Expected value and linear expectation of a discrete random variable

Which calculation begins the expected-value working?

Vary the probability of minus three while keeping the middle probability at one half. Increasing the mass at minus three takes the same mass away from seven. Track the signed weighted contributions and the mean. When that probability is zero or one half, a listed zero-mass value is not in the positive support

Why does the mean decrease when p increases?

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

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 \(\displaystyle \operatorname{E}(3X-2)=3\left(\frac{9}{4}\right)-2=\frac{19}{4}\)

Complete distribution

\(\displaystyle \begin{array}{c|ccc}x&0&2&5\\\Pr(X=x)&1/4&1/2&1/4\end{array}\)

For values 0, 2 and 5 with probabilities 1/4, 1/2 and 1/4, calculate the contribution from X = 5.

Complete distribution

\(\displaystyle \begin{array}{c|ccc}x&0&2&5\\\Pr(X=x)&1/4&1/2&1/4\end{array}\)

Add all three weighted contributions to find E(X)

Complete distribution

\(\displaystyle \begin{array}{c|ccc}x&0&2&5\\\Pr(X=x)&1/4&1/2&1/4\end{array}\)

Ignoring the probabilities. One of these lines is right. Which one?

Keep signed values in the weighted calculation

The table gives the values and their probabilities for a finite score model. The probabilities sum to one. Negative scores are permitted; this is not a model of a non-negative count

Distribution and requested quantity

\(\displaystyle x=-3,\quad\Pr(X=-3)=\frac16\)

Find the contribution from X=−3.

Distribution and requested quantity

\(\displaystyle x=2,\quad\Pr(X=2)=\frac12\)

Find the contribution from X=2.

Distribution and requested quantity

\(\displaystyle x=7,\quad\Pr(X=7)=\frac13\)

Find the contribution from X=7.

Distribution and requested quantity

\(\displaystyle -\frac12,\quad 1,\quad\frac73\)

Add all three weighted contributions to find E(X)

Score distribution

\(\displaystyle \begin{array}{c|ccc}x&-3&2&7\\\Pr(X=x)&1/6&1/2&1/3\end{array}\)

The most likely score is 2. Which statement about the expected value is correct?

Separate a supported value from a theoretical average

The expected value 17/6 lies between −3 and 7 but is not one of the three scores. An individual trial still produces one supported score. The probability-weighted mean is not a promise of a particular outcome

Retain the sign of a negative multiplier

Define Y=−2X+5 for the same score model. The expectation rule multiplies the mean by −2 and adds 5 once. Independence is not relevant to applying a deterministic affine rule to X.

Distribution and requested quantity

\(\displaystyle \operatorname{E}(X)=\frac{17}{6},\quad Y=-2X+5\)

Find E(Y) for Y=−2X+5.

Check the linear rule by weighting corresponding values

The three X values −3,2,7 map to Y values 11,1,−9 with their original probabilities. Their weighted sum agrees with the linear expectation rule. The constant contributes 5 because the probabilities sum to one

Recover an unknown support value from the mean

A table has values u,2,7 and probabilities 1/6,1/2,1/3. Its stated mean is 17/6. Only one value is unknown; the probability in that row is non-zero, so the resulting linear equation determines u.

Distribution and requested quantity

\(\displaystyle 2\left(\frac12\right)+7\left(\frac13\right)\)

Find the sum of the two known weighted contributions

Distribution and requested quantity

\(\displaystyle \frac{u}{6}+\frac{10}{3}=\frac{17}{6}\)

Find u from the stated mean

Substitute the recovered value back into the table

Using u=−3 gives weighted total −1/2+10/3=17/6. The recovered value is different from 2 and 7, and the original probabilities remain non-negative and total one

Evaluate the stated probability rule before weighting

Substitute each stated value into the probability rule. The resulting masses sum to one. The probability rule supplies weights, not the random-variable values themselves

Distribution and requested quantity

\(\displaystyle p(x)=\frac{x+2}{10},\quad x=-1\)

Find Pr(X=−1) from the stated rule

Distribution and requested quantity

\(\displaystyle \begin{array}{c|rrrr}x&-1&0&1&2\\\Pr(X=x)&1/10&2/10&3/10&4/10\end{array}\)

Find E(X) for this four-value distribution

Do not add the probabilities to calculate the mean

The probability sum is 1 because it normalises the distribution. The expected value happens to be 1 in this example for a different reason: it is the weighted sum of the four values

Distinguish a realised average from the model expectation

Return to the score model −3,2,7 with masses 1/6,1/2,1/3. A particular run of 60 trials records these scores 12,36,12 times. Those observed frequencies need not equal the model proportions. Calculate the realised average from the observed frequencies

Distribution and requested quantity

\(\displaystyle \overline{x}=\frac{(-3)(12)+2(36)+7(12)}{60}\)

Find the realised average score for those 60 trials

Distribution and requested quantity

\(\displaystyle \operatorname{E}(X)=\frac{17}{6},\qquad T=X_1+\cdots+X_{60}\)

Find the expected total score across 60 trials governed by the same model

Does the model require every run of 60 trials to have total score 170?

Add expectations without inventing independence

Let Z=2X, so Z depends on X. Still E(X+Z)=E(X)+E(Z)=3E(X). The additivity rule for finite expectations does not require independence. Do not transfer that claim to a variance rule

Distribution and requested quantity

\(\displaystyle \operatorname{E}(X)=\frac{17}{6},\quad Z=2X\)

Find E(X+Z) when Z=2X for the original score model

Write the complete unknown-value and expectation argument, ungraded

A new table has values −2,1,u with masses 1/5,2/5,2/5 and mean 2. Write every weighted term, solve for u and substitute it back. Then find E(3X−4) and explain why a trial need not equal the expected value. The whole written derivation and interpretation are self-reviewed, not graded by the scalar checks

Compare the equation, substitution and interpretation

The known contributions −2/5 and 2/5 cancel, so 2u/5=2 gives u=5. Substitution gives mean 2. The transformed mean is 3(2)−4=2. Neither mean being 2 makes 2 a possible original value: the support is −2,1,5.

Check what the question asks you to average

Start with the displayed distribution, retain signs in each weighted term, and distinguish a mean from a mode or an observed average. For an affine rule add the constant once; for an unknown value write the full mean equation