Deriving The Investor Return Equation From Account Balances

From Account Balances to an Investor Return

An account balance changes for two different reasons: the investments earn or lose value, and the investor adds or removes money. A dollar-weighted return must distinguish these effects. A deposit increases the account balance, but it is not investment earnings; a withdrawal reduces the balance, but it is not an investment loss. The balance equation keeps those components separate and provides a direct route to the return equation.

The derivation below assumes equally spaced periods, external cash flows made at period ends, and one constant effective return per period. Those timing assumptions matter: a contribution at a period’s end does not earn a return during that period. Cash flows made at other times require a different timing treatment.

Definition: Let $B_k$ be the account balance immediately after the external cash flow at the end of period $k$, with $B_0$ the initial balance. Let $C_k$ be the net external cash flow into the account at the end of period $k$: a contribution is positive, and a withdrawal is negative. Let $j$ be the constant effective investment return per period. The dollar-weighted return is the rate that makes the account’s initial value, external cash flows, and ending value consistent under this timing model.

The model assumes that during period $k$, the opening balance $B_{k-1}$ earns $jB_{k-1}$. There are no external cash flows during the period; the flow $C_k$ occurs after that period’s investment earnings have been credited. Fees, taxes, or other non-investment changes must either be excluded or represented explicitly as cash flows according to the chosen model.

The One-Period Balance Equation

Separate the change in the account into investment earnings and the external flow. If $G_k$ denotes the investment earnings in period $k$, then

B_k=B_{k-1}+G_k+C_k.

Under the constant-return assumption, investment earnings are $G_k=jB_{k-1}$. Substitution gives the balance recurrence

B_k=B_{k-1}(1+j)+C_k.

This is the essential accounting identity for the derivation. The return applies to the balance invested during the period, while the end-of-period cash flow is added afterward. Rearranging also shows how to isolate the period’s earnings from observed balances and a known flow:

G_k=B_k-B_{k-1}-C_k.

A balance increase by itself is therefore not a return: first remove the amount contributed, or add back the amount withdrawn, to identify investment earnings.

Deriving the Multi-Period Return Equation

Apply the recurrence repeatedly. After the first period, $B_1=B_0(1+j)+C_1$. After the second, the entire first-period balance earns another period’s return, while $C_2$ is added only at the end of period 2:

B_2=\bigl(B_0(1+j)+C_1\bigr)(1+j)+C_2 =B_0(1+j)^2+C_1(1+j)+C_2.

Continuing in the same way, the opening balance earns $n$ periods of return, and a flow made at the end of period $k$ earns only the remaining $n-k$ periods. Thus, after $n$ periods,

B_n=B_0(1+j)^n+\sum_{k=1}^{n}C_k(1+j)^{n-k}.

This formula follows directly from the balance recurrence: each earlier balance is multiplied by $1+j$ once per subsequent period, while each end-of-period flow enters at its own date. For $k=n$, the exponent is zero, so the final-period flow is included without earning a period’s return.

The same relation can be written at time 0 by dividing both sides by $(1+j)^n$. This gives

B_0+\sum_{k=1}^{n}\frac{C_k}{(1+j)^k} =\frac{B_n}{(1+j)^n}.

From the investor’s perspective, a contribution is money paid into the account and is therefore a negative investor cash flow; a withdrawal is money received and is positive. Moving all values to one side gives the equivalent zero-present-value equation:

-B_0-\sum_{k=1}^{n}\frac{C_k}{(1+j)^k} +\frac{B_n}{(1+j)^n}=0.

Here $C_k$ retains its account perspective: positive $C_k$ is a contribution, so its investor-perspective cash flow is $-C_k$. A negative $C_k$ is a withdrawal, so $-C_k$ is positive. This is the connection between an account-balance calculation and the dollar-weighted return equation for investor cash flows.

Choose a consistent perspective.
Use positive $C_k$ for money entering the account and negative $C_k$ for money leaving it.

Apply the period recurrence.
Multiply the opening balance by $1+j$, then add the end-of-period flow.

Accumulate each flow to the ending date.
A flow at the end of period $k$ is multiplied by $(1+j)^{n-k}$, not by $(1+j)^n$.

Check the result.
Substitute the proposed return into the recurrence period by period and confirm the stated ending balance.

Worked Examples

Worked Example: Separating Earnings From External Flows

An account opens with $\$5{,}000$. It earns a constant effective return of $6\%$ per period. The investor contributes $\$800$ at the end of period 1, withdraws $\$400$ at the end of period 2, and contributes $\$250$ at the end of period 3. Find the balance after period 3, and identify the investment earnings in each period.

Method. Use $B_k=B_{k-1}(1+j)+C_k$, with $j=0.06$. The flow is applied after each period’s earnings.

Period 1. The opening balance earns $0.06(5{,}000)=\$300$. Then the $\$800$ contribution is added:

B_1=5{,}000(1.06)+800=5{,}300+800=\$6{,}100.

Period 2. Earnings are $0.06(6{,}100)=\$366$. The withdrawal is $C_2=-\$400$:

B_2=6{,}100(1.06)-400=6{,}466-400=\$6{,}066.

Period 3. Earnings are $0.06(6{,}066)=\$363.96$. The contribution is then added:

B_3=6{,}066(1.06)+250=6{,}429.96+250=\$6{,}679.96.

Check using the accumulated-flow equation. The initial balance earns three periods, the period-1 contribution earns two, the period-2 withdrawal is carried forward one period, and the final contribution earns no period:

\begin{aligned} B_3 &=5{,}000(1.06)^3+800(1.06)^2-400(1.06)+250\\ &=5{,}955.08+898.88-424+250\\ &=\$6{,}679.96. \end{aligned}

The two calculations agree. The common mistake this method avoids is treating the whole balance increase as investment earnings; the $\$800$ and $\$250$ deposits and the $\$400$ withdrawal are external flows, not returns.

Worked Example: Finding the Constant Return From Balances

An account begins with $\$1{,}000$, receives a $\$100$ contribution at the end of period 1, and has a $\$200$ withdrawal at the end of period 2. Its ending balance immediately after the withdrawal is $\$1{,}007.50$. Find the constant effective return per period.

Method. Accumulate each value to the end of period 2. The initial balance earns two periods, the first-period contribution earns one, and the period-2 withdrawal earns none:

1{,}000(1+j)^2+100(1+j)-200=1{,}007.50.

Expand the left side and collect terms:

\begin{aligned} 1{,}000(1+2j+j^2)+100+100j-200&=1{,}007.50,\\ 1{,}000j^2+2{,}100j+900&=1{,}007.50,\\ 1{,}000j^2+2{,}100j-107.50&=0. \end{aligned}

The quadratic factors as

$$ (j-0.05)(1{,}000j+2{,}150)=0. $$

The roots are $j=0.05$ and $j=-2.15$. The second root is below $-1$, so it cannot be an effective return for a standard account whose invested balance is multiplied by $1+j$ each period. The admissible return is therefore $5\%$ per period.

Check by reconstructing the balances. After period 1, the balance is $1{,}000(1.05)+100=\$1{,}150$. Period-2 earnings are $0.05(1{,}150)=\$57.50$, so after the withdrawal the balance is $1{,}150+57.50-200=\$1{,}007.50$, as stated. The common mistake avoided here is assigning the contribution two periods of growth even though it arrived at the end of period 1.

Worked Example: Translating the Balance Equation to Investor Cash Flows

An account starts with $\$2{,}000$. The investor contributes $\$250$ at the end of year 1, contributes another $\$300$ at the end of year 2, and withdraws $\$500$ at the end of year 3. If the constant annual effective return is $4\%$, determine the ending balance and write the equivalent investor-perspective return equation.

Method. First use account signs: $C_1=250$, $C_2=300$, and $C_3=-500$. The balance equation is

B_3=2{,}000(1.04)^3+250(1.04)^2+300(1.04)-500.

Compute the factors and terms:

\begin{aligned} (1.04)^2&=1.0816, & (1.04)^3&=1.124864,\\ 2{,}000(1.124864)&=2{,}249.728,\\ 250(1.0816)&=270.40,\\ 300(1.04)&=312.00. \end{aligned}

Therefore,

B_3=2{,}249.728+270.40+312.00-500 =\$2{,}332.128.

The investor’s cash flows are $-\$2{,}000$ at time 0, $-\$250$ at year 1, $-\$300$ at year 2, and $+\$500$ at year 3 from the withdrawal. The terminal account value is also received by the investor at year 3. Thus the return equation is

-2{,}000-\frac{250}{1.04} -\frac{300}{(1.04)^2} +\frac{500}{(1.04)^3} +\frac{2{,}332.128}{(1.04)^3}=0.

This equation is equivalent to the accumulated-balance equation because discounting that balance equation by $(1.04)^3$ moves every value to time 0. As a numerical check on the balances, $B_1=2{,}000(1.04)+250=\$2{,}330$, $B_2=2{,}330(1.04)+300=\$2{,}723.20$, and $B_3=2{,}723.20(1.04)-500=\$2{,}332.128$. The common mistake avoided is giving the year-3 withdrawal a negative investor sign: it is money received by the investor, so its investor cash flow is positive.

Timing and Sign Traps

The exponent in the accumulated-balance equation records how long each flow remains invested. A contribution at the end of period 1 in a three-period model earns two periods, so its factor is $(1+j)^2$. A contribution at the end of period 3 earns none, so its factor is $(1+j)^0=1$. A withdrawal is negative when written as an account flow, but positive when written as an investor receipt.

For example, in the two-period calculation above, an incorrect equation that lets the period-1 contribution earn two periods would be

1{,}000(1+j)^2+100(1+j)^2-200=1{,}007.50.

At $j=5\%$, its left side is $1{,}102.50+110.25-200=\$1{,}012.75$, not $\$1{,}007.50$. The contribution was made after period 1’s earnings, so it has only one remaining period in which to grow. The correct term is $100(1+j)$.

The recurrence also provides a useful diagnostic when balances and flows are given. For each period, compute $B_k-B_{k-1}-C_k$. Under the constant-return model, this amount must equal $jB_{k-1}$. If the implied rates differ across periods, the balances do not support one constant per-period return under the stated timing and flow assumptions.

Key takeaway: With end-of-period external flows and a constant effective return per period, the account evolves by $B_k=B_{k-1}(1+j)+C_k$. Repeating this identity gives the investor return equation; the flow sign and the number of periods remaining determine each term.

Check Your Understanding

Use end-of-period cash-flow timing and the account sign convention: contributions are positive and withdrawals are negative.

An account opens with $\$1{,}800$, earns $3\%$ per period, receives $\$250$ at the end of period 1, and pays a $\$100$ withdrawal at the end of period 2. Find its ending balance. (Answer: $\$2{,}067.12$)
An account opens with $\$4{,}000$, earns $2\%$ per period, has a $\$300$ withdrawal at the end of period 1 and a $\$500$ contribution at the end of period 2. Find its period-2 ending balance. (Answer: $\$4{,}355.60$)
An account opens with $\$1{,}600$, receives $\$200$ at the end of period 1, and has a $\$100$ withdrawal at the end of period 2. The ending balance is $\$1{,}874$. Find the admissible constant effective return per period. (Answer: $5\%$)
An account opens with $\$1{,}000$, earns $10\%$ per period, receives $\$100$ at the end of period 1, has no flow at the end of period 2, and withdraws $\$200$ at the end of period 3. Find the ending balance. (Answer: $\$1{,}252.00$)
An account opens with $\$5{,}000$, receives $\$400$ at the end of period 1, and pays a $\$600$ withdrawal at the end of period 2. Write the accumulated-balance equation for an unknown constant return $j$, and find the ending balance if $j=4\%$. (Answer: $B_2=5{,}000(1+j)^2+400(1+j)-600$; at $4\%$, $B_2=\$5{,}224.00$)

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