Discrete Probability Distributions

Precise definition

A discrete random variable $X$ has a probability mass function $p(x)=P(X=x)$ on a finite or countable support, with $p(x)\ge0$ and $\sum_xp(x)=1$. Its cumulative distribution is $F(x)=P(X\le x)=\sum_{t\le x}p(t)$.

Notation and mathematical language

The expectation is $E[X]=\sum xp(x)$ when absolutely convergent; variance is $E[(X-\mu)^2]=E[X^2]-\mu^2$. Standard deviation is the nonnegative square root. Expectation is a long-run or probability-weighted centre, not necessarily a possible outcome.

Conceptual picture

A mass function places probability at individual support points. Bar heights are probabilities, while cumulative values add all mass to the left. Transformations use $E[g(X)]=\sum g(x)p(x)$ without needing the distribution of $g(X)$ first.

Conditions and key results

Every listed probability and total must be valid. Infinite-series formulas need convergence. A model such as binomial, geometric, or Poisson carries additional assumptions; matching a support shape does not establish them.

A reliable strategy

1. List the support and define what one random outcome means.
2. Verify nonnegative masses sum to 1; solve any unknown normalizing constant.
3. Compute requested events by summing exactly the relevant points.
4. Calculate expectation and variance, then interpret units and assess whether model assumptions fit the process.

Fully worked example

Interpretation and application

Discrete distributions model counts, claims, arrivals, and games. Expected profit summarizes repeated risk under a model; it does not guarantee any one outcome or capture risk preference by itself.

Common mistakes

Verification and reasonableness

• Sum the mass function and event complements.
• Compute variance by both $E[(X-\mu)^2]$ and $E[X^2]-\mu^2$.
• Check units: variance has squared units and standard deviation original units.

Practice

1. If $P(X=0)=0.4$ and $P(X=1)=0.6$, find $E[X]$.
2. What must a PMF sum to?
3. Can an expectation be outside the support points?

Further deduction

For independent discrete variables, $E[X+Y]=E[X]+E[Y]$ even without independence, while $\operatorname{Var}(X+Y)=\operatorname{Var}X+\operatorname{Var}Y$ requires zero covariance and is guaranteed by independence. Keeping those hypotheses separate prevents a common mistake when aggregating risk.

A geometric distribution models the trial number of the first success under independent Bernoulli trials with constant success probability $p$. Its mass is $(1-p)^{k-1}p$ for $k=1,2,\ldots$ and mean $1/p$. If the success chance changes or trials depend on history, the geometric formula is not justified even when outcomes are discrete.

A probability generating function $G(s)=E[s^X]=\sum_{k\ge0}P(X=k)s^k$ packages a nonnegative integer-valued distribution. Where differentiation is justified, $G'(1)=E[X]$ and $G''(1)=E[X(X-1)]$. Independent sums multiply generating functions, turning convolution into algebra. The device is powerful but its domain and differentiability conditions matter; it supplements rather than replaces checking that probabilities are nonnegative and total one.

Related topics

• Random Variables
• Expected Value
• Variance