Random Variables, Mean and Variance
Here a discrete random variable is given by a probability table, and the task is to find an unknown probability, the expected value, the variance or the standard deviation. The work is short but every arithmetic slip is penalised, so lay out the table of x, p(x), xp(x) and x²p(x).
Distribution and expectation
For a discrete random variable X with values x₁, x₂, …, the probabilities satisfy p(xᵢ) ≥ 0 and Σp(xᵢ) = 1; this condition usually determines an unknown k. Then E(X) = Σxᵢp(xᵢ), E(X²) = Σxᵢ²p(xᵢ), Var(X) = E(X²) − [E(X)]² and SD = √Var.
Properties
E(aX + b) = aE(X) + b, so adding a constant shifts the mean. Var(aX + b) = a²Var(X), so adding a constant changes nothing about spread and multiplying by a scales variance by a². Variance is in squared units; SD returns to the original units.
Worked example
X takes 0, 1, 2 with probabilities 0.2, 0.5, 0.3. The total 0.2 + 0.5 + 0.3 = 1, so the distribution is valid. E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 1.1. E(X²) = 0(0.2) + 1(0.5) + 4(0.3) = 1.7. Var(X) = 1.7 − (1.1)² = 1.7 − 1.21 = 0.49, so SD = 0.7. The variable averages 1.1 with a typical deviation of about 0.7.
Traps
- Computing Var as E(X²) − E(X) rather than subtracting the square of the mean.
- Never using Σp = 1, so an unknown k remains unfound.
- Writing Σx²p², squaring the probability as well as the value.
- Giving the variance when the question asks for the standard deviation.