4. Duration and Convexity

Overview

Duration is an important concept that helps us to compute, quickly and approximately, how much the price of a bond should change in response to the yield change
- Expressed in years, but it’s not the same as time to maturity
- A sensitivity measure
  - High duration: bond price sensitive to interest rate changes
  - Low duration: insensitive, bond price stable
Duration also summarizes a portfolio of bonds to one number, making it easy to hedge

Bond Pricing Review

(Time $t=0$ price of) zero coupon bond: $P(0, T) = e^{-T\times R(0, T)}$
Write in a simplified notation:

\[B = e^{-TR}\]
Zero coupon bond with notional $\mathcal N$:

\[B = \mathcal N e^{-TR}\]
Non-zero coupon bond that pays a fixed coupon $c$ at times $\{T_1, T_2, \ldots, T_N\}$ and notional $\mathcal N$ at maturity $T_N$:

\[B = \mathcal N e^{-T_NR_N} + \sum_{j=1}^N c~e^{-T_j R_j},\]

where $R_j = R(0, T_j)$
Despite the seemingly involved notations, it’s simply sum of discounted future cash flows

Bond Pricing Review (Cont.)

\begin{align*} B &= \mathcal N e^{-T_NR_N} + \sum_{j=1}^N c~e^{-T_j R_j} \\ &= c~e^{-T_1 R_1} + c~e^{-T_2 R_2} + \cdots + c~e^{-T_N R_N} + \mathcal N e^{-T_NR_N} \end{align*}

Example from previous set’s Bond Pricing: Exercise I:

\[2.5 e^{-0.5\times 0.0385} + 2.5 e^{-1\times 0.0365} + 2.5 e^{-1.5\times 0.0358} + 102.5 e^{-2\times 0.0351} = \$102.78\]
Here we have notional $\mathcal N = 100$ and
- $c = 2.5$,
- $T_1 = 0.5$, $R_1 = 3.85\%$,
- $T_2 = 1.0$, $R_2 = 3.65\%$,
- $T_3 = 1.5$, $R_3 = 3.58\%$
- $B = 102.78$

Sensitivity of ZCB Price to Its Yield Change

Imagine, in a small period of time, the yield changes from $R$ to $R+\Delta$ for some small $\Delta$. The ZCB price before and after the yield change are respectively

\[B(0) = e^{-TR} \quad\text{and}\quad B(\Delta) = e^{-T(R+\Delta)}\]

We are treating $B$ as a function of the yield change $\Delta$. Now find its first derivative at $\Delta = 0$: \begin{align*} &B'(0) = -Te^{-TR} \approx \frac{B(\Delta) - B(0)}{\Delta} \\ \iff& -TB(0) \approx \frac{B(\Delta) - B(0)}{\Delta} \\ \iff& -T\Delta \approx \frac{B(\Delta) - B(0)}{B(0)} \end{align*}

Right hand side: percentage change of the ZCB price
Left hand side: Term times yield change in the opposite direction
Hence the negative sign
In the case of ZCB, the term $T$ is the sensitivivity and is called the (Macaulay) duration of this ZCB

ZCB Duration Example

\[-T\Delta \approx \frac{B(\Delta) - B(0)}{B(0)}\]

Duration of a ZCB is its term $T$
On a day when the yield drops $0.06\% = 6\text{ bps}$ across the entire curve, we have
- 1Y ZCB price up $0.06\% = 0.06\% \times 1$
- 5Y ZCB price up $0.3\% = 0.06\% \times 5$
- 10Y ZCB price up $0.6\% = 0.06\% \times 10$
- 30Y ZCB price up $1.8\% = 0.06\% \times 30$
The longer-term the bond is, the more sensitive it is to the yield change
- This is true for nonzero coupon bonds too, as we will see soon

$1\text{ bp}$ (basis point) is $0.01\%$, a common unit of measurement in the rates market
Does the yield change by the same amount across the entire curve?
- Yes, more or less, most of the time, as we will see when we get to Principal Component Analysis (PCA)
- Yield changing by the same amount across the entire curve is so common it has a name “parallel shift of the curve”

Sensitivity of Coupon Bond Price to Yield Change

We repeat the same exercise but for coupon bonds
Imagine, in a small period of time, the yields at all tenors change from $R_j$ to $R_j+\Delta$ for the same small amount $\Delta$. The bond price before and after the yield change are respectively \begin{align*} B(0) &= \mathcal N e^{-T_NR_N} + \sum_{j=1}^N c~e^{-T_j R_j} \\ &= c~e^{-T_1 R_1} + c~e^{-T_2 R_2} + \cdots + c~e^{-T_N R_N} + \mathcal N e^{-T_NR_N} \end{align*} and \begin{align*} B(\Delta) &= \mathcal N e^{-T_N(R_N + \Delta)} + \sum_{j=1}^N c~e^{-T_j (R_j + \Delta)} \\ &= c~e^{-T_1 (R_1 + \Delta)} + c~e^{-T_2 (R_2 + \Delta)} + \cdots + c~e^{-T_N (R_N + \Delta)} + \mathcal N e^{-T_N(R_N + \Delta)} \end{align*}
Next we find the first derivative of $B(\Delta)$ at $\Delta = 0$

Sensitivity of Coupon Bond Price to Yield Change (Cont.)

\begin{align*} B(0) &= c~e^{-T_1 R_1} + c~e^{-T_2 R_2} + \cdots + c~e^{-T_N R_N} + \mathcal N e^{-T_NR_N}, \\ B(\Delta) &= c~e^{-T_1 (R_1 + \Delta)} + c~e^{-T_2 (R_2 + \Delta)} + \cdots + c~e^{-T_N (R_N + \Delta)} + \mathcal N e^{-T_N(R_N + \Delta)}, \end{align*}

We find the first derivative of $B(\Delta)$ at $\Delta = 0$ to be \begin{align*} B'(0) &= {\color{red}-\color{red}T_1}c~e^{-T_1 R_1} {\color{red}-\color{red}T_2} c~e^{-T_2 R_2} - \cdots {\color{red}-\color{red}T_N} c~e^{-T_N R_N} {\color{red}-\color{red}T_N} \mathcal N e^{-T_N R_N} \approx \frac{B(\Delta) - B(0)}{\Delta} \end{align*}

To obtain the percentage change of bond price, multiply by $\Delta/B(0)$ on both sides: \begin{align*} {\color{red}-}\left(\frac{{\color{red}T_1}c~e^{-T_1 R_1} + {\color{red}T_2} c~e^{-T_2 R_2} + \cdots + {\color{red}T_N} c~e^{-T_N R_N} + {\color{red}T_N} \mathcal N e^{-T_N R_N}}{B(0)}\right)\Delta \approx \frac{B(\Delta) - B(0)}{B(0)} \end{align*}
The quantity in the parentheses is the duration, denoted by $D$ hereafter, but it can be further simplified

Duration of Coupon Bond

Yield change $\Delta$ and percentage change of the bond price has the relation \begin{align*} -D\Delta \approx \frac{B(\Delta) - B(0)}{B(0)} \end{align*} where $D$ is the duration
Again a negative sign because left hand side is yield change which right hand side is bond price change

Duration of Coupon Bond (Cont.)

\begin{align*} B(0) = c~e^{-T_1 R_1} + c~e^{-T_1 R_1} + \cdots + c~e^{-T_N R_N} + \mathcal N e^{-T_NR_N} \end{align*}

Yield change $\Delta$ and percentage change of the bond price has the relation \begin{align*} -D\Delta \approx \frac{B(\Delta) - B(0)}{B(0)}, \end{align*} where the duration $D$ is

\[D = \frac{{\color{red}T_1}c~e^{-T_1 R_1} + {\color{red}T_2} c~e^{-T_2 R_2} + \cdots + {\color{red}T_N} c~e^{-T_N R_N} + {\color{red}T_N} \mathcal N e^{-T_N R_N}}{B(0)}\]

\[= T_1 w_1 + T_2 w_2 + \cdots + T_N w_N + T_N w_N^*,\]

where \begin{align*} w_j = \frac{c~e^{-T_jR_j}}{B(0)}, \quad\forall j=1, 2, \ldots, N, \quad\text{ and }\quad w_N^* = \frac{\mathcal Ne^{-T_NR_N}}{B(0)}, \end{align*} which satisfy $w_1 + w_2 + \cdots + w_N + w_N^* = 1$

Duration: Properties & Implications

\begin{align*} -D\Delta \approx \frac{B(\Delta) - B(0)}{B(0)}, \end{align*}

The duration $D = T_1 w_1 + T_2 w_2 + \cdots + T_N w_N + T_N w_N^*,$ where \begin{align*} w_j = \frac{c~e^{-T_jR_j}}{B(0)}, \quad\forall j=1, 2, \ldots, N, \quad\text{ and }\quad w_N^* = \frac{\mathcal Ne^{-T_NR_N}}{B(0)}, \end{align*} which satisfy $w_1 + w_2 + \cdots + w_N + w_N^* = 1$
Duration is a weighted average of all payment dates (the terms)

Recall that fair price of a bond is sum of all cash flows discounted
Weights of duration are percentages of each discounted cash flow in the bond price

In the case of ZCB $(c = 0)$, this formula reduces to $D=T_N$, consistent with previous result

Duration: Properties & Implications (Cont.)

\begin{align*} -D\Delta \approx \frac{B(\Delta) - B(0)}{B(0)}, \end{align*}

The duration $D = T_1 w_1 + T_2 w_2 + \cdots + T_N w_N + T_N w_N^*,$ where \begin{align*} w_j = \frac{c~e^{-T_jR_j}}{B(0)}, \quad\forall j=1, 2, \ldots, N, \quad\text{ and }\quad w_N^* = \frac{\mathcal Ne^{-T_NR_N}}{B(0)}, \end{align*} which satisfy $w_1 + w_2 + \cdots + w_N + w_N^* = 1$
Duration is a weighted average of all payment dates (the terms)

Since $w_N^* \gg w_j,$ the duration is still somewhat close to the bond’s term $T_N$

Previous conclusion “the longer the bond is, the more sensitive it is to yield change” is still valid

Since $T_1 < T_2 < \cdots < T_N$, we know $D$ being a weighted average means $D \le T_N$
- Coupons pulls the duration of a bond lower
- The higher the coupon rate is, the lower the duration ($w_j$’s larger)

Duration of Coupon Bond: Example

Example from previous set’s Bond Pricing: Exercise I: \begin{align*} B &= c~e^{-T_1 R_1} + c~e^{-T_2 R_2} + \cdots + c~e^{-T_N R_N} + \mathcal N e^{-T_NR_N}\\\\ \$102.78 &= 2.5 e^{-0.5\times 0.0385} + 2.5 e^{-1\times 0.0365} + 2.5 e^{-1.5\times 0.0358} + 102.5 e^{-2\times 0.0351} \end{align*}
Duration is \begin{align*} D &= \left(T_1 c~e^{-T_1 R_1} + T_2 c~e^{-T_2 R_2} + \cdots + T_N c~e^{-T_N R_N} + T_N \mathcal N e^{-T_NR_N}\right)/B \end{align*} \begin{align*} &= \left(0.5\times 2.5 e^{-0.5\times 0.0385} + 1\times 2.5 e^{-1\times 0.0365} + 1.5\times 2.5 e^{-1.5\times 0.0358} + 2\times 102.5 e^{-2\times 0.0351}\right)/102.78 = 1.9293 \text{ (years)} \end{align*}
If, today, the treasury yield is down 2 bps across the entire yield curve, the price of this bond will be up approximately $0.02\%\times 1.9293 = 0.039\%$
If, today, the treasury yield is down 2 bps only for all tenors $\le$ 2y, the bond price will still be up $0.039\%$. (Why?)

Duration of Coupon Bond: Example (Cont.)

$D = 1.9293$ is close to but shorter than the term 2y
The weights are \begin{align*} w_1 &= 2.5 e^{-0.5\times 0.0385}/102.78 = 0.02386,\\ w_2 &= 2.5 e^{-1\times 0.0365}/102.78 = 0.02345,\\ w_3 &= 2.5 e^{-1.5\times 0.0358}/102.78 = 0.02305,\\ w_4 &= 2.5 e^{-2\times 0.0351}/102.78 = 0.02267,\\ w_4^* &= 100 e^{-2\times 0.0351}/102.78 = 0.90699 \end{align*}
$w_4^* \gg w_1 > w_2 > w_3 > w_4$

Duration of Coupon Bond: Exercise

Example from previous set’s Bond Pricing: Exercise II: \begin{align*} B &= c~e^{-T_1 R_1} + c~e^{-T_2 R_2} + \cdots + c~e^{-T_N R_N} + \mathcal N e^{-T_NR_N}\\\\ \$100.56 &= 2 e^{-0.5\times 0.0385} + 2 e^{-1\times 0.0365} + 102 e^{-1.5\times 0.0358} \end{align*}
What’s its duration?
If, today, the treasury yield is up 5 bps across the entire yield curve, what’s the percentage change of the bond price?

Duration of Coupon Bond: Answer

What’s its duration?

\begin{align*} w_1 &= 2 e^{-0.5 \times 0.0385} / 100.56,\\ w_2 &= 2 e^{-1 \times 0.0365} / 100.56,\\ w_3 &= 102 e^{-1.5 \times 0.0358} / 100.56,\\ D &= 0.5 w_1 + 1 w_2 + 1.5 w_3 \approx 1.47 \end{align*}

If, today, the treasury yield is up 5 bps across the entire yield curve, what’s the percentage change of the bond price?

\begin{align*} \frac{B(\Delta) - B(0)}{B(0)} &\approx - D \Delta \\ &= -1.47 \times 0.05 \% \\ &= -0.0735\% \end{align*}

Dollar Duration

Recall that, with the Macaulay duration $D$, we have

\[-D \Delta \approx \frac{B(\Delta) - B(0)}{B(0)}\]
The $\Delta$ on the left hand side is the absolute change of the yield, while the right hand side is the percentage change of the bond price
To make the right hand side absolute change of the price too, multiply through by $B(0)$ and define the dollar duration

\[D^{\$} = DB(0)\]
We then obtain

\[-D^{\$} \Delta \approx B(\Delta) - B(0)\]
It will be clear just a bit later why sometimes absolute changes are preferred

DV01

A related sensitivity measure is DV01, the dollar value of 1 bp, also known as PV01, the present value of 1 bp
DV01 is the price change from a +1 bp paralle yield curve shift
For a bond, plugging $\Delta = 1 \text{bp} = 10^{-4}$ into $-D^{\$} \Delta \approx B(\Delta) - B(0)$, we have

\[\text{DV01} = -D^{\$}\times 10^{-4} \approx B(\Delta) - B(0)\]

Duration of a Bond Portfolio

We repeat the same exercise for a portfolio of bonds
If I hold $M$ bonds whose prices are \begin{align*} B_1 &= c_1e^{-T_1 R_1} + c_1e^{-T_2 R_2} + \cdots + c_1e^{-T_N R_N} + \mathcal N_1 e^{-T_{N_1}R_{N_1}}, \\ B_2 &= c_2e^{-T_1 R_1} + c_2e^{-T_2 R_2} + \cdots + c_2e^{-T_N R_N} + \mathcal N_2 e^{-T_{N_2}R_{N_2}}, \\ % B_3 &= c_3e^{-T_1 R_1} + c_3e^{-T_2 R_2} + \cdots + c_3e^{-T_N R_N} + \mathcal N_3 e^{-T_{N_3}R_{N_3}}, \\ &\qquad\qquad\qquad\qquad\qquad\qquad\vdots\\ B_M &= c_Me^{-T_1 R_1} + c_Me^{-T_2 R_2} + \cdots + c_Me^{-T_N R_N} + \mathcal N_M e^{-T_{N_M}R_{N_M}}, \\ \end{align*} which sum up to be the portfolio value $P = \sum_{j=0}^M B_j$
If, in a small period of time, the yields at all tenors change from $R_j$ to $R_j+\Delta$ for the same small amount $\Delta$. The bond prices become \begin{align*} B_1(\Delta) &= c_1e^{-T_1 (R_1 + \Delta)} + c_1e^{-T_2 (R_2 + \Delta)} + \cdots + c_1e^{-T_{N_1} (R_{N_1} + \Delta)} + \mathcal N_1 e^{-T_{N_1}(R_{N_1} + \Delta)},\\ B_2(\Delta) &= c_2e^{-T_1 (R_1 + \Delta)} + c_2e^{-T_2 (R_2 + \Delta)} + \cdots + c_2e^{-T_{N_2} (R_{N_2} + \Delta)} + \mathcal N_2 e^{-T_{N_2}(R_{N_2} + \Delta)}, \\ &\qquad\qquad\qquad\qquad\qquad\qquad\vdots\\ B_M(\Delta) &= c_Me^{-T_1 (R_1 + \Delta)} + c_Me^{-T_2 (R_2 + \Delta)} + \cdots + c_Me^{-T_{N_M} (R_{N_M} + \Delta)} + \mathcal N_M e^{-T_{N_M}(R_{N_M} + \Delta)}, \end{align*} which sum up to be the portfolio value $P(\Delta) = \sum_{j=0}^M B_j(\Delta)$

Duration of a Bond Portfolio (Cont.)

Find the first derivative of $P(\Delta)$ at $\Delta=0$: \begin{align*} P'(0) =& \sum_{j=0}^M B_j'(0)\approx\frac{P(\Delta) - P(0)}{\Delta}\\ =& -T_1c_1e^{-T_1 R_1} -T_2 c_1e^{-T_2 R_2} - \cdots -T_{N_1} c_1e^{-T_{N_1} R_{N_1}} -T_{N_1} \mathcal N_1 e^{-T_{N_1}R_{N_1}} \\ &-T_1 c_2e^{-T_1 R_1} -T_2 c_2e^{-T_2 R_2} - \cdots -T_{N_2} c_2e^{-T_{N_2} R_{N_2}} -T_{N_2} \mathcal N_2 e^{-T_{N_2}R_{N_2}} \\ &\qquad\qquad\qquad\qquad\qquad\qquad\vdots\\ &- T_1 c_Me^{-T_1 R_1} -T_2 c_Me^{-T_2 R_2} - \cdots -T_N c_Me^{-T_N R_N} -T_{N_M} \mathcal N_M e^{-T_{N_M}R_{N_M}} \end{align*}
Multiply by $\Delta/P(0)$ on both side to get \begin{align*} -\Delta&\left[\left(T_1c_1e^{-T_1 R_1} + T_2 c_1e^{-T_2 R_2} + \cdots + T_{N_1} c_1e^{-T_{N_1} R_{N_1}} + T_{N_1} \mathcal N_1 e^{-T_{N_1}R_{N_1}} \right.\right.\\ &+T_1 c_2e^{-T_1 R_1} + T_2 c_2e^{-T_2 R_2} + \cdots + T_{N_2} c_2e^{-T_{N_2} R_{N_2}} + T_{N_2} \mathcal N_2 e^{-T_{N_2}R_{N_2}} \\ &\qquad\qquad\qquad\qquad\qquad\qquad\vdots\\ &\left.\left. +T_1 c_Me^{-T_1 R_1} + T_2 c_Me^{-T_2 R_2} + \cdots + T_N c_Me^{-T_N R_N} +T_{N_M} \mathcal N_M e^{-T_{N_M}R_{N_M}}\right)/P(0)\right] \approx \frac{P(\Delta) - P(0)}{P(0)} \end{align*}

Duration of a Bond Portfolio (Cont..)

\begin{align*} -\Delta&\left[\left(T_1c_1e^{-T_1 R_1} + T_2 c_1e^{-T_2 R_2} + \cdots + T_{N_1} c_1e^{-T_{N_1} R_{N_1}} + T_{N_1} \mathcal N_1 e^{-T_{N_1}R_{N_1}} \right.\right.\\ &+T_1 c_2e^{-T_1 R_1} + T_2 c_2e^{-T_2 R_2} + \cdots + T_{N_2} c_2e^{-T_{N_2} R_{N_2}} + T_{N_2} \mathcal N_2 e^{-T_{N_2}R_{N_2}} \\ &\qquad\qquad\qquad\qquad\qquad\qquad\vdots\\ &\left.\left. +T_1 c_Me^{-T_1 R_1} + T_2 c_Me^{-T_2 R_2} + \cdots + T_N c_Me^{-T_N R_N} +T_{N_M} \mathcal N_M e^{-T_{N_M}R_{N_M}}\right)/P(0)\right] \\&\approx \frac{P(\Delta) - P(0)}{P(0)} \end{align*}

The quantity in the square brackets is the duration $D$ of the portfolio, and again we write \begin{align*} -\Delta D\approx \frac{P(\Delta) - P(0)}{P(0)} \end{align*}
Since, all $c_i e^{-T_jR_j}$ and $\mathcal N_i e^{-T_{N_i}R_{N_i}}$ terms sum up to be the portfolio value $P(0)$, the duration $D$ is again a weighted average of all payment dates (terms) of future cash flows, with weights the corresponding discounted cash flows

Portfolio Duration: Properties & Implications

In general, the duration of a portfolio of future cash flows is the weighted average of all payment dates, with weights the corresponding discounted cash flows
Like the coupon bond case, since $\mathcal N_i e^{-T_{N_i}R_{N_i}} \gg c_i e^{-T_jR_j}$, portfolio duration is somewhat close to a weighted average of the terms of the bonds in the portfolio
But coupon pulls the duration lower, and again the higher the coupon rate is, the lower the duration

Portfolio Dollar Duration

The portfolio version of dollar duration is defined as

\[D^{\$} = DP(0)\]

and we have

\[-D^{\$} \Delta \approx P(\Delta) - P(0)\]
This is saying, for small parallel shift of the curve, the value change of a portfolio is approximately equal to its dollar duration times the yield change, but in the opposite direction

Hedging by Duration

You can hedge a portfolio of bonds by making its dollar duration 0
If you want to hedge a portfolio of bonds with dollar duration $D^{\$}$, add (buy or short sell) one more bond to it with dollar duration $-D^{\$}$
Adding this one trade to the book, the entire portfolio will now have zero dollar duration and hence

\[P(\Delta) - P(0) \approx -D^{\$}\Delta = 0\]
Why are we matching the dollar duration, not the duration?

Cost of Hedging

\[\]

\[D^{\$} = DB(0)\]

\[\]

We have a set amount for $-D^{\$}$, but should we go high on $D$ or $B(0)$?

Hedging by Duration (Cont.)

Adding one trade to the book, your portfolio now has zero dollar duration and is immune to small parallel shifts in the yield curve:

\[P(\Delta) - P(0) \approx -D^{\$}\Delta = 0\]
It is still exposed to shifts that are either large or nonparallel. How to hedge those?
- Large parallel shift: By convexity
- Nonparallel shift: By PCA

Treasury ETFs

etfdb.com’s screener is a convenient tool to look up exchange-traded funds (ETFs)
Treasury ETFs can be found in: etfdb.com > ETF Screener > Asset Class > Bonds > Treasuries

\[\]

Ticker	Fund	Duration (yrs)
SGOV	iShares 0-3 Month Treasury Bond ETF	0.10
SHY	iShares 1-3 Year Treasury Bond ETF	1.85
IEI	iShares 3-7 Year Treasury Bond ETF	4.28
IEF	iShares 7-10 Year Treasury Bond ETF	7.05
TLH	iShares 10-20 Year Treasury Bond ETF	12.16
TLT	iShares 20+ Year Treasury Bond ETF	15.72

\[\]

Each ETF will have their own product page with all details, expense ratio, duration (for bond ETFs), holdings and more
Long term Treasury bonds market data can be downloaded from: TLT product page > Holdings > Detailed Holdings and Analytics

Duration of TLT: Example

From 9/17 to 9/18 close, treasury yield was up 6 bps
- Search “treasury yield today” to find this link
Form 9/17 to 9/18 close, TLT price dropped from 90.12 to 89.19, or down $1.03\%$
- Search TLT on finance.yahoo.com > click Historical Data on the left
  
  \[0.06\%\times 15.72 = 0.9432\%\]

Duration and Volatility

\begin{align*} -D\Delta \approx \frac{B(\Delta) - B(0)}{B(0)} \end{align*}

Short duration means bond price is stable; It does not change much no matter how much the yield curve moves
- An extreme example, the price of a bond maturing tomorrow (duration 1/365) should be very close to the notional principal, no matter how the market moves
Longer duration means bond price is more sensitive to yield change, so

\[\]

\[\text{Long Duration} = \text{Volatile Bond Price}\]

Duration and Volatility (Cont.)

\[\]

\[\text{Long Duration} = \text{Volatile Bond Price}\]

\[\]

Previous example, on a day when the yield drops $0.06\% = 6\text{ bps}$ across the entire curve:
- 1Y duration bond price up $0.06\% = 0.06\% \times 1$
- 5Y duration bond price up $0.3\% = 0.06\% \times 5$
- 10Y duration bond price up $0.6\% = 0.06\% \times 10$
- 30Y duration bond price up $1.8\% = 0.06\% \times 30$

Backtest Portfolio

portfoliovisualizer.com > See all the Analytics Tools > Backtest Portfolio > Edit Portfolio

\[\]

Modified Duration

So far we’ve derived the formulas under the assumption that the yield $R_j$’s are expressed with continuous compounding
A different definition of the yield will lead to different formula
Consider the below definition of yield $R$: \begin{align*} B = \mathcal N \left(1+\frac{R}{m}\right)^{-mT_N} + \sum_{j=1}^N c\left(1+\frac{R}{m}\right)^{-mT_j} \end{align*}
Compared the above to the bond pricing formula \begin{align*} B = \mathcal N e^{-T_NR_N} + \sum_{j=1}^N c~e^{-T_j R_j}, \end{align*} in the new definition, the same yield $R$ is used for all terms, and it’s expressed with a compounding frequency of $m$ times per year

Modified Duration (Cont.)

With this new definition of $R$, we have

\[-D^* \Delta = \frac{B(\Delta) - B(0)}{B(0)},\]

where

\[D^* = \frac{D}{1+R/m},\]

with $D$ the discounted cash flow weighted terms as before

Modified Duration (Cont..)

Given \begin{align*} B(\Delta) &= \mathcal N \left(1+\frac{R+\Delta}{m}\right)^{-mT_N} + \sum_{j=1}^N c\left(1+\frac{R+\Delta}{m}\right)^{-mT_j}, \end{align*} find the derivative $B'(0)$: \begin{align*} B'(0) &= \mathcal N \left(1+\frac{R}{m}\right)^{-mT_N}\left(-\frac{T_N}{1+\frac{R}{m}}\right) + \sum_{j=1}^N c\left(1+\frac{R}{m}\right)^{-mT_j}\left(-\frac{T_j}{1+\frac{R}{m}}\right)\\ &= -\left[{\color{red}T_N}\mathcal N \left(1+\frac{R}{m}\right)^{-mT_N} + \sum_{j=1}^N {\color{red}T_j} c\left(1+\frac{R}{m}\right)^{-mT_j}\right]/\left(1+\frac{R}{m}\right)\\ &\approx\frac{B(\Delta) - B(0)}{\Delta} \end{align*}

Modified Duration (Cont…)

We have \begin{align*} -\left[{\color{red}T_N}\mathcal N \left(1+\frac{R}{m}\right)^{-mT_N} + \sum_{j=1}^N {\color{red}T_j} c\left(1+\frac{R}{m}\right)^{-mT_j}\right]/\left(1+\frac{R}{m}\right) \approx\frac{B(\Delta) - B(0)}{\Delta} \end{align*}
Multiply by $\Delta/B(0)$ on both sides to obtain \begin{align*} -\Delta\left[\frac{{\color{red}T_N}\mathcal N \left(1+\frac{R}{m}\right)^{-mT_N} + \sum_{j=1}^N {\color{red}T_j} c\left(1+\frac{R}{m}\right)^{-mT_j}}{\mathcal N \left(1+\frac{R}{m}\right)^{-mT_N} + \sum_{j=1}^N c\left(1+\frac{R}{m}\right)^{-mT_j}}\right]/\left(1+\frac{R}{m}\right) \approx\frac{B(\Delta) - B(0)}{B(0)}, \end{align*} where the quantity in the square brackets is $D$ and we’ve used the fact \begin{align*} B(0) = \mathcal N \left(1+\frac{R}{m}\right)^{-mT_N} + \sum_{j=1}^N c\left(1+\frac{R}{m}\right)^{-mT_j} \end{align*}

Convexity of ZCB

The approximation

\[-D \Delta \approx \frac{B(\Delta) - B(0)}{B(0)}\]

is only accurate when $\Delta$ is small. For large yield curve movement, we need 2nd order approximation
Consider the ZCB case \begin{align*} B(\Delta) &= e^{-(R+\Delta)T} \approx B(0) + \Delta B'(0) + \frac{\Delta^2}{2} B''(0)\\ &= B(0) - T\Delta B(0) + T^2\frac{\Delta^2}{2} B(0) \end{align*}
Divide through by $B(0)$ to obtain \begin{align*} \frac{B(\Delta) - B(0)}{B(0)} \approx - T\Delta + T^2\frac{\Delta^2}{2}\\ \end{align*}
The 2nd term on the right hand side is an adjustment to the original approximation. $T^2$ is called the convexity, denoted by $C$ hereafter

Convexity of Coupon Bond

Taylor expansion of $B(\Delta)$ at $\Delta = 0$ gives

\[B(\Delta) \approx B(0) + \Delta B'(0) + \frac{\Delta^2}{2} B''(0),\]

so

\[\frac{B(\Delta) - B(0)}{B(0)} \approx \Delta \frac{B'(0)}{B(0)} + \frac{\Delta^2}{2} \frac{B''(0)}{B(0)}\]
We will define

\[\frac{B'(0)}{B(0)} = -D, \qquad\frac{B''(0)}{B(0)} = C\]

and write

\[\frac{B(\Delta) - B(0)}{B(0)} \approx -D \Delta + \frac{1}{2}C\Delta^2\]

Convexity of Coupon Bond (Cont.)

\[\]

\[\frac{B(\Delta) - B(0)}{B(0)} \approx \Delta \frac{B'(0)}{B(0)} + \frac{\Delta^2}{2} \frac{B''(0)}{B(0)}\]

We have \begin{align*} B(\Delta) &= \mathcal N e^{-T_N(R_N+\Delta)} + \sum_{j=1}^N c~e^{-T_j (R_j+\Delta)}, \\ \frac{B'(0)}{B(0)} &= -\frac{{\color{red}T_N}\mathcal N e^{-T_NR_N} + \sum_{j=1}^N {\color{red}T_j}c~e^{-T_j R_j}}{\mathcal N e^{-T_NR_N} + \sum_{j=1}^N c~e^{-T_j R_j}} = -D, \\ \frac{B''(0)}{B(0)} &= \frac{{\color{red}T_N^2}\mathcal N e^{-T_NR_N} + \sum_{j=1}^N {\color{red}T_j^2}c~e^{-T_j R_j}}{\mathcal N e^{-T_NR_N} + \sum_{j=1}^N c~e^{-T_j R_j}} = C \end{align*}

Convexity: Properties & Implications

\[\]

\begin{align*} D &= \frac{{\color{red}T_N}\mathcal N e^{-T_NR_N} + \sum_{j=1}^N {\color{red}T_j}c~e^{-T_j R_j}}{\mathcal N e^{-T_NR_N} + \sum_{j=1}^N c~e^{-T_j R_j}}, \\ C &= \frac{{\color{red}T_N^2}\mathcal N e^{-T_NR_N} + \sum_{j=1}^N {\color{red}T_j^2}c~e^{-T_j R_j}}{\mathcal N e^{-T_NR_N} + \sum_{j=1}^N c~e^{-T_j R_j}} \end{align*}

Duration is a weighted average of all payment date terms
Convexity is the weighted average of all payment date terms squared
Same weights as duration

Since $w_N^* \gg w_j,$ the duration is close to the bond’s term squared: $T_N^2$
For ZCB $(c=0)$, the convexity is equal to the term squared $T^2$
Like duration, high coupon rate pulls convexity lower

Dollar Convexity

Recall that

\[\frac{B(\Delta) - B(0)}{B(0)} \approx -D \Delta + \frac{1}{2}C\Delta^2\]
One side is absolute change but the other is the percentage change
Define the dollar duration and dollar convexity as

\[D^{\$} = DB(0), \qquad C^{\$} = CB(0),\]

respectively. We then obtain

\[B(\Delta) - B(0) \approx -D^{\$} \Delta + \frac{1}{2}C^{\$}\Delta^2\]

Portfolio Convexity and Portfolio Dollar Convexity

The duration and convexity of a portfolio of $M$ bonds are \begin{align*} D &= \frac{\sum_{i=1}^M{\color{red}T_{N_M}}\mathcal N_M e^{-T_{N_M}R_{N_M}} + \sum_{i=1}^M\sum_{j=1}^N {\color{red}T_j}c_i~e^{-T_j R_j}}{\sum_{i=1}^M\mathcal N_M e^{-T_{N_M}R_{N_M}} + \sum_{i=1}^M\sum_{j=1}^N c_i~e^{-T_j R_j}}, \\ C &= \frac{\sum_{i=1}^M{\color{red}T_{N_M}^2}\mathcal N_M e^{-T_{N_M}R_{N_M}} + \sum_{i=1}^M\sum_{j=1}^N {\color{red}T_j^2}c_i~e^{-T_j R_j}}{\sum_{i=1}^M\mathcal N_M e^{-T_{N_M}R_{N_M}} + \sum_{i=1}^M\sum_{j=1}^N c_i~e^{-T_j R_j}}, \end{align*} respectively
And its dollar duration and dollar convexity are defined as

\[D^{\$} = DP(0), \qquad C^{\$} = CP(0),\]

respectively
We have

\[P(\Delta) - P(0) \approx -D^{\$} \Delta + \frac{1}{2}C^{\$}\Delta^2\]

Hedging by Duration and Convexity

We have shown that, if you make a portfolio’s dollar duration zero, it’s immune to small parallel shifts in the yield curve
To make it immune to large parallel shift, you need to make dollar convexity zero too, so that

\[P(\Delta) - P(0) \approx -D^{\$} \Delta + \frac{1}{2}C^{\$}\Delta^2 = 0\]
Suppose you are hedging a portfolio with dollar duration $D^{\$}_P$ and dollar convexity $C^{\$}_P$. If you could add a bond with dollar duration $-D^{\$}_P$ and dollar convexity $-C^{\$}_P$, you are done

But in general this cannot be achieved by one single bond
With two equations ($-D^{\$} = 0$ and $-C^{\$} = 0$) we need two variables

Hedging by Duration and Convexity (Cont.)

Suppose we have two bonds $B_1$ and $B_2$:
- $B_1$ has dollar duration $D^{\$}_1$ and dollar convexity $C^{\$}_1$
- $B_2$ has dollar duration $D^{\$}_2$ and dollar convexity $C^{\$}_2$
Buy $h_1$ units of $B_1$ and $h_2$ units of $B_2$ so that \begin{align*} &h_1 D^{\$}_1 + h_2 D^{\$}_2 = -D^{\$}_P\\ &h_1 C^{\$}_1 + h_2 C^{\$}_2 = -C^{\$}_P \end{align*}
Solve the system to obtain hedge ratio

Linear Systems for Hedging

It’s not uncommon we need so solve a linear system to figure out the hedge ratio
Steps to hedge
1. Characterize factors (reasons) of the portfolio value change, in our example $\Delta$ and $\Delta^2$
2. Write portfolio value change in terms of the factor changes
3. Find hedge (trades, in our example, bonds) and hedge ratio to make coefficients ($D$ and $C$ in our example) zero
We will see linear system for hedging again in PCA

Hedging by Duration and Convexity (Cont..)

After all the hard work we have

\[P(\Delta) - P(0) \approx -D^{\$} \Delta + \frac{1}{2}C^{\$}\Delta^2 = 0\]
The portfolio is now immune to large parallel shifts in the yield curve
But still exposed to nonparallel shifts, which can be hedged by PCA