This is a problem from Marcel Finan's Exam FM/2 course. It is not homework but I am studying for the FM exam and trying to get through this.
You are given $\frac{d}{dt}\bar{s}_t$ = $(1.02)^{2t}$. Calculate $\delta$.
So we will integrate to and get $\bar{s}_t$ = $\int{(1.02)^{2t}}dt$ = $\frac{(1.02)^{2t}}{2ln(1.02)}$ = $\frac{(1.02)^{2t}}{ln(1.02)^2}$.
This is where I get confused. I can expand $\bar{s}_t$ since it equals $\frac{(1+i)^t-1}{\delta}$, but i don't have a value for $i$ explicitly defined in the problem. I don't think $i$ = .02, so where do i go from here?
To solve this, it is perhaps easier to use the fact that $$ \bar{s}_t = \frac{e^{t\delta} -1}{\delta}, $$ so that $$\frac{d}{dt}\bar{s}_t = e^{t\delta}= (1.02)^{2t}.$$ Taking natural log of both sides should give that $\delta = 2\log(1.02)$.