Derivation of Continuous Compounding

The derivation of the formula for continuous compounding starts from interest calculation with discrete compounding. As compounding intervals become shorter the resulting exponential growth converges to an exponential function. Because of the infinitesimally small intervals continuous compounding results in an instantaneous rate of return.

While continuous compounding is next to irrelevant in private banking, it’s of crucial importance in financial calculus when pricing derivatives.

Discrete Compounding

For an account earning fixed interest at discrete intervals the development of the balance B(t) over time is given by the following formula:

\begin{alignat*}{2}
B(t)&&=&B_0 \cdot (1+r\frac{1}{m})^{t \cdot m}\\
\\
B(t):&& &\mathrm{balance \space at \space time} \space t\\
B_0: && &\mathrm{balance \space at \space time \space zero}\\
t: && & \mathrm{time \space in \space years}\\
r: && & \mathrm{interest \space rate}\\
m: && & \mathrm{compounding}\\
 && & \mathrm{frequency}
\end{alignat*}

With discrete compounding interest earnings are credited to the account B at fixed intervals of duration Δt=1/m and in turn earn interest in the next interval. For instance, $100 earning a fixed nominal interest rate of 10% over 10 years grow to the following amounts at different compounding intervals Δt:

In the derivation of continuous compounding the periods Δt become infinitesimally small. As decreasing differences between end balances in Table 1 suggest, interest calculation with an infinite number of compounding steps yields finite results. This shall subsequently be proven.

Compound Interest as a Recursion

In order to find a solution for compounding frequency going to infinity, the first step is to rewrite the compound interest formula as a recursion:

\tag{1}B_{t+1} = B_t \cdot (1 + r \frac{1}{m})

Equation (1) has investment B change in discrete time steps Δt=1/m. Now subtract B_t on both sides to get a difference equation:

B_{t+1} - B_t = B_t \cdot r \frac{1}{m}

Finally, with Δt=1/m:

\tag{2}\Delta B_t = B_t \cdot r \Delta t

Differential Calculus

Since time steps Δt tend to zero for continuous compounding, the derivation of a formula for continuous compounding follows from differential calculus. Infinitesimally small Δt turn Equation (2) into a differential equation with instantaneous return r(t).

\tag{3}dB(t) = r(t)~B(t) dt

In order to get an easy solution by integration, divide by B(t):

\tag{4}\frac{1}{B(t)} dB(t)=r(t)dt

Subsequently, integrate the interval from t to t+τ:

\begin{gather*}
\tag{5}\int_{B(t)}^{B(t+\tau)}\frac{1}{B(t)} dB(t)\\=\int_{t}^{t+\tau} r(s)ds
\end{gather*}

The primitive function to the integral on the left hand side of Equation (5) is the natural logarithm, so that the following solution results:

\tag{6}\begin{align}
&\int_{B(t)}^{B(t+\tau)}\frac{1}{B(t)} dB(t)\\
&\quad\begin{aligned}&=ln(B(t+\tau))\\
&\quad-ln(B(t))
\end{aligned}\\
&\quad=ln(\frac{B(t+\tau)}{B(t)})
\end{align}

Substituting Equation (6) into Equation (5):

\tag{7}ln(\frac{B(t+\tau)}{B(t)})\\=\int_{t}^{t+\tau} r(s)ds

Applying the exponential function to both sides of equation (7) yields the general formula for continuous compounding:

\tag{8}B(t+\tau)=B(t)~ e^{(\int_{t}^{t+\tau} r(s)ds)}

Since in the general case the instantaneous rate of return r(t) is a function of time, Equation (8) has an integral averaging over the time period τ. But if r is constant, the integral, which is also the right side of Equation (7), simplifies:

\tag{9}\int_{t}^{t+\tau} r ds=r \tau

Inserting Equation (9) into Equation (8) produces the formula for a constant instantaneous rate of return, which may look more familiar.

\tag{10}B(t+\tau)=B(t) \cdot e^{r \tau}

Finally, setting t=0 and naming B(0)=B₀ simplifies Equation (10) to the usual formula for continuous compounding with a constant rate r.

\tag{11}B(\tau)=B_0 \cdot e^{r \tau}

References

Compound Interest, Wikipedia.org

Differential Calculus, Wikipedia.org