I have a question for the financial part of my course which I am struggling to answer as i am not sure my answer makes sense.
Question:
Time is counted from the present t = 0 in years. Suppose for the first 12 years the force of interest is 5%. After that it changes to 3% with probability 0.25, remains unchanged with probability 0.5 and increases to 7% with probability 0.25.
Find the expected present value of a continuous payment stream of £190 per annum for 20 years, beginning at time 0.
Solution:
So i have that $v(t)= exp(-\int_{0}^{t}0.05ds)$ for $t \leq 12$
and $v(t)= exp(-\int_{0}^{12}0.05ds -\int_{12}^{20} \delta_{12} ds )=exp(-0.6-8 \delta_{12})$ for $t > 12$
Hence, the present value is given by
$\int_{0}^{20} v(t)dt=\int_{0}^{12}exp(-0.05t)dt+\int_{12}^{20}(-0.6-8\delta_{12})dt$
$=\frac{1-exp(-0.6)}{0.05}+8exp(-0.6-8\delta_{12})$
where $\delta_{12} = 0.03, 0.05, 0.07$,
This is all fine (I think!) but when i actually put the percentages in, i get that:
$=\frac{1-exp(-0.6)}{0.05}+8exp(-0.6-8*0.03)=12.47745*190$
$=\frac{1-exp(-0.6)}{0.05}+8exp(-0.6-8*0.05)=11.838747*190$
$=\frac{1-exp(-0.6)}{0.05}+8exp(-0.6-8*0.07)=11.53165673*190$
Here is what i am confused about!! Why does the present value go down as the interest rate goes up? Have i done something wrong here?
I understand that to find the actual present value taking probabilities into account, i would need to do
$\frac{1}{4}12.47745*190+\frac{1}{2}11.838747*190+\frac{1}{4}11.53165673*190$
and this would be my present value but I am concerned that I have gone wrong, on similar examples the present value gets higher the higher the interest.
Any help much appreciated
Best Answer
To begin with, the best way I would choose to solve this is to find the present value in each of the three cases as:
$$PV = 190\times(\bar{a}_{12}+v^{12}\bar{a}_{8}) = 190\times(\dfrac{1-e^{-0.05\times12}}{0.05}+e^{-0.05\times12}\dfrac{1-e^{-\delta\times8}}{\delta})$$
Which gives, for $i=0.03$, $190\times12.9271$, for $i=0.05$, $190\times12.6424$, and for $i=0.07$, $190\times12.3856$. If you were to use the integration method as you did, then your formula should be:
$$PV=\int_0^{20}\rho(t)v(t)dt=190\times\int_0^{20}v(t)dt$$
Then $v(t)=e^{-0.05t}$ for $t<12$, and for $\delta=0.03$, $0.05$, or $0.07$, it's $e^{-0.05\times12}\times e^{-\delta (t-12)}$. Therefore, our formula becomes
$$PV=190\times(\int_0^{12}v(t)dt + e^{-0.6}\int_{12}^{20}v(t)dt)$$ $$=190\times(\int_0^{20}e^{-0.05t}dt + e^{-0.6}\int_{0}^{8}e^{-\delta t}dt)$$
Which yields the same values as above.
I've put in full working just in case, but your mistake seems to be integrating from $12$ to $t$ instead of from $12$ to $20$, which would give you $\delta(t-12)$ as I've shown above (instead of $8\delta$). As a reference check, remember that $v(t)$ should (almost) always vary with $t$, since it's the present value of 1 at time $t$. The final answer should be £2403.38.