Exam FMMeasurement of interestFree to read

Force of interest

The force of interest δ(t) is the instantaneous rate of growth of an accumulation function — the continuous-compounding limit that makes every other rate a special case.

The formulas

Definition

$${\delta }_t=\frac{a^{\prime }(t)}{a(t)}=\frac{d}{dt}\ln a(t)$$

Accumulation

$$a(t)=\exp \left({\int }_0^t{\delta }_{s}ds\right)$$

Constant force

$$\delta =\ln (1+i),i=e^{\delta }-1$$

Discounting

$$v^t=e^{-\delta t}$$

Where it comes from

1. Over a tiny interval $dt$ the fund grows by $a^{\prime }(t)dt$, which as a PROPORTION of the fund is $\frac{a^{\prime }(t)}{a(t)}dt$ — that proportion per unit time is the force.
2. $\frac{a^{\prime }(t)}{a(t)}=\frac{d}{dt}\ln a(t)$, so integrating both sides from 0 to $t$ and using $a(0)=1$ gives $\ln a(t)={\int }_0^t{\delta }_{s}ds$.
3. If the force is constant then $a(t)=e^{\delta t}$, and matching to $(1+i{)}^t$ gives $\delta =\ln (1+i)$.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• δ = ln(1 + i). Everything continuous is that one substitution.
• δ < d⁽ᵐ⁾ … < δ < … < i⁽ᵐ⁾ < i is the standard ordering: d < d⁽²⁾ < … < δ < … < i⁽²⁾ < i.

Traps

• Using a constant-force formula when δ(t) varies with t — then you must integrate.
• Forgetting that ∫₀ᵗ δₛ ds appears inside an exponential, not multiplied by it.

Related

• Nominal rates convertible m-thly
• Effective interest and discount rates
• Continuously payable annuities

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