Expected Return as a Long-Run Weighted Mean

Expected Return as a Weighted Mean

An investment return can vary from one period to another because different outcomes are possible. A single average across those outcomes is useful only if it gives the outcomes the right weights. When outcomes have different probabilities, the appropriate average weights each return by its probability. This expected return is also the long-run average return under a stable model when the observed frequencies of outcomes approach their probabilities.

The calculation uses the simple return for one period, expressed either as a decimal or a percentage. For example, a return of 6% is $0.06$ in a formula. The expected return describes the probability model; it is not a promise that a particular period, or a short sequence of periods, will earn that rate.

Definition: Suppose a one-period return $R$ can take the distinct values $r_1,r_2,\ldots,r_k$, with corresponding probabilities $p_1,p_2,\ldots,p_k$. The probabilities satisfy $p_j\geq 0$ for every $j$ and $\sum_{j=1}^{k}p_j=1$. The expected return is the probability-weighted mean $$ E[R]=\sum_{j=1}^{k}p_jr_j. $$ The return values must use a consistent unit, such as all decimals or all percentages.

The probabilities—not the number of scenario labels—determine the weights. If an outcome has probability 0.50, it receives half the weight even if it is just one of three listed outcomes. Negative returns are included with their negative signs.

Why the Weighted Mean Is the Long-Run Mean

Consider repeated periods generated by the same return model. In $n$ periods, let $N_j(n)$ be the number of times outcome $r_j$ occurs. The arithmetic average of the $n$ realized returns can be grouped by outcome: each return value is multiplied by the fraction of periods in which it occurred. Thus,

\overline{R}_n =\frac{1}{n}\sum_{t=1}^{n}R_t =\sum_{j=1}^{k}\frac{N_j(n)}{n}r_j.

For the long-run interpretation, assume each period is an independent draw from the same discrete distribution and the possible returns are finite. Under these assumptions, the relative frequency $N_j(n)/n$ approaches $p_j$ as the number of periods grows. Substituting those limiting frequencies into the grouped average gives the expected-return formula:

\overline{R}_n \longrightarrow \sum_{j=1}^{k}p_jr_j =E[R].

This argument explains why probability is the weight: in the long run, an outcome with probability $p_j$ occupies approximately that fraction of periods. The statement is a long-run result, not a claim that every finite sample has exactly the model probabilities. If the probabilities change over time, or if the model does not describe the return process, this long-run interpretation need not apply.

Check the outcomes and probabilities.
List each distinct return once, pair it with its probability, and verify that the probabilities sum to 1.

Multiply each return by its probability.
Keep gains positive and losses negative; use consistent units throughout.

Add the weighted returns.
The result is the model's expected return for one period.

Worked Calculations

Worked Example: Unequal Probabilities Across Three Outcomes

A hypothetical investment has a one-year return of $-8\%$ with probability $0.20$, $5\%$ with probability $0.50$, or $17\%$ with probability $0.30$. Find its expected one-year return. State why simply averaging the three returns would be inappropriate.

Method. Use the probability-weighted mean, first checking that the scenarios form a complete distribution: $$ 0.20+0.50+0.30=1.00. $$ The probabilities are valid. Convert the percentage returns to decimals: $-8\%=-0.08$, $5\%=0.05$, and $17\%=0.17$.

Multiply each return by its probability: $$ 0.20(-0.08)=-0.016,\qquad 0.50(0.05)=0.025,\qquad 0.30(0.17)=0.051. $$ Add the contributions: $$ E[R]=-0.016+0.025+0.051=0.060=6.0\%. $$

The check is that the weighted contributions sum to $0.060$, or $6.0\%$, with the loss lowering the result. The common mistake is to assign each of the three outcomes weight $1/3$, giving $(-8\%+5\%+17\%)/3=4.67\%$ (rounded to two decimal places). That calculation assumes equal probabilities, contrary to the stated model. The expected return is 6.0%.

Worked Example: Matching Frequencies Over Repeated Periods

A hypothetical asset has the same one-year return distribution in each of 30 independent years: $-4\%$ with probability $0.20$, $4\%$ with probability $0.50$, and $12\%$ with probability $0.30$. In a particular 30-year sequence, the outcomes occur 6, 15, and 9 times, respectively. Calculate the model's expected return and the arithmetic average for this sequence.

Model expected return. The probabilities add to $0.20+0.50+0.30=1.00$. Weight each return by its probability: $$ E[R]=0.20(-4\%)+0.50(4\%)+0.30(12\%) =-0.8\%+2.0\%+3.6\%=4.8\%. $$

Sequence average. The observed relative frequencies are $6/30=0.20$, $15/30=0.50$, and $9/30=0.30$. Therefore, the average return in this sequence is $$ \overline{R}_{30} =\frac{6}{30}(-4\%)+\frac{15}{30}(4\%)+\frac{9}{30}(12\%) =-0.8\%+2.0\%+3.6\%=4.8\%. $$

The two calculations agree because the sequence's frequencies exactly match the model probabilities. This is a particular finite sequence, not a general guarantee about 30 periods. A common mistake is to average the three outcome values equally, yielding $4.0\%$; that ignores that the outcomes occurred at different frequencies and have different probabilities. The correct model expectation and sequence average here are both 4.8%.

Worked Example: Expected Return and Expected Dollar Gain

A hypothetical investor places $ \$20{,}000 $ in an investment for one year. Its return is $-12\%$ with probability $0.25$, $6\%$ with probability $0.50$, or $18\%$ with probability $0.25$. Calculate the expected percentage return, the dollar gain in each outcome, and the expected ending value.

Step 1: Find the expected return. The probabilities sum to $0.25+0.50+0.25=1.00$. Thus, $$ E[R]=0.25(-12\%)+0.50(6\%)+0.25(18\%) =-3.0\%+3.0\%+4.5\%=4.5\%. $$

Step 2: Convert each outcome to dollars. A one-period dollar gain is the initial amount multiplied by that outcome's return: $$ \$20{,}000(-0.12)=-\$2{,}400,\quad \$20{,}000(0.06)=\$1{,}200,\quad \$20{,}000(0.18)=\$3{,}600. $$ The expected dollar gain is $$ 0.25(-\$2{,}400)+0.50(\$1{,}200)+0.25(\$3{,}600) =-\$600+\$600+\$900=\$900. $$

Step 3: Check using ending values. The three possible ending values are $\$17{,}600$, $\$21{,}200$, and $\$23{,}600$. Their probability-weighted mean is $$ 0.25(\$17{,}600)+0.50(\$21{,}200)+0.25(\$23{,}600) =\$4{,}400+\$10{,}600+\$5{,}900=\$20{,}900. $$ This is the initial $\$20{,}000$ plus the expected gain of $\$900$. The common mistake is to treat the $4.5\%$ expected return as a guaranteed dollar gain. It is a probability-weighted result; the possible one-year gains are $-\$2{,}400$, $\$1{,}200$, and $\$3{,}600$. The expected return is 4.5% and the expected ending value is $\$20{,}900$.

Interpretation and Common Traps

Expected return is useful for summarizing the center of a specified return distribution and comparing modeled outcomes on a consistent basis. Its interpretation depends on the probability model being appropriate. A calculated expectation does not say which outcome will occur next, and a short run of returns can differ substantially from the expected value. The long-run frequency argument applies when the return process has the stable, independent structure stated above; it should not be used automatically for changing conditions.

Do not give scenarios equal weights unless their probabilities are equal. The probability, not the scenario count, sets the weight.
Keep losses negative. Replacing a negative return with its absolute value changes the expected return.
Verify probabilities sum to one. If they do not, the listed outcomes may be incomplete or the probabilities may need correction before calculating.
Distinguish an expectation from a guaranteed outcome. Expected return is a weighted mean across possible outcomes, not a return promised in every period.

Key takeaway: For a discrete return distribution, expected return is $\sum p_jr_j$. It is the long-run arithmetic average when repeated outcomes occur with relative frequencies approaching their probabilities under a stable model.

Check Your Understanding

For each problem, use the stated probabilities and keep losses negative. Answers are provided for checking.

A hypothetical asset returns $-5\%$ with probability $0.40$ and $10\%$ with probability $0.60$. What is its expected return? (Answer: $4.0\%$)
A return is $-6\%$, $3\%$, or $15\%$ with probabilities $0.20$, $0.50$, and $0.30$, respectively. Calculate the expected return. (Answer: $4.8\%$)
In 40 independent periods, returns of $-8\%$, $4\%$, and $12\%$ occur 10, 20, and 10 times. The model probabilities are $0.25$, $0.50$, and $0.25$. Find both the model expected return and the 40-period arithmetic average. (Answer: $3.0\%$ for each)
A hypothetical investment of $\$8{,}000$ earns $-10\%$, $5\%$, or $15\%$ with probabilities $0.10$, $0.60$, and $0.30$. Find its expected percentage return and expected dollar gain. (Answer: $6.5\%$ and $\$520$)
A model assigns probabilities $0.15$, $0.55$, and $0.30$ to returns of $-20\%$, $4\%$, and $16\%$, respectively. A colleague equally averages the three returns. Find the correct expected return and state why the colleague's method is wrong. (Answer: $4.0\%$; equal weighting is wrong because the probabilities are unequal.)

What You'll Learn