Arithmetic Versus Geometric Average Returns

Two Averages Answer Different Questions

An arithmetic average summarizes the returns observed in separate periods. A geometric average identifies the constant per-period return that would produce the same total growth as those returns. Both are useful, but they answer different questions. The arithmetic average describes the average of the period-by-period rates; the geometric average describes the rate of compounded growth.

Suppose an investment earns a simple return $r_t$ in period $t$, with each period having the same length. Its gross return factor is $1+r_t$. With no contributions or withdrawals, the factors multiply: after $n$ periods, initial wealth $W_0$ becomes $W_n=W_0\prod_{t=1}^{n}(1+r_t)$. The comparison below assumes returns are measured consistently and that the investment’s returns are reinvested. In the usual unlevered setting, each return is at least $-100\%$; the standard geometric average is defined when every gross factor is positive.

Definition: For returns $r_1,\ldots,r_n$, the arithmetic average return is their sum divided by the number of periods. The geometric average return is the constant per-period return that produces the same compounded wealth as the observed return series.

\bar r_A=\frac{1}{n}\sum_{t=1}^{n}r_t, \qquad \bar r_G=\left(\prod_{t=1}^{n}(1+r_t)\right)^{1/n}-1

The subscript $A$ denotes arithmetic and $G$ denotes geometric. The formula for the geometric average follows by setting the final wealth under a constant return equal to the actual final wealth:

W_0(1+\bar r_G)^n = W_0\prod_{t=1}^{n}(1+r_t).

Cancel $W_0$, take the positive $n$th root, and subtract 1 to obtain the geometric-average formula. The geometric average is also called the compound annual growth rate when the periods are years. It describes the realized growth rate, not a guaranteed future return.

Why the Geometric Average Cannot Be Larger

Write each period’s gross return factor as $x_t=1+r_t$. When all factors are positive, the arithmetic-geometric mean inequality says that the geometric mean of positive numbers cannot exceed their arithmetic mean. Applying it to the gross factors gives:

\left(\prod_{t=1}^{n}(1+r_t)\right)^{1/n} \leq \frac{1}{n}\sum_{t=1}^{n}(1+r_t) = 1+\bar r_A.

Subtracting 1 from both sides gives $\bar r_G\leq\bar r_A$. Equality holds exactly when every period has the same return. If the returns differ, the inequality is strict: the geometric average is lower.

One justification for the inequality uses the natural logarithm. The logarithm is concave because its second derivative is $-1/x^2<0$ for $x>0$. Concavity implies that the average of the logarithms of positive numbers is no greater than the logarithm of their arithmetic average. Apply this to $x_1,\ldots,x_n$:

\frac{1}{n}\sum_{t=1}^{n}\ln(1+r_t) \leq \ln\left(\frac{1}{n}\sum_{t=1}^{n}(1+r_t)\right).

Exponentiating preserves the inequality. The left side becomes the geometric mean of the gross factors, and the right side becomes their arithmetic mean. Subtracting 1 establishes the claimed result for returns. This proof also shows why the result concerns gross factors, rather than taking the geometric mean of the returns themselves: returns can be zero or negative, whereas the logarithm argument must be positive.

Worked Examples

Worked Example: Three Returns and the Actual Growth Rate

A hypothetical investment account begins with $50,000 and earns returns of $10\%$, $-10\%$, and $10\%$ in three successive years. Find the arithmetic average, the geometric average, and the ending value. Compare the two averages.

Method. Average the three returns for the arithmetic average. Multiply their gross factors for the actual wealth change, then take the cube root to find the geometric average.

Arithmetic average.

\bar r_A=\frac{0.10-0.10+0.10}{3} =\frac{0.10}{3} =0.033333\ldots

Thus, the arithmetic average is $3.33\%$ per year when rounded to two decimal places.

Compounded wealth and geometric average. The three gross factors are $1.10$, $0.90$, and $1.10$. Their product is:

$$ 1.10(0.90)(1.10)=1.089. $$

The account therefore ends at $50{,}000(1.089)=\$54{,}450.00$. The constant annual rate producing that same factor satisfies $(1+\bar r_G)^3=1.089$, so:

\bar r_G=1.089^{1/3}-1 \approx 0.02883 =2.88\%.

The check is that $50{,}000(1+0.02883)^3$ is approximately $54,450.00, subject to rounding the displayed rate. The geometric average, $2.88\%$, is below the arithmetic average, $3.33\%$. A common mistake is to treat the arithmetic average as the rate that actually compounded the account; the product of the yearly gross factors, not the arithmetic average, determines ending wealth.

Worked Example: Same Arithmetic Average, Different Variability

Two hypothetical funds each receive an initial deposit of $20,000 and are held for three years without additional cash flows. Fund A earns $10\%$ in each year. Fund B earns $40\%$, $0\%$, and $-10\%$. Calculate each arithmetic average, ending value, and geometric average. Which fund compounds faster over these three years?

Arithmetic averages. Fund A’s average is $(10\%+10\%+10\%)/3=10\%$. Fund B’s average is $(40\%+0\%-10\%)/3=30\%/3=10\%$. The arithmetic averages are equal.

Fund A. Its gross factor over the full period is $1.10^3=1.331$. Ending wealth is $20{,}000(1.331)=\$26{,}620.00$. Since its return is the same each year, its geometric average is exactly $10\%$.

Fund B. Its gross factors are $1.40$, $1.00$, and $0.90$. Multiplying them gives $1.40(1.00)(0.90)=1.26$. Ending wealth is $20{,}000(1.26)=\$25{,}200.00$. Its geometric average is:

\bar r_G=1.26^{1/3}-1 \approx 0.08008 =8.01\%.

The check is that $20{,}000(1.26)=\$25{,}200.00$, and a constant return of approximately $8.01\%$ for three years produces approximately the same factor. Fund B has the same arithmetic average as Fund A but ends with $1,420.00 less. Its returns vary more, and the geometric average captures the lower compounded growth. A common mistake is to conclude that equal arithmetic averages imply equal ending wealth; that conclusion ignores the different products of the gross factors.

Worked Example: An Arithmetic-Average Projection Is Not the Realized Compound Result

A hypothetical reserve account starts with $40,000 and earns $25\%$, $-20\%$, and $25\%$ over three successive years, with no cash flows. Calculate its actual ending balance and both averages. Then compare the actual balance with a projection that applies the arithmetic average every year.

Arithmetic average.

\bar r_A=\frac{0.25-0.20+0.25}{3} =\frac{0.30}{3} =0.10=10\%.

Actual result. The gross factors are $1.25$, $0.80$, and $1.25$. Their product is $1.25(0.80)(1.25)=1.25$, so the ending balance is $40{,}000(1.25)=\$50{,}000.00$. The geometric average is:

\bar r_G=1.25^{1/3}-1 \approx 0.07722 =7.72\%.

Projection using the arithmetic average. Applying $10\%$ each year would give:

40{,}000(1.10)^3 =40{,}000(1.331) =\$53{,}240.00.

That projection is $3,240.00 above the actual ending balance. The check is that the actual gross factor, $1.25$, equals $(1+0.07722)^3$ to the displayed precision. The arithmetic average is a valid summary of the individual annual returns, but it is not the constant compound rate for this account. The geometric average gives that rate.

How Variability Reduces Compounded Growth

The examples illustrate a general mechanism: a loss and a gain of the same percentage do not cancel in wealth. For example, a $10\%$ gain followed by a $10\%$ loss multiplies wealth by $1.10(0.90)=0.99$, a net loss of $1\%$. The loss is calculated from the higher balance reached after the gain. This asymmetry helps explain why fluctuating returns can produce a geometric average below the arithmetic average.

A useful approximation makes the connection to variability explicit. Let $\bar r_A$ be the arithmetic average of the observed returns, and define their population variance over this particular series as $s_r^2=\frac{1}{n}\sum_{t=1}^n(r_t-\bar r_A)^2$. For returns that are small enough for a second-order approximation to be reasonable, the expansion $\ln(1+r)\approx r-r^2/2$ gives:

\frac{1}{n}\sum_{t=1}^{n}\ln(1+r_t) \approx \bar r_A-\frac{1}{2n}\sum_{t=1}^{n}r_t^2 = \bar r_A-\frac{1}{2}\left(s_r^2+\bar r_A^2\right).

The left side is the average log gross return. Since the geometric gross return is the exponential of this average, applying a further small-change approximation yields:

\bar r_G\approx\bar r_A-\frac{1}{2}s_r^2.

This is an approximation, not an exact formula. It is most useful as a qualitative guide when period returns are not large: with the same arithmetic average, greater dispersion tends to lower the geometric average. Large returns, severe losses, or strongly asymmetric return patterns can make the approximation inaccurate. The exact geometric average always comes from multiplying the gross factors. The exact difference is influenced by the whole return pattern, not determined by variability alone.

Return order does not affect either average or the compounded result when the set of period returns is unchanged and there are no intervening cash flows. Multiplication is commutative, so rearranging the gross factors leaves their product unchanged. The timing of external deposits and withdrawals would change the wealth calculation; the formulas here compare the returns on an investment without those cash flows.

Common Mistakes and Interpretation

Using an arithmetic average as a compound growth rate. It does not generally reproduce ending wealth. Use the geometric average for the constant rate equivalent to the realized compound result.
Taking a geometric average of the returns themselves. Use the gross factors $1+r_t$, take their geometric mean, and subtract 1.
Assuming the averages must be equal. They are equal only when every period has the same return, provided the gross factors are positive.
Claiming that greater variability alone gives an exact numerical answer. The volatility-drag approximation is helpful intuition, not a replacement for calculating the exact product.

For an investment held over multiple periods, the geometric average is generally the appropriate descriptive measure of realized compound growth. The arithmetic average remains useful for describing the average of individual period returns. Distinguishing these interpretations is important in return measurement and is aligned with the SOA VEE Accounting and Finance syllabus.

Check Your Understanding

For each problem, use gross return factors to find compounded growth and distinguish it from the arithmetic average.

A fund earns $12\%$ in year one and $-8\%$ in year two. Find the arithmetic and geometric average returns, rounding the geometric average to two decimal places. (Answer: arithmetic average $=2.00\%$; geometric average $\approx1.51\%$.)
An investment earns $10\%$ followed by $-10\%$. Find its two-year gross growth factor and geometric average return, rounding to two decimal places. (Answer: gross growth factor $=0.99$; geometric average $\approx-0.50\%$ per year.)
A hypothetical account begins with $15,000 and earns $6\%$ in each of four years. Find the arithmetic average, geometric average, and ending balance to the cent. (Answer: both averages $=6.00\%$; ending balance $=\$18{,}937.15$.)
A hypothetical account earns $20\%$, $-10\%$, and $5\%$ over three years. Find the arithmetic average, total gross growth factor, and geometric average to two decimal places. (Answer: arithmetic average $=5.00\%$; gross growth factor $=1.134$; geometric average $\approx4.28\%$ per year.)

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