Assuming that the force of interest is doubled. Show that the effective annual interest rate is more than doubled.
I tried to prove this question by using arbitrary numbers but I am not sure that this is the right way to prove it.
Assume that the annual interest rate is 10% and if we double that we get 20%. The effective annual interest rate for the same I for 2 years is $(1.10)^2-1=0.21$ and since $0.21 > 0.20$ than it is more than the doubled value.
Is there another way to prove this? thank you
The force of interest is defined as $$ \delta(t)=\frac{a'(t)}{a(t)} $$ where $a(t)$ is the accumulation factor $$ a(t)=a(t-1)\mathrm{e}^{\int_{t-1}^t\delta(\tau)\,\mathrm d \tau} $$ and the effective interest rate is $$ i(t)=\frac{a(t)-a(t-1)}{a(t-1)}=\frac{a(t)}{a(t-1)}-1>0 $$ Assume that $\delta(t+1)=2\delta(t)$, so we have $$ \frac{a(t+1)}{a(t)}=\mathrm{e}^{\int_t^{t+1}\delta(\tau)\,\mathrm d \tau}=\mathrm{e}^{\int_{t-1}^{t}\delta(\tau+1)\,\mathrm d \tau}=\mathrm{e}^{\int_{t-1}^{t}2\,\delta(\tau)\,\mathrm d \tau}=\left[\frac{a(t)}{a(t-1)}\right]^2 $$ and then $$ \frac{i(t+1)}{i(t)}=\frac{\frac{a(t+1)}{a(t)}-1}{\frac{a(t)}{a(t-1)}-1}=\frac{a(t)}{a(t-1)}+1=2+i(t)>2 $$ that is $i(t+1)>2i(t)$.