Problem on force of interest accumulation with continuous payment stream

Hi,

Can someone help me with the following:

Question:

The force of interest per unit time at time t, δ(t), is given by:

δ(t) = 0.1-0.005t for t<6
δ(t) = 0.07 for t>6


(ii) Calculate the present value at time 0 of a continuous payment stream at the rate £50e0.05t per unit time received between time 12 and time 15.

Solution:

In the solution you integrate the rate of payment with limits 15,12, multiplied by V(t). What I would like to know is, for the V(t) you use 'exp integrate 0.07 for t>6, with limits t,6'

Why is the upper limit t?? In my answer I put the upper limit as 15

Thanks

Nik

 

First thing draw a time line.The function when 't' is between 0-6yrs and 'greater' than 6yrs is given.We know that ''t'' is somewhere between 12 and 15, because there is a continuous payment. So just mark ''t'' on your timeline. Now this ''t'' is definitely greater than 6yrs. So when we are discounting we integrate the corresponding function 0.07 with the upper limit 't' and lower limit 6. And we need to further discount this to time 0 and so we integrate 0.1-0.005 with the limits 6 and 0.

Please find attachment to understand what I'm trying to say. I'm not very good with words.

Hope this helps.