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.


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

$\int_{0}^{20} v(t)dt=\int_{0}^{12}exp(-0.05t)dt+\int_{12}^{20}(-0.6-8\delta_{12})dt$


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:


$=\frac{1-exp(-0.6)}{0.05}+8exp(-0.6-8*0.05)=11.838747*190$ $=\frac{1-exp(-0.6)}{0.05}+8exp(-0.6-8*0.07)=11.53165673*190$

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


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

  • $\begingroup$ I haven't checked through the rest of it, but the present value should be decreasing when the interest rate increases. Think about it this way - if the interest rate was 5%, and it increases to 10%, you would need less money today to have the same amount of money in the future. Therefore the present value of that future amount of money is less today. This comes from the definition of present value. $\endgroup$
    – mardat
    Dec 9, 2014 at 4:25
  • $\begingroup$ Ah thank you!! I actually thought that too, but looking at another example with full solutions and it is not the case, so thought i must be tired hah! Does the rest of the method (finding present value with percentages) look okay? Thank you! $\endgroup$ Dec 9, 2014 at 4:27
  • $\begingroup$ Not exactly the PV, we need to add the PV of the first $12$ year chunk. $\endgroup$ Dec 9, 2014 at 4:35
  • $\begingroup$ have i not done that with the 0-12 integral? $\endgroup$ Dec 9, 2014 at 4:37
  • $\begingroup$ I have the present value as $\int_{0}^{20} v(t)dt=\int_{0}^{12}exp(-0.05t)dt+\int_{12}^{20}(-0.6-8\delta_{12})dt$, would $\int_{0}^{12}exp(-0.05t)dt$ not be the PV for the first 12 years? Sorry I'm a little confused $\endgroup$ Dec 9, 2014 at 4:56

1 Answer 1


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:


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.

  • $\begingroup$ Wow, thank you so much I have been working on this for a long time, and finally can see how to get the correct answer! Thank you, your help is perfect, Im not sure why I integrated from 12 to t actually, i was basing my answer on a similar example but didnt quite understand it. Thanks again it makes much more sense now and that is in fact the correct answer too :) $\endgroup$ Dec 9, 2014 at 8:14

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