1
$\begingroup$

How do I find the value of bonds with continuous coupon yields and interest rates that are both a function of time?

The bond has a redemption of 2000 at time $t=2$ and pays continuous coupon payments of $K(t)=100e^{-t}$. The spot interest rate is $r(t)=\frac{2}{50-t}$.

$\endgroup$
4
  • 1
    $\begingroup$ Can you post an example ISIN please? $\endgroup$ Jan 19, 2021 at 22:12
  • $\begingroup$ I'm sorry but what is an ISIN? $\endgroup$
    – Podski
    Jan 19, 2021 at 22:19
  • 3
    $\begingroup$ ISIN is a kind of an identifier for a particular bond issue. Can you please point to an example of such a bond that exists in practice? $\endgroup$ Jan 20, 2021 at 0:00
  • 1
    $\begingroup$ I don't know if it's a real life example but the problem i'm trying to solve is A bond V(t) has a redemption of Z = £1000 at time t = 2. The interest rate is r = 2/(50−t) and the bond pays a continuous coupon yield of K(t) = 100 exp(−t). Determine the value of the bond at time t = 0. But I'm getting to a point where I have an integral that I can't solve using methods I've been taught, so I'm inclined to think that i'm wrong. $\endgroup$
    – Podski
    Jan 20, 2021 at 2:01

1 Answer 1

2
$\begingroup$

To echo on @Dimitri Vulis comment on your question, this kind of bond structure is rather contrived, but nevertheless let me try to give you some starting pointers.

I am assuming that your interest rate is a continuously compounded zero rate, i.e. the discount factor for time t equals $D(t)=e^{-r(t)t}$. Then the present value fo the bond equals discounted (one-time) redemption payment and discounted coupon income stream, i.e.

$$V(0)=B\times D(T) + \int_{s=0}^{T}K(s)D(s)\,\mathrm{d}s$$

Inserting your data, we get

$$V(0)=2000e^{-\frac{1}{12}} + 100\int_{s=0}^{T}e^{\frac{2}{50-s}s^2}\,\mathrm{d}s$$

The integral on the RHS seems to have no closed form solution (at least thats what Wolfram Alpha tells me).

HTH?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.