# Calculate duration of zero coupon bond

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?

• Which step of the derivation do you not understand ? That would help to answer you. – siou0107 Jan 15 '20 at 10:57
• I am getting confused with the general way they derive it i guess, so the: $\frac{d F}{d x}=A \times a \times e^{a x}=a \times F(x)$ – mbih Jan 15 '20 at 11:02
• So I am confused of how we get the $\left[-(T-t) \times P_{z}(t, T)\right]$ from $\left[\frac{d P_{z}(t, T)}{d r}\right]$ – mbih Jan 15 '20 at 11:10
• 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 $y = ax$. – siou0107 Jan 15 '20 at 11:13
• 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... – siou0107 Jan 15 '20 at 12:46

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