Discrete Univariate Distributions

Study guide: Exam P-1

A Discrete Univariate Distribution describes the probability law of a single Random Variable $X$ that takes on a countable set of values.

The probability mass function (PMF) $f(k)$ specifies the probability that the variable takes a particular distinct value $k$
The PMF must satisfy $f(k) \geq 0$ for all $k$ and $\sum_{k} f(k) = 1$ :

$P(X = k) = f(k), \quad k \in \{x_1, x_2, \ldots\}$

The cumulative distribution function (CDF) specifies the probability that the variable is at most $x$ :

$F(x) = P(X \leq x)$

$= \sum_{k \leq x} f(k)$

Choosing the right one

Most of the difficulty is identification, not computation. Read the question for what is being counted and whether the trial count is fixed:

Successes in a fixed number of independent trials → Binomial Distribution $(n,p)$ , with $E[X] = np$
Successes in a fixed number of draws taken without replacement from a finite pool → Hypergeometric Distribution $(N,K,n)$ , with $E[X] = nK/N$
Trials until the first success, the trial count not fixed in advance → Geometric Distribution $(p)$ , with $E[X] = 1/p$
Trials until the $r$ -th success → Negative Binomial Distribution $(r,p)$ , with $E[X] = r/p$
Events occurring over a fixed interval of time or space, with no trial count at all → Poisson Distribution $(\lambda)$ , with $E[X] = \lambda$
One of the integers $1, \ldots, n$ , each equally likely → Discrete Uniform, with $E[X] = (n+1)/2$ (on $a, \ldots, b$ the mean is $(a+b)/2$ )
Binomial vs. hypergeometric turns entirely on replacement. When the population is large relative to the sample the two nearly agree, and the binomial is the intended shortcut.
Binomial vs. Poisson: a fixed number of trials points to binomial; "per year", "per hour", "per 100 policies" with no trial count points to Poisson.
The variance-to-mean ratio is a fast identification check: it is $< 1$ for binomial, $= 1$ for Poisson, and $> 1$ for the negative binomial counted as failures before the $r$ -th success ( $\text{Var}/E = 1/p$ ). The trials count above has ratio $(1-p)/p$ , below 1 whenever $p > 1/2$ , so apply the test to the failures form.
For "at least one", reach for the complement $1 - P(X=0)$ rather than summing the tail.

PMF Verification for a Simple Discrete Distribution A random variable $X$ has PMF $f(k) = c \cdot k$ for $k = 1, 2, 3, 4$ . Find $c$ and compute $P(X \leq 3)$ .

Answer For $f$ to be a valid PMF we need $\sum_{k=1}^{4} c \cdot k = 1$ , so $c(1 + 2 + 3 + 4) = 10c = 1 \implies c = \frac{1}{10}$ Then $P(X \leq 3) = F(3) = \frac{1}{10}(1+2+3) = \frac{6}{10} = 0.6$