0
$\begingroup$

I am currently studying interest rate risk management, and i can't seem to get the derivation right, and I would like to do all of the steps, to be sure that I understand what is going on.

Let Pz (t, T ) be the price of a zero coupon bond at time t with maturity T and continuously compounded interest rate r.

Duration = $-\frac{1}{P} \frac{d P}{d r}$

Let A and a be two constants and x be a variable. Let $F(x)=A \times e^{a x}$ be a function of x. Then, the first derivative of F with respect to x, denoted by $\frac{d F}{d x}$, is given by

Derivative of F(x) with respect to $x=\frac{d F}{d x}=A \times a \times e^{a x}=a \times F(x)$

The book shows (duration of zero coupon bond): $D_{z, T}=-\frac{1}{P_{z}(t, T)}\left[\frac{d P_{z}(t, T)}{d r}\right]$

$=-\frac{1}{P_{z}(t, T)} \times\left[-(T-t) \times P_{z}(t, T)\right]$

$=T-t$

Because I know the theory this makes total sense, but I cannot derive it. Does someone know how to do this?

Improve this question
$\endgroup$
13
  • $\begingroup$ Which step of the derivation do you not understand ? That would help to answer you. $\endgroup$
    – siou0107
    Jan 15, 2020 at 10:57
  • $\begingroup$ I am getting confused with the general way they derive it i guess, so the: $\frac{d F}{d x}=A \times a \times e^{a x}=a \times F(x)$ $\endgroup$
    – mbih
    Jan 15, 2020 at 11:02
  • $\begingroup$ So I am confused of how we get the $\left[-(T-t) \times P_{z}(t, T)\right]$ from $\left[\frac{d P_{z}(t, T)}{d r}\right]$ $\endgroup$
    – mbih
    Jan 15, 2020 at 11:10
  • $\begingroup$ This is a mere application of a calculus rule called chain rule. I let you check Wikipedia but is states that $\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{\mathrm{d}z}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x}$. Here $z = e^y$ and $y = ax$. $\endgroup$
    – siou0107
    Jan 15, 2020 at 11:13
  • 2
    $\begingroup$ You just differentiate $P = e^{-r\left(T - t\right)}$ using the chain rule and multiply that derivative by $-\frac{1}{P}$, I cannot imagine a simpler explanation... $\endgroup$
    – siou0107
    Jan 15, 2020 at 12:46

1 Answer 1

Reset to default
1
$\begingroup$

$$P_z\left(r, t, T\right) = e^{-r\left(T - t\right)} \Rightarrow \partial_r P_z = -\left(T - t\right)P \Rightarrow D_{z, T} = T - t$$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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