)) \\& = B * \pi - c \end{aligned}$$`

---

# Cumulative mass function

* Defines the the cumulative probability `$F_X(x)$` up to the value of `$x$`
* For a discrete random variable `$X$`

`$$F_X(x) = \Pr (X \leq x) = \sum_{k \leq x} p_X(k)$$`

* All random variables with a PMF have a CMF

---

# Cumulative mass function

* `$F_X$` is monotonically non-decreasing - if `$x \leq y$`, then `$F_X(x) \leq F_X(y)$`
* `$F_X(x)$` tends to `$0$` as `$x \rightarrow -\infty$`, and to `$1$` as `$x \rightarrow \infty$`
* `$F_X(x)$` is a piecewise constant function of `$x$`
* If `$X$` is discrete and takes integer values, the PMF and the CMF can be obtained from each other by summing or differencing:

`$$F_X(k) = \sum_{i = -\infty}^k p_X(i),$$`
    `$$p_X(k) = \Pr (X \leq k) - \Pr (X \leq k-1) = F_X(k) - F_X(k-1)$$`
    
    for all integers `$k$`
    
---

# Bernoulli CMF

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# Binomial CMF

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# Binomial CMF

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# Binomial CMF

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# Binomial CMF

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# Geometric CMF

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# Poisson CMF

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# RECAP:

* Random variables: `$P(X=x)$`
* DISCRETE VARIABLES: **different from continuous**
* PMF: probability mass function
  * Probability for a particular value
* CDF: cumulative distribution function
  * ADD UP VALUES (akin to a sum)
* **Distributions**: 
  * Bernoulli: 
      * pmf: `$1-p$` for k = 0; `$p$` for `$k = 1$`
      * mean `$p$`, var `$p(1-p)$`
  * Binomial: 
      * pmf: `$p_X(k) = {N\choose{k}}\pi^{k} (1- \pi)^{n-k}$`
      * mean `$np$`, var `$np(1-p)$` (also written `$npq$`)
  * Geometric: 
      * pmf: `$p_X(k) = (1 - p)^{k-1}p, \quad k = 1, 2, \ldots$`
      * mean `$\frac{1}{p}$` , var `$\frac{(1-p)}{p^2}$`
  * Poisson: 
      * pmf: `$p_X(k) = e^{-\lambda} \frac{\lambda^{k}}{k!}, \quad k = 0,1,2,\ldots$`
      * mean `$\lambda$`, var `$\lambda$`