Discrete Random Variables

01 / UNDERSTAND & APPLY

1. A random variable maps outcomes to numbers

A random variable X assigns a numerical value to each outcome of a probability experiment. If two coins are tossed, the outcomes HH,HT,TH,TT are not themselves numbers. The variable X = number of heads maps them to 2,1,1,0. Different outcomes can produce the same value, so the probabilities of those outcomes must be added when constructing the distribution of X.

A discrete random variable takes finitely many or countably many values. Its probability mass function, or pmf, is p(x) = P(X = x). A valid pmf has p(x) ≥ 0 for every value and Σp(x) = 1. We use the term pmf to distinguish these point masses from a continuous probability density, even though some elementary texts call both a probability distribution function.

The support is the set of values with positive probability. It is not necessarily an interval and need not contain the mean. A variable counting successes has integer support, but its expectation can be fractional. For example, an average of 1.5 successes over many repetitions does not mean a single experiment can produce half a success.

When a formula for p(x) includes an unknown constant, first normalize the total and then check nonnegativity. A constant that makes the probabilities sum to one can still produce negative probabilities if the proposed formula or parameter range is invalid. Both requirements are essential.

02 / UNDERSTAND & APPLY

2. Cumulative probability is a step function

The cumulative distribution function is F(t) = P(X ≤ t). For a discrete variable, it adds the masses at values no greater than t. It is nondecreasing and right-continuous, with limits zero at the far left and one at the far right. A jump at x has size P(X = x). The distinction between ≤ and < matters exactly at a support value.

For the distribution on 0,1,2 with masses 1/6,1/3,1/2, F(t) is 0 for t < 0, 1/6 for 0 ≤ t < 1, 1/2 for 1 ≤ t < 2, and 1 for t ≥ 2. A graph uses horizontal segments and open/closed endpoint markers to show which step contains each support value.

If Y = g(X), several X-values may map to the same Y-value. To find P(Y = y), add P(X = x) over every x satisfying g(x) = y. Squaring a variable is a common example: x and −x have the same square. Transforming the values without combining their probabilities can produce a duplicated or incorrect distribution.

For an interval event, use the endpoint convention carefully. P(a < X ≤ b) = F(b) − F(a). If the left endpoint is included, subtract P(X < a) instead. On integer support, P(a ≤ X ≤ b) can often be written F(b) − F(a − 1) when a and b are integers.

03 / UNDERSTAND & APPLY

3. Expectation is a probability-weighted average

For a finite discrete variable, E[X] = Σx p(x). Each possible value is weighted by its probability. If an infinite distribution is used, the corresponding convergence conditions matter; for a finite expectation of a signed variable, absolute convergence is the standard safe condition. All numerical examples here have finite support or a geometric distribution with finite moments.

For a function g, E[g(X)] = Σg(x)p(x). In general this is not g(E[X]). Squaring an average differs from averaging squares unless the variable has no variability. This distinction will lead directly to the variance formula.

Expectation is linear: E[aX + bY + c] = aE[X] + bE[Y] + c whenever the relevant expectations exist. Independence is not needed. To see why, expand the weighted sum over the joint outcomes and distribute addition. Independence is a sufficient condition for the product identity E[XY] = E[X]E[Y] when the expectations are finite; it is not necessary for that equality and is not needed for the basic sum rule.

An indicator I_A is 1 when event A occurs and 0 otherwise. Its expectation is P(A). Writing a count as a sum of indicators often gives its mean without first finding the full distribution. For example, the expected number of red balls in n draws from an urn with red fraction R/N is nR/N even without replacement, because every draw position has the same marginal red probability.

04 / UNDERSTAND & APPLY

4. Variance measures spread around the mean

Variance is Var(X) = E[(X − μ)²], where μ = E[X]. Expanding the square and using linearity gives Var(X) = E[X²] − μ². It is nonnegative because the defining squared deviation is nonnegative. A negative computed variance signals arithmetic or modelling error, apart from possible tiny numerical rounding in machine calculations.

For constants a,b, E[aX + b] = aE[X] + b and Var(aX + b) = a²Var(X). Adding b shifts all outcomes and their mean together, leaving deviations unchanged. Multiplying by a multiplies each deviation by a, so its square gains a². Standard deviation scales by |a|, not by a when a is negative.

For two variables, Var(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y), where Cov(X,Y) = E[XY] − E[X]E[Y]. Independent variables have zero covariance, so their variances add. Without independence or another zero-covariance argument, omitting the covariance term is unjustified.

A simple warning example is Y = X. Then Var(X + Y) = Var(2X) = 4Var(X), not 2Var(X). The variables are perfectly linked. By contrast, E[X + Y] = 2E[X] still holds, illustrating again that expectation linearity is less demanding than variance addition.

05 / UNDERSTAND & APPLY

5. Four conditions create a binomial count

A binomial experiment has a fixed number n of trials, two categories called success and failure on each trial, a constant success probability p, and independence across trials. The words success and failure are labels, not judgments of desirability. The random variable X counts successes, so its possible count values are 0,1,…,n; these form its support when 0<p<1.

One specified sequence with r successes and n−r failures has probability pʳ(1−p)^(n−r) by independence. There are C(n,r) ways to choose the success positions. These sequences are disjoint, so adding them gives P(X=r) = C(n,r)pʳ(1−p)^(n−r). The binomial theorem confirms that all these probabilities sum to (p + 1−p)ⁿ = 1.

For p=0 or p=1, the distribution is degenerate at 0 or n. Handle these cases directly instead of relying on ambiguous-looking powers such as 0⁰ in an endpoint formula. For 0<p<1, every support value has positive probability.

Trials can have more than two physical outcomes and still produce a binomial count if each is classified into success versus not-success. Rolling a fair die n independent times and counting sixes has p=1/6. But sampling without replacement from a finite urn usually violates independence and constant conditional probability, so the exact count distribution is hypergeometric rather than binomial.

06 / UNDERSTAND & APPLY

6. Mean, variance and the most likely count are different summaries

Write a binomial count as X = I₁ + ⋯ + Iₙ, where each indicator is one on success. Each has mean p and variance p(1−p). Linearity gives E[X]=np. Independence makes covariances zero, giving Var(X)=np(1−p). The standard deviation is √[np(1−p)]. These derivations explain why the moment formulas depend on the trial assumptions.

The mean need not be an integer and need not be the most likely count. The mode maximizes the pmf over integer r. For 0<p<1, compare adjacent probabilities: P(X=r+1)/P(X=r) = [(n−r)/(r+1)]p/(1−p). The probabilities rise while this ratio exceeds one and fall after it drops below one.

If (n+1)p is not an integer, the unique mode is floor((n+1)p). If (n+1)p is an integer m, the two modes are m−1 and m. The boundary cases p=0 or p=1 are handled separately. This is an exact discrete statement; rounding the mean np can fail to identify a tie.

As p moves from zero toward one-half for fixed n, the variance grows; beyond one-half it decreases symmetrically. The uncertainty is largest at p=1/2, not at the largest mean. A near-certain success count can have a high mean but small spread.

07 / UNDERSTAND & APPLY

7. How many trials are needed for a target chance?

For independent trials with success probability p, the probability of no successes in n trials is (1−p)ⁿ. Therefore the chance of at least one is 1−(1−p)ⁿ. If a target probability q is specified with 0<p<1 and 0<q<1, solve (1−p)ⁿ ≤ 1−q.

Taking logarithms gives n ln(1−p) ≤ ln(1−q). Since ln(1−p) is negative, dividing reverses the inequality: n ≥ ln(1−q)/ln(1−p). The smallest allowed integer n is the ceiling of that ratio, followed by a direct check of n and n−1. Floating-point rounding near an integer boundary should not decide the answer by itself.

At least r successes is a binomial tail sum, often easier through its complement of fewer than r successes. Exactly r is a single pmf value. At most r includes the endpoint r. These word distinctions determine whether a term is included in the sum and cannot be repaired by using the same formula with an approximate threshold.

A fixed probability p and independence are essential for the exponential no-success expression. If probabilities differ across independent trials, no-success probability is the product of their individual failure probabilities. If trials are dependent, conditional failure probabilities must be used instead.

08 / UNDERSTAND & APPLY

8. A waiting time is not a fixed-trial count

Let T be the trial number of the first success in independent trials with constant p, where 0<p≤1. For 0<p<1, T has support 1,2,3,… and P(T=k)=(1−p)^(k−1)p. For p=1, T=1 with probability one, so its support is only {1}. The event requires k−1 failures followed by a success in that specified order, so there is no binomial coefficient.

The tail probability is P(T>n)=(1−p)ⁿ for nonnegative integer n. Consequently the geometric distribution is memoryless: P(T>m+n|T>m)=(1−p)ⁿ, provided the conditioning probability is positive. After observing m failures, the chance of another n failures is unchanged because the future trials remain independent with the same p.

The mean waiting time is E[T]=1/p and the variance is (1−p)/p². The mean follows, for example, from the tail-sum identity E[T]=Σₙ₌₀∞P(T>n)=Σₙ₌₀∞(1−p)ⁿ. The geometric series converges for p>0. If p=0, a success never occurs under the model and there is no finite waiting-time mean.

Some conventions define a geometric variable as the number of failures before the first success, starting at zero. That variable is T−1, with mean (1−p)/p and the same variance. State which convention you use before quoting a formula. Also, the first success on trial k is not the same event as exactly one success among k trials.

09 / UNDERSTAND & APPLY

9. Original practice: pmfs and moments

Normalize first, then calculate moments. When a variable is transformed, either use a justified moment identity or combine probabilities for equal transformed values.

10 / UNDERSTAND & APPLY

10. Original practice: count or wait?

Identify whether the experiment fixes the number of trials or stops at an event. Then check independence and whether the success probability remains constant.

11 / UNDERSTAND & APPLY

11. Match the distribution to the experiment

A random variable is a numerical map, its pmf assigns probability masses, its CDF accumulates those masses, its expectation is a weighted average, and its variance measures squared spread. These objects answer different questions. An expected count need not be a possible count, and a most likely count need not equal the expectation.

For repeated trials, ask four questions before saying binomial: is n fixed, are outcomes classified into two categories, is p constant, and are trials independent? For a first-success waiting time, use the geometric model and state whether you count trials or failures. Without replacement, inspect dependence rather than applying a familiar formula by appearance.

The common errors are forgetting to combine transformed values, replacing E[X²] by E[X]², adding variances without a dependence check, omitting a binomial coefficient for an unordered success count, adding one to a waiting-time formula under the wrong convention, and reversing a logarithmic inequality incorrectly.

Final checks include total probability one, nonnegative variance, a mean within finite support bounds, and exact enumeration for a small case. For a minimum-trials answer, verify both the claimed minimum and the preceding integer. These checks test the boundary cases where otherwise correct formulas are most often misapplied.