Maths › Further Statistics 1 › Discrete random variables and expectation
Discrete random variables and expectation
A probability distribution is a set of weights; the mean is where they balance and the variance is how far they scatter. Both follow from one sum, and both survive being scaled and shifted in predictable ways.
Builds on Probability and Venn diagrams and Measures of location and spread.
IN THIS TOPIC
- Compute E(X) and Var(X) from a probability distribution table.
- Evaluate E(g(X)) for functions such as X² and aX + b.
- Use the mean and variance to judge whether a model fits observed data.
WHAT YOU PROBABLY THINK
The expected value is the outcome you should expect to see most often.
Where the weights balance
For a discrete random variable, the expectation weights each value by its probability:
E(X) is the balance point of the distribution, not its most likely value, and often not even a possible value: a fair die has mean 3.5, which no face shows. That is where the opening claim goes wrong. The mode is the commonest outcome; the mean is the centre of gravity.
WORKED EXAMPLE
A distribution, end to end
X takes values 1, 2, 3, 4 with P(X = x) = x/10. Find E(X), E(X²) and Var(X).
The probabilities 0.1, 0.2, 0.3, 0.4 sum to 1, so the model is valid.
E(X) = (1 + 4 + 9 + 16)/10 = 3.
E(X²) = (1 + 8 + 27 + 64)/10 = 10, so Var(X) = 10 − 3² = 1. Note the mode is 4 while the mean is 3: the distribution leans right, and the balance point trails the peak.
Functions, scaling and shifting
E(g(X)) applies g to each value before weighting, so E(X²) = Σx²P(X = x), which is why the variance formula needs it. Linear functions behave tidily: E(aX + b) = aE(X) + b, and Var(aX + b) = a²Var(X). Shifting moves the balance point without changing the spread; scaling stretches the spread by the square of the factor, because variance is measured in squared units.
WORKED EXAMPLE
Transforming the same variable
For the X above, find E(2X + 3) and Var(2X + 3).
E(2X + 3) = 2(3) + 3 = 9.
Var(2X + 3) = 2² × 1 = 4: the +3 contributes nothing to the spread.
Checking the long way, by listing the values 5, 7, 9, 11 with the same probabilities, gives the same two numbers.
YOUR TURN
Judging a model
A shop models daily sales of a rare item with P(X = 0) = 0.5, P(X = 1) = 0.3, P(X = 2) = 0.2. Find the mean and variance, and comment on whether a model with mean equal to variance would fit.
Show the working
E(X) = 0 + 0.3 + 0.4 = 0.7.
E(X²) = 0 + 0.3 + 0.8 = 1.1, so Var(X) = 1.1 − 0.49 = 0.61.
The variance is below the mean, so a Poisson model, which forces them equal, would overstate the day-to-day variability here.
THE EXAM BIT
- Check the probabilities sum to 1 before anything else; an invalid table makes every later answer worthless.
- Var(X) = E(X²) − [E(X)]², never E(X²) − E(X); the square goes on the mean.
- Var(aX + b) uses a², and b vanishes. State that explicitly rather than recomputing from scratch.
- When asked to comment on a model, compare the mean and variance you found with what the proposed model demands.
CHECK YOURSELF
X has E(X) = 4 and Var(X) = 9. Find E(3X − 2) and Var(3X − 2).
Show a hint
Means shift and scale; variances only scale, by the square.
Show the answer
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E(X) = Σ x P(X = x) is the balance point; Var(X) = E(X²) − μ² measures the scatter round it.
E(aX + b) = aE(X) + b, while Var(aX + b) = a²Var(X): shifting moves the centre, scaling squares into the spread.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Compute E(X) and Var(X) from a probability distribution table.
- Evaluate E(g(X)) for functions such as X² and aX + b.
- Use the mean and variance to judge whether a model fits observed data.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.