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Expected value and linear expectation of a discrete random variable

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Expected value of a discrete random variable

The expected value is a probability-weighted average of the possible values. It describes the long-run average and does not have to be one of the values that the random variable can take

\[\operatorname{E}(X)=\sum_xx\Pr(X=x),\quad\operatorname{E}(aX+b)=a\operatorname{E}(X)+b\]
  1. 1Weight each value Multiply each value by the probability in the same row \(\displaystyle 5\times\frac{1}{4}=\frac{5}{4}\)
  2. 2Add the weighted terms Add every value-probability product, including any negative products \(\displaystyle 0\times\frac{1}{4}+2\times\frac{1}{2}+5\times\frac{1}{4}=\frac{9}{4}\)
  3. 3Interpret the mean Read the expected value as the long-run average over repeated trials \(\displaystyle \operatorname{E}(X)=\frac{9}{4}=2.25\notin\{0,2,5\}\)
  4. 4Transform the expectation Keep the multiplier on the mean and add the constant once \(\displaystyle \operatorname{E}(3X-2)=3\left(\frac{9}{4}\right)-2=\frac{19}{4}\)
Year 12, VCE Units 3 & 4, Chapter 10, 4 difficulty levels, No calculatorBuild a worksheet

Expected value of a discrete random variable

The expected value is a probability-weighted average of the possible values. It describes the long-run average and does not have to be one of the values that the random variable can take

\[\operatorname{E}(X)=\sum_xx\Pr(X=x),\quad\operatorname{E}(aX+b)=a\operatorname{E}(X)+b\]
  1. Weight each value Multiply each value by the probability in the same row \(\displaystyle 5\times\frac{1}{4}=\frac{5}{4}\)
  2. Add the weighted terms Add every value-probability product, including any negative products \(\displaystyle 0\times\frac{1}{4}+2\times\frac{1}{2}+5\times\frac{1}{4}=\frac{9}{4}\)
  3. Interpret the mean Read the expected value as the long-run average over repeated trials \(\displaystyle \operatorname{E}(X)=\frac{9}{4}=2.25\notin\{0,2,5\}\)
  4. Transform the expectation Keep the multiplier on the mean and add the constant once \(\displaystyle \operatorname{E}(3X-2)=3\left(\frac{9}{4}\right)-2=\frac{19}{4}\)

Worked example

A discrete random variable has the values and probabilities shown. Find both expected values

\[\Pr(X=1)=\frac{1}{2},\quad\Pr(X=3)=\frac{1}{3},\quad\Pr(X=6)=\frac{1}{6},\qquad\operatorname{E}(X),\quad\operatorname{E}(2X-1)\]
  1. \(\displaystyle 1\left(\frac{1}{2}\right)=\frac{1}{2},\quad3\left(\frac{1}{3}\right)=1,\quad6\left(\frac{1}{6}\right)=1\)Multiply each possible value by its probability
  2. \(\displaystyle \operatorname{E}(X)=\frac{1}{2}+1+1=\frac{5}{2}\)Add the three weighted products to find the expected value
  3. \(\displaystyle \operatorname{E}(2X-1)=2\operatorname{E}(X)-1=2\left(\frac{5}{2}\right)-1=4\)Apply the linear expectation rule and add the constant with its sign

Common mistakes

Ignoring the probabilities
✕\(\displaystyle \operatorname{E}(X)=0+2+5=7\)✓\(\displaystyle \operatorname{E}(X)=0\left(\frac{1}{4}\right)+2\left(\frac{1}{2}\right)+5\left(\frac{1}{4}\right)=\frac{9}{4}\)
Expected value is a weighted sum, so each value must be multiplied by its probability
Dividing by the number of values
✕\(\displaystyle \operatorname{E}(X)=\frac{0+2+5}{3}=\frac{7}{3}\)✓\(\displaystyle \operatorname{E}(X)=0\left(\frac{1}{4}\right)+2\left(\frac{1}{2}\right)+5\left(\frac{1}{4}\right)=\frac{9}{4}\)
Dividing by the number of values is an unweighted average and ignores the different probabilities
Scaling the constant
✕\(\displaystyle \operatorname{E}(3X+2)=3\operatorname{E}(X)+6\)✓\(\displaystyle \operatorname{E}(3X+2)=3\operatorname{E}(X)+2\)
The multiplier applies to E(X), while the constant is added once

Difficulty levels

introCalculate a direct expected value from a small table of integer values
standardUse a mass-function source or apply a linear expectation rule with a negative constant
challengeRecover an unknown value from a stated mean or use a sum of expectations
examInterpret an expected value and select or calculate a linear expectation result

Questions about this topic

What does expected value mean?+

It is the theoretical probability-weighted mean. For independent repeated trials governed by the same distribution, the sample average tends to settle near it; no finite run is guaranteed to have exactly that average

Can an expected value be a value that X cannot take?+

Yes. A weighted average can lie between listed values without being one of the possible values

How is E(aX + b) calculated?+

Multiply E(X) by a and then add b once, using E(aX + b) = aE(X) + b.

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 10: Discrete random variables

In this chapter

Validate, complete and normalise a discrete probability distributionFinite events, probability assignments and complementsDistribution of a finite random variableRepresent a multi-stage finite sample spaceExpected value and linear expectation of a discrete random variableVariance, standard deviation and linear variance of a discrete random variableInterval, conditional and μ ± 2σ probabilities from a discrete distributionDistribution of a linear transformation of a discrete random variableConditional probability from exact representations