J
Jonyy
Member
I came across the following question in CT1 2016 Sept Q12 part iv exam (I slightly modified the question on purpose for clarity):
Suppose we know that a continuous payment stream is paid at the rate of r(t) = e^(–0.04t) per annum for 1 year with a constant force of interest δ = 0.03. What is the present value of the payment stream from t=0 to t=1?
As I understand that following the original solution, the current solution it shoud be :
Now, I don't fully understand why by multiplying the continuous payment stream rate with the v(t) = e^(-0.03t) and then integrating it you will result in the PV? I'm not sure why does it have to be multiply by a the rate before the integral. But my logic tells me that if you want to get the PV of "one" of the discreet payment (instantaneous payment) at any time t out of the continuous payment stream where 0<t<1, I got:
PV = v(t) [R(t+dt) - R(t)]
Where dt is close to 0. And the term after v(t) is the instantaneous payment at time t found by integrating r(t) and then substracting it's value of t+dt with it's value of t.
Can anyone explain how does the last equation fits in the solution when integrating to?
Suppose we know that a continuous payment stream is paid at the rate of r(t) = e^(–0.04t) per annum for 1 year with a constant force of interest δ = 0.03. What is the present value of the payment stream from t=0 to t=1?
As I understand that following the original solution, the current solution it shoud be :
Now, I don't fully understand why by multiplying the continuous payment stream rate with the v(t) = e^(-0.03t) and then integrating it you will result in the PV? I'm not sure why does it have to be multiply by a the rate before the integral. But my logic tells me that if you want to get the PV of "one" of the discreet payment (instantaneous payment) at any time t out of the continuous payment stream where 0<t<1, I got:
PV = v(t) [R(t+dt) - R(t)]
Where dt is close to 0. And the term after v(t) is the instantaneous payment at time t found by integrating r(t) and then substracting it's value of t+dt with it's value of t.
Can anyone explain how does the last equation fits in the solution when integrating to?
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