Why the Choice of Random Variable Matters
A multi-stage experiment can produce several different quantities of interest. A device might first pass or fail an inspection and, if it passes, then receive a continuous measurement. The number of successes in a sequence of attempts is different from the value recorded on the final attempt. Before calculating a probability or moment, specify which quantity is being recorded and which values that quantity can take.
The support is not a decorative detail. It identifies impossible values, reveals whether a model is discrete, continuous, or mixed, and provides a check on later calculations. A formula can be evaluated correctly and still answer the wrong question if the random variable or its support was misread.
In a discrete model, support values are the values with positive probability. In a continuous model, the support describes the region over which the variable can occur; a single point usually has probability zero. A variable is mixed when it has both one or more point probabilities and a continuous part. When describing a support, include only values the model permits, not every value between the minimum and maximum if there are gaps.
From Experiment Outcomes to a Variable and Its Support
Let \(\Omega\) be the set of possible outcomes of an experiment. Once the recorded quantity is defined as \(X=g(\omega)\), the support is determined by the outcomes that can produce each value. To find it, work forward through the experiment: list the possible branches, apply the rule defining \(X\) on each branch, and combine the resulting sets of values. For a continuous stage, include the full attainable interval, including its endpoints when appropriate.
The link between an event about \(X\) and an event about the experiment is exact:
This identity explains the basic method. First identify which experiment outcomes make the stated condition on \(X\) true; then calculate the probability of those outcomes. If the experiment has distinct stages or branches, use their stated probabilities and conditional rules rather than assuming independence.
State in words what one value of \(X\) represents, including the units if relevant.
Record what can happen at each stage and which later stages occur on each branch.
Apply the definition of \(X\) separately to each branch and combine the attainable values.
Confirm that impossible values are excluded and that the probabilities or moments use the right branch conditions.
Worked Examples
Worked Example: A Discrete Count From Sequential Draws
A container holds 3 red and 2 blue tokens. Two tokens are drawn sequentially without replacement. Let \(X\) be the number of red tokens drawn. Find the support, the probability that \(X=1\), and \(E[X]\) and \(\operatorname{Var}(X)\).
Method and assumptions. Each pair of tokens is selected without replacement, so the two colors are dependent. Each of the \(\binom{5}{2}=10\) unordered pairs is equally likely. Since two tokens are drawn, \(X\) can only be 0, 1, or 2; each value is attainable. Thus the support is \(\{0,1,2\}\).
There is 1 way to select 0 red tokens: choose both blue tokens. There are \(\binom{2}{2}=1\) such pairs, so \(P(X=0)=1/10\). There are \(\binom{3}{2}=3\) pairs with 2 red tokens, so \(P(X=2)=3/10\). For 1 red token, choose 1 of the 3 red and 1 of the 2 blue tokens, giving \(3\cdot2=6\) pairs and \(P(X=1)=6/10\).
Therefore \(P(X=1)=0.6\). The expectation and second moment are
Using \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\), \(\operatorname{Var}(X)=1.8-(1.2)^2=1.8-1.44=0.36\). As a check, the probabilities sum to 1, the mean 1.2 lies between 0 and 2, and the variance is positive.
Common mistake avoided: Treating the draws as independent with replacement would use the wrong model and probability for two red tokens. The problem specifies no replacement; the counts of token pairs above respect that restriction.
Worked Example: Support of a Continuous Two-Stage Quantity
A sensor generates two independent readings \(U\) and \(V\), each uniformly distributed on \([0,1]\). Let \(T=U+V\). Find the support, \(P(T\leq1)\), \(E[T]\), and \(\operatorname{Var}(T)\).
Method and assumptions. Both readings can take any value from 0 to 1, and they are independent. The smallest possible sum is \(0+0=0\); the largest is \(1+1=2\). Every value between these extremes is attainable, so the support of \(T\) is \([0,2]\).
The event \(T\leq1\) consists of points \((u,v)\) in the unit square satisfying \(u+v\leq1\). The square has area 1. The qualifying region is a right triangle with base 1 and height 1, so its area is \(1/2\). The joint density is 1 throughout the unit square, and hence \(P(T\leq1)=1/2\).
For a uniform variable on \([0,1]\), \(E[U]=E[V]=1/2\) and \(\operatorname{Var}(U)=\operatorname{Var}(V)=1/12\). By linearity of expectation, \(E[T]=E[U]+E[V]=1\). Independence gives \(\operatorname{Cov}(U,V)=0\), so
The check is consistent: \(T\) lies between 0 and 2, its mean is 1, and the probability of the lower half of its support is \(1/2\) by symmetry of the two readings around their combined midpoint. The method uses the support of a sum; it does not assume that \(T\) is uniform.
Common mistake avoided: Adding the two ranges correctly gives \([0,2]\), but it does not make the sum uniform on that interval. The probability calculation uses the area of the appropriate region in the joint support, not a uniform model for \(T\).
Worked Example: A Mixed Variable With a Gap in Its Support
A hypothetical screening procedure rejects an item with probability \(0.25\). If the item is rejected, its recorded score is \(X=0\). If accepted, which occurs with probability \(0.75\), a number \(U\) is generated uniformly on \([0,4]\), independently of the screening outcome, and the score is \(X=2+U\). Find the support, \(P(X\leq3)\), and \(E[X]\).
Method and assumptions. The rejection branch produces the single value 0. The acceptance branch produces scores from 2 through 6. Thus
The score has a point probability \(P(X=0)=0.25\) and a continuous range on \([2,6]\), so it is mixed. Values strictly between 0 and 2 cannot occur; the interval \([0,6]\) would incorrectly include them.
For \(X\leq3\), rejection always qualifies. On acceptance, \(2+U\leq3\) is equivalent to \(U\leq1\). Since \(U\) is uniform on an interval of length 4, \(P(U\leq1)=1/4\). Therefore
The conditional mean score on acceptance is \(E[2+U]=2+2=4\), because a uniform variable on \([0,4]\) has mean \((0+4)/2=2\). Rejection contributes score 0. Consequently,
As a check, \(0\leq E[X]\leq6\), and the event probability includes the full rejection probability plus only one quarter of the acceptance branch. The calculation follows the two branches in the problem statement rather than treating all scores between 0 and 6 as equally likely.
Common mistake avoided: Using the entire interval \([0,6]\) as the support loses the gap and hides the atom at zero. The support must preserve both the isolated value and the continuous interval.
Why Support Errors Change the Answer
Support is a constraint on every calculation involving a random variable. For a discrete variable, the probabilities assigned to its support values must be nonnegative and sum to 1. Its expectation is calculated by weighting only those values by their probabilities. For a continuous variable with density \(f_X\), probabilities are integrals over the portion of its support that satisfies the event. A value outside the support contributes no probability.
A useful justification for computing probabilities by translating events is simply that the two events contain exactly the same experiment outcomes. For example, if \(X=g(\omega)\), then an outcome belongs to \(\{X\leq c\}\) precisely when \(g(\omega)\leq c\). Therefore the probability can be found from the experiment’s outcome model, even if \(X\) itself is not a basic stage of that experiment.
This reasoning also explains why assumptions matter. Independence permits a product rule for stage probabilities; it cannot be presumed just because stages are described separately. Sampling without replacement changes the probabilities at later draws. A conditional stage occurs only on the branch where its condition is met. State these details before using a calculation.
Common Traps and a Final Check
- Defining the experiment but not the recorded variable. “Two tokens are drawn” does not say whether \(X\) is the number of red tokens, the color of the first token, or a numerical score.
- Using a range that includes impossible values. The smallest and largest attainable values do not guarantee that every value between them is possible.
- Calling a continuous interval’s endpoint impossible. A continuous variable can have an endpoint in its support even though the probability of equaling that endpoint is zero.
- Assuming independence between stages. Check the experiment’s rules. Sequential draws without replacement are dependent.
- Using a continuous calculation at a point mass. A continuous density alone does not account for a separate positive probability at a single value.
Check Your Understanding
For each problem, identify the support before calculating the requested quantity.
- A fair six-sided die is rolled, followed by a fair coin. Let \(X\) equal the die result if the coin is heads and 0 if it is tails. Find the support and \(P(X\geq4)\). (Answer: Support \(\{0,1,2,3,4,5,6\}\); \(P(X\geq4)=0.2500\).)
- Two independent readings are uniform on \([0,2]\) and \([1,3]\), respectively. Let \(Z\) be their sum. Find its support and \(P(Z\leq3)\). (Answer: Support \([1,5]\); \(P(Z\leq3)=0.5000\).)
- A box contains 4 green and 3 yellow tiles. Two are selected without replacement, and \(N\) is the number of green tiles selected. Find the support and \(P(N=2)\). (Answer: Support \(\{0,1,2\}\); \(P(N=2)=2/7=0.2857\).)
- A hypothetical test produces a score of 1 with probability 0.20. Otherwise, with probability 0.80, it produces \(X=3+U\), where \(U\) is uniform on \([0,2]\). Find the support and \(P(X\leq4)\). (Answer: Support \(\{1\}\cup[3,5]\); \(P(X\leq4)=0.6000\).)
- Two independent uniform values \(A\) and \(B\) are generated on \([0,1]\). A system records \(Q=2A+B\). Find the support and \(E[Q]\). (Answer: Support \([0,3]\); \(E[Q]=1.5\).)