2
$\begingroup$

I know the YTM of a coupon bond is the interest rate $i$ which verifies

$ P =\frac{C}{(1+i)} + \frac{C}{(1+i)^2} + ...+ \frac{C}{(1+i)^n} + \frac{F}{(1+i)^n} $

where $P$ is price, $C$ is the coupon payment and $F$ is face value.

I don't understand why $i = C/F$ when $P=F$. In words: I can't grasp why the yield to maturity equals the coupon rate when the bond is priced at face value.

On the one hand I can't solve that equation above so that this fact is verified, but I might need some tools I don't have yet to do so. On the other hand it doesn't make intuitive sense to me on a conceptual level.

What am I missing?

$\endgroup$
4
$\begingroup$

Let $P$ denote the dirty price, $F$ the face value and $i$ the YTM. Using the geometric sum we get

\begin{align} P &= \sum_{j=1}^n \frac{C}{{(1+i)}^j} + \frac{F}{(1+i)^n}\\ &= C\frac{1-\frac{1}{{(1+i)}^n} }{i} + \frac{F}{(1+i)^n} \end{align}

and thus

\begin{align} P=F \Leftrightarrow & F= C\frac{1-\frac{1}{{(1+i)}^n} }{i} + \frac{F}{(1+i)^n} \\ \Leftrightarrow & C\left(1-\frac{1}{{(1+i)}^n}\right) =i \left( F- \frac{F}{(1+i)^n} \right)\\ \Leftrightarrow & \frac{C}{F} = i \end{align}

$\endgroup$
0
$\begingroup$

The price of a bond is determined by the sum of the discounted cashflows plus the discounted face value of the bond. Intuitively and academically, a bond cannot be worth more than the sum of the future cashflows plus future value. In the case of yield equaling coupon rate, the price is equal to par because the rate at which you are discounting makes it so that the sum of the discounted cashflows and discounted par equal present par. Understanding this, by looking at the equation you should be able to convince yourself that is the case. Here is a simple example using a 3 year note with a 3% coupon:

$$ 100 = \frac{3}{(1+0.03)^1} + \frac{3}{(1+0.03)^2} + \frac{100+3}{(1+0.03)^3}$$ $$ 100 = 2.912621 \ + \ 2.827788 \ + \ 94.25959 $$

Let me know if that helps.

$\endgroup$
  • $\begingroup$ Well, actually he split the fraction for the nth period in C and F so it's actually fine and should result in the correct value. $\endgroup$ – David Duarte Nov 27 '18 at 15:28
  • $\begingroup$ Ooops, missed that. Revised to reflect. $\endgroup$ – PlantFox Nov 27 '18 at 15:29

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.