Random Variables

A random variable assigns a number to an outcome

A random variable X maps outcomes of a random process to numerical values. It is not the list of numbers already observed. For three coin tosses, X might count heads; different toss sequences can give the same X.

A discrete variable has a finite or countably infinite set of possible values. A continuous model assigns probability through a density over intervals. Choose the model for the measurement and process, not simply because the spreadsheet contains decimals.

Worked example: a hypothetical reward

Suppose a random reward X in credits has this stated distribution:

x	P(X = x)	xP(X = x)	x²P(X = x)
0	0.50	0	0
2	0.30	0.60	1.20
5	0.20	1.00	5.00
Total	1.00	1.60	6.20

The probability mass function gives the probability at each discrete value. Its entries are nonnegative and sum to 1.

$E[X]=\sum_x xP(X=x)=1.6.$ $\operatorname{Var}(X)=E[X^2]-E[X]^2=6.2-1.6^2=3.64.$

The standard deviation is √3.64 ≈ 1.9079 credits. The expected value 1.6 is a probability-weighted average, not a possible reward on one trial and not a guaranteed result. Under suitable independent repeated trials, the average reward tends toward it.

The CDF answers “at most” questions

The cumulative distribution function F(x) = P(X ≤ x) works for both discrete and continuous variables. For the reward distribution, F(2) = 0.50 + 0.30 = 0.80. Between 2 and 5 it stays at 0.80; at 5 it jumps to 1.

For an interval with an open lower bound:

$P(a<X\le b)=F(b)-F(a).$

For discrete outcomes, changing < to ≤ can change the answer. Here P(X < 2) = 0.50, whereas P(X ≤ 2) = 0.80.

Density is not probability at a point

For a continuous model with density f, interval probability is the area under f over that interval. A density value may exceed 1 because it has units of inverse measurement; only total probability must be 1. Under an absolutely continuous model, an exact value has probability 0.

For a uniform variable on [0, 0.5], the density is 2 on its support. The probability between 0.1 and 0.2 is 2 × 0.1 = 0.2. Density 2 is valid and does not mean a 200% probability.

Practice

Using the reward distribution, find P(X > 0) and E[2X + 3].

Show the explanation

P(X > 0) = 0.30 + 0.20 = 0.50. By linearity, E[2X + 3] = 2E[X] + 3 = 6.2. Linearity of expectation does not require independent variables; multiplying probabilities usually does require a condition such as independence.

Continue

Explore discrete and continuous models. NIST defines probability distributions and their functions.