Discrete random variables

\[\newcommand{\E}{\mathrm{E}} \newcommand{\Var}{\mathrm{Var}} \newcommand{\Cov}{\mathrm{Cov}}\]

Basic concepts

• Random variable
• Function of a random variables
• Averages of interest
• Conditioning
• Independence

Random variable

• A random process or variable with a numerical outcome

• A random variable \(X\) that is a function of the sample space

  \[ \begin{eqnarray} X:\text{Sample Space} \rightarrow \mathcal{R} \end{eqnarray} \]

• 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)

Examples of random variables

• \(X\) = 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)

Discrete random variables

• A random variable with a finite or countably infinite range
• A random variable with an uncountably infinite number of values

Probability mass functions

• Probability of the values that it can take
• Probability mass function (PMF)

Probability mass functions

• \(P(C, T, C) = P(C)P(T)P(C) = \frac{1}{2}\frac{1}{2}\frac{1}{2} = \frac{1}{8}\)

• This applies to all outcomes

  \[ \begin{eqnarray} p(X = 0) & = & P(C, C, C) = \frac{1}{8}\\ p(X = 1) & = & P(T, C, C) + P(C, T, C) + P(C, C, T) = \frac{3}{8} \\ p(X = 2) & = & P(T, T, C) + P(T, C, T) + P(C, T, T) = \frac{3}{8} \\ p(X = 3) & = & P(T, T, T) = \frac{1}{8} \end{eqnarray} \]

• \(p(X = a) = 0\), for all \(a \notin (0, 1, 2, 3)\)

Probability mass functions

Probability mass functions

• Probability Mass Function: For a discrete random variable \(X\), define the probability mass function \(p_X(x)\) as

  \[ \begin{eqnarray} p(x) & = & P(X = x) \end{eqnarray} \]
• Upper vs. lower-case letters

• Note that

  \[\sum_x p_{X}(x) = 1\]

Probability mass functions

• Can also add probabilities for smaller sets \(S\) of possible values of \(XS\)

  \[\Pr(X \in S) = \sum_{x \in S} p_X (x)\]

• For example, if \(X\) is the number of heads obtained in two independent tosses of a fair coin, the probability of at least one head is

  \[\Pr (X > 0) = \sum_{x=1}^2 p_X (x) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}\]

Calculate the PMF of a random variable \(X\)

• For each possible value \(x\) of \(X\)
  1. Collect all possible outcomes that give rise to the event \(\{ X=x \}\)
  2. Add their probabilities to obtain \(p_X (x)\)

Bernoulli random variable

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

  \[ \begin{eqnarray} p_X(k) & = & \pi^{k} (1- \pi)^{1 - k} \nonumber \end{eqnarray} \]

  for \(k \in \{0,1\}\) and \(p(k) = 0\) otherwise.

• We will (equivalently) say that

  \[ \begin{eqnarray} Y & \sim & \text{Bernoulli}(\pi) \nonumber \end{eqnarray} \]

Bernoulli random variable

• Suppose we flip a fair coin and \(Y = 1\) if the outcome is Heads.

  \[ \begin{eqnarray} Y & \sim & \text{Bernoulli}(1/2) \nonumber \\ p(1) & = & (1/2)^{1} (1- 1/2)^{ 1- 1} = 1/2 \nonumber \\ p(0) & = & (1/2)^{0} (1- 1/2)^{1 - 0} = (1- 1/2) \nonumber \end{eqnarray} \]

Bernoulli random variable

Binomial random variable

• 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,

  \[ \begin{eqnarray} p_X(k) & = & {{N}\choose{k}}\pi^{k} (1- \pi)^{n-k} \nonumber \end{eqnarray} \]

  for \(k \in \{0, 1, 2, \ldots, N\}\) and \(p(k) = 0\) otherwise, and \(\binom{N}{k} = \frac{N!}{(N-k)! k!}\). Equivalently,

  \[ \begin{eqnarray} Y & \sim & \text{Binomial}(N, \pi) \nonumber \end{eqnarray} \]

Binomial random variable

• Recall our experiment example
• \(P(T) = P(C) = 1/2\)

• \(Z =\) number of units assigned to treatment

  \[ \begin{eqnarray} Z & \sim & \text{Binomial}(1/2)\\ p(0) & = & {{3}\choose{0}} (1/2)^{0} (1- 1/2)^{3-0} = 1 \times \frac{1}{8}\\ p(1) & = & {{3}\choose{1}} (1/2)^{1} (1 - 1/2)^{2} = 3 \times \frac{1}{8} \\ p(2) & = & {{3}\choose{2}} (1/2)^{2} (1- 1/2)^1 = 3 \times \frac{1}{8} \\ p(3) & = & {{3}\choose{3}} (1/2)^{3} (1 - 1/2)^{0} = 1 \times \frac{1}{8} \end{eqnarray} \]

Poisson random variable

• 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 license
• Generally referred to as event counts

Poisson random variable

• 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,

  \[ \begin{eqnarray} p_X(k) & = & e^{-\lambda} \frac{\lambda^{k}}{k!}, \quad k = 0,1,2,\ldots \end{eqnarray} \]

  for \(k \in \{0, 1, \ldots, \}\) and \(0\) otherwise.

• \(X\) follows a Poisson distribution with rate parameter \(\lambda\)

  \[ \begin{eqnarray} X & \sim & \text{Poisson}(\lambda) \nonumber \end{eqnarray} \]

Poisson random variable

• 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 or more threats?

  \[ \begin{eqnarray} P(X = 10) & = & e^{-\lambda} \frac{5^{10}}{10!} \end{eqnarray} \]

Poisson random variable

Approximating a binomial random variable

• The Poisson PMF with parameter \(\lambda\) is a good approximation for a binomial PMF with parameters \(n\) and \(p\)

  \[e^{-\lambda} \frac{\lambda^{k}}{k!} \approx {{N}\choose{k}}\pi^{k} (1- \pi)^{n-k}, \quad \text{if } k \ll n\]

  • Provided \(\lambda = np\), \(n\) is very large, and \(p\) is very small

• Sometimes using the Poisson PMF results in simpler models and easier calculations

Approximating a binomial random variable

• \(n = 100\) and \(p = 0.01\)

• Using the binomial PMF

  \[\frac{100!}{95! 5!} \times 0.01^5 (1 - 0.01)^{95} = 0.00290\]

• Using the Poisson PMF with \(\lambda = np = 100 \times 0.01 = 1\)

  \[e^{-1} \frac{1}{5!} = 0.00306\]

Functions of random variables

• Given a random variable \(X\), you may wish to create a new random variable \(Y\) using transformations of \(X\)
• Linear transformation \(Y = g(X) = aX + b\)
• Logarithmic transformation \(g(X) = \log(X)\)
• 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 several numbers, the probabilities of all possible values of \(X\)
• Often desirable to summarize this information into a single representative number
• Expectation of \(X\) - weighted average of the possible values of \(X\)

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?

Expectation

• Expected value (known as expectation or the mean) of a random variable \(X\), with PMF \(p_X\) is

  \[ \begin{eqnarray} \E[X] & = & \sum_{x:p(x)>0} x p(x) \end{eqnarray} \]

  where \(\sum_{x:p(x)>0}\) is all values of \(X\) with probability greater than 0

Moments

• 1st moment: \(\E[X^1] = \E[X]\)
• 2nd moment: \(\E[X^2]\)
• \(n\)th moment: \(\E[X^n]\)

Variance

• \(\Var(X)\)

• Defined as the expected value of the random variable \((X - \E[X])^2\)

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

  • Since \((X - \E[X])^2\) can only take non-negative values, variance is always non-negative
• Measure of dispersion of \(X\) around its mean

• We will define the standard deviation of \(X\), \(\sigma_X = \sqrt{\Var(X)}\)

Calculating variance of a random variable

• We could generate the PMF of the random variable \((X - \E[X])^2\), then 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

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

• Rewrite our variance formula:

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

Binomial

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

\[ \begin{eqnarray} E[Z] & = & E[Y_{1} + Y_{2} + Y_{3} + \ldots + Y_{N} ] \\ & = & \sum_{i=1}^{N} E[Y_{i} ] \\ & = & N \pi \\ \text{var}(Z) & = & \sum_{i=1}^{N} \text{var}(Y_{i}) \\ & = & N \pi (1-\pi) \end{eqnarray} \]

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

Cumulative distribution function

• Defines the the cumulative probability \(F_X(x)\) up to the value of \(x\)

• For a discrete random variable \(X\), \(F_X\) provides the probability \(\Pr (X \leq x)\). For every \(x\)

  \[F_X(x) = \Pr (X \leq x) = \sum_{k \leq x} p_X(k)\]

• All PMFs have a CDF

Cumulative distribution function

• \(F_X\) is monotonically nodecreasing – 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 CDF 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 \(k\)