Force of Interest Under Simple and Compound Accumulation

Summary

The document explains how to derive the force of interest, defined as the proportional rate of change in an accumulation function. For simple interest at annual rate i, the accumulation function is linear in time, so its logarithmic derivative varies with time. For compound interest, accumulation is exponential in time, and rewriting it with the exponential function makes the differentiation straightforward.

The provided answers show that the force of interest under compound accumulation is constant and equals the natural logarithm of one plus the annual effective rate. The simple-interest expression is included for comparison. This is a basic actuarial finance derivation rather than a trading strategy, and the document does not discuss market data, risk, or empirical evidence.

Key ideas

• The force of interest is the accumulation function’s derivative divided by its value.
• Under simple interest, the force of interest changes with time.
• Under compound interest, the accumulation function can be expressed exponentially.
• The force of interest for compound accumulation is the natural logarithm of one plus the annual rate.

Tags

# Force of Interest Compounding at Annual Rate i


# Force of Interest Compounding at Annual Rate i












I'm working through some actuarial practice and am lost as to what's going on with the differentiation here (it's been a while since I've had calc):

Derive an expression for $\delta_t$ if accumulation is based on:

(a) simple interest at annual rate i, and

(b) compound interest at annual rate i.

I understand the answer to (a), which is $\delta_t$ = $\frac{ A'(t)}{A(t)}$ = $\frac{i}{1+i \cdot t}$.

However, I don't understand their method of getting $\ln(1+i)$ for part (b).

Any help would be much appreciated!

(p.s. this is Example 1.13 p.40 from Broverman's Mathematics of Investment and Credit 5th ed)

## Answer by meh (score 1, accepted)

https://quant.stackexchange.com/a/19579

Compound interest is $A(t) = (1+i)^t$. So then $\delta_t = \frac{d}{dt}ln(A(t)) = \frac{d}{dt}t*ln(1+i) = ln(1+i)$

## Answer by Bjørn Kjos-Hanssen (score 0)

https://quant.stackexchange.com/a/36609

From rewriting $$A(t)=(1+i)^t = e^{t\ln(1+i)}$$ we get by the Chain Rule that $$\delta_t = \frac{A'(t)}{A(t)}=\frac{\ln(1+i)\cdot e^{t\ln(1+i)}}{A(t)}=\ln(1+i).$$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.