MathsFurther 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 come out of 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 proposed model fits observed data.

COMMON MISCONCEPTION

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)=ΣxP(X=x)\text{E}(X) = Σ \, x \, \text{P}(X = x)IN THE FORMULAE BOOKLET
Var(X)=Σx2P(X=x)-μ2\text{Var}(X) = Σ \, x^{2} \, \text{P}(X = x) - μ^{2}IN THE FORMULAE BOOKLET

Both are printed in the booklet under Discrete distributions, along with E(g(X)) = Σ g(x)P(X = x), so the marks come from setting the table out correctly rather than from recall. E(X) is the balance point of the distribution. It need not be the commonest value, and it need not even be a possible one. A fair die has mean 3.5 and no face shows 3.5. The mode is the commonest outcome. The mean is the centre of gravity.

The distribution P(X = x) = x/10 balanced on its mean 3, with the mode at 4: the pivot need not sit under a barx = 1x = 2x = 3x = 40.10.20.30.4mean 3mode 4
FIG. 1A probability distribution as weights on a beam: the mean is the pivot that balances them, whether or not any weight sits there.

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. The mode is 4 while the mean is 3. This 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) is Σx²P(X = x), and the variance formula depends on exactly that. Linear functions behave tidily. You get E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X).

Shifting moves the balance point and leaves the spread alone. Scaling stretches the spread by the square of the factor, because variance is measured in squared units. Doubling every value quadruples the variance but only doubles the standard deviation.

X against 2X + 3: the mean moves from 3 to 9 and the spacing doubles, so the variance goes from 1 to 4mean 3X: mean 3, variance 1mean 92X + 3: mean 9, variance 4shifting slides the centre; scaling squares into the spread
FIG. 2Shifting a distribution slides the mean and leaves the spread alone; doubling the values doubles the mean and quadruples the variance.

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.

Two numbers that test a model

Once you have a mean and a variance you can ask whether a named model could have produced the data. Each family leaves its own fingerprint. A Poisson distribution forces the variance to equal the mean. A binomial forces the variance below the mean, since np(1 − p) is always less than np. A geometric distribution has a variance well above its mean once p is small.

Questions that say 'comment on the suitability of the model' want that comparison spelled out, with the two numbers you computed set against what the proposed family demands. A sentence naming the direction of the mismatch earns more than a bare verdict.

GUIDED PRACTICE

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 Poisson model 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 sits below the mean. Poisson forces the two equal, so it would overstate the day-to-day variability here.

ASSESSMENT FOCUS

  • 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. Say so on the page instead of recomputing from a new table.

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

E(3X − 2) = 12 − 2 = 10. Var(3X − 2) = 9 × 9 = 81. The −2 has no effect on spread.

E(X) = Σ x P(X = x) is the balance point, and 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.

6 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the discrete random variables and expectation questions page.

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 proposed model fits observed data.

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