Exam FM — Duration, Convexity & Immunization Flashcards

$D_{mac}=\dfrac{\sum_{t} t\,v^{t}\,CF_{t}}{\sum_{t} v^{t}\,CF_{t}}$, the present-value-weighted **average time** of the cash flows. Each flow's weight is the fraction of total price contributed by its discounted value $v^{t}CF_{t}$, so $D_{mac}$ is measured in the same time units (usually years) as $t$.

Exactly $n$. With only one cash flow, the PV-weighted average time is just its own payment date: $D_{mac}=\dfrac{n\,v^{n}C}{v^{n}C}=n$. This is why a zero-coupon bond maturing in $n$ years has Macaulay duration $n$ — the longest duration available for a given maturity.

$D_{mod}=-\dfrac{1}{P}\dfrac{dP}{di}$, the negative percentage sensitivity of price to a change in the periodic yield. It connects to Macaulay duration by $D_{mod}=\dfrac{D_{mac}}{1+i}$. (Under a force of interest / continuous compounding the two coincide: $D_{mod}=D_{mac}$.)

$D_{mod}=\dfrac{D_{mac}}{1+i}=\dfrac{7.5}{1.04}\approx 7.2115$ years. Modified duration is always slightly **smaller** than Macaulay duration (you divide by $1+i>1$).

Dollar duration $=-\dfrac{dP}{di}=P\cdot D_{mod}$ — the **absolute** (not percentage) change in price per unit change in yield. Unlike $D_{mac}$ and $D_{mod}$, which are unitless time/elasticity measures, dollar duration carries currency units and is additive across a portfolio in dollars.

Dollar duration $=P\cdot D_{mod}=5000\times 4=20000$. For $\Delta i=+0.0025$: $\Delta P\approx -(\text{dollar duration})\cdot\Delta i=-20000\times 0.0025=-50$. The bond loses about $\$50$, falling to roughly $\$4950$.

Price: $P=\dfrac{80}{1.06}+\dfrac{80}{1.06^{2}}+\dfrac{1080}{1.06^{3}}\approx 1053.46$. Weighted-time numerator: $\dfrac{1\cdot80}{1.06}+\dfrac{2\cdot80}{1.06^{2}}+\dfrac{3\cdot1080}{1.06^{3}}\approx 2938.24$. $D_{mac}=\dfrac{2938.24}{1053.46}\approx 2.789$ years. (Note $D_{mac}<3=$ maturity, as always for a coupon bond.)

$D_{mod}=\dfrac{D_{mac}}{1+i}=\dfrac{2.789}{1.06}\approx 2.631$ years. Interpretation: a $1\%$ ($100$ bp) rise in yield drops the price by roughly $2.631\%$ to first order.

Maturity reflects only the final redemption date, but duration is a PV-weighted average of **all** payment dates. The intermediate coupons land earlier than maturity and pull the average time down. Only a zero-coupon bond (no intermediate flows) has duration equal to its maturity.

For a perpetuity paying $1$ at the end of each period forever, $D_{mac}=\dfrac{1+i}{i}$. At $i=0.08$: $D_{mac}=\dfrac{1.08}{0.08}=13.5$ years. (This special case is frequently misremembered — memorize $\frac{1+i}{i}$.)

PV $=100\sum_{t=1}^{5}1.07^{-t}\approx 410.02$. Weighted-time numerator $=100\sum_{t=1}^{5} t\cdot1.07^{-t}\approx 1174.69$. $D_{mac}=\dfrac{1174.69}{410.02}\approx 2.865$ years. (The duration sits a bit **before** the middle date $t=3$ because the earlier level payments carry more present-value weight.)

Portfolio duration is the **PV-weighted average** of the component durations: $D_{P}=\dfrac{\sum_{k} P_{k}\,D_{k}}{\sum_{k} P_{k}}$, where $P_{k}$ is the present value (market value) of asset $k$ and $D_{k}$ its duration. Weight by market value, not by face amount or by count of bonds.

$D_{P}=\dfrac{4000\times3+6000\times8}{4000+6000}=\dfrac{12000+48000}{10000}=\dfrac{60000}{10000}=6$ years. The answer leans toward Asset B because B holds the larger share of the portfolio's value.

Set the weighted average to $5$: $2(1-w)+10w=5$. $2+8w=5\Rightarrow 8w=3\Rightarrow w=0.375$. So $37.5\%$ of portfolio value in the long bond and $62.5\%$ in the short bond gives $D_{mod}=5$.

$\Delta P\approx -P\cdot D_{mod}\cdot\Delta i$, equivalently $P(i+\Delta i)\approx P(i)\,[\,1-D_{mod}\,\Delta i\,]$. A rate **increase** ($\Delta i>0$) gives a price **decrease** ($\Delta P<0$). This is the tangent-line (linear) estimate of the true, convex price curve.

$\Delta i=+0.005$. $\Delta P\approx -P\,D_{mod}\,\Delta i=-1053.46\times2.631\times0.005\approx -13.86$. Estimated new price $\approx 1053.46-13.86=1039.60$. (The exact reprice is $1039.73$; the linear estimate slightly **understates** the price because it ignores convexity.)

Macaulay form: $P(i+\Delta i)\approx P(i)\left[\dfrac{1+i}{1+i+\Delta i}\right]^{D_{mac}}$. Modified form: $P(i+\Delta i)\approx P(i)\,[1-D_{mod}\,\Delta i]$. Both are first-order; the Macaulay form is multiplicative and uses $D_{mac}$ directly, the modified form is additive and uses $D_{mod}=D_{mac}/(1+i)$. **Do not** divide by $(1+i)$ twice by mixing them.

$P(0.065)\approx 1053.46\left[\dfrac{1.06}{1.065}\right]^{2.789}$. $\dfrac{1.06}{1.065}\approx 0.995305$; raised to $2.789$: $\approx 0.98696$. $P\approx 1053.46\times0.98696\approx 1039.72$. (Essentially the exact reprice of $1039.73$ — the Macaulay form tracks the curve more closely than the linear modified form here.)

$\Delta i=-0.003$. $\Delta P\approx -P\,D_{mod}\,\Delta i=-2{,}000{,}000\times9\times(-0.003)=+54{,}000$. A yield **drop** raises the present value of the liabilities by about $\$54{,}000$ — longer-duration liabilities are more exposed to falling rates.

$C=\dfrac{1}{P}\dfrac{d^{2}P}{di^{2}}=\dfrac{\sum_{t} t(t+1)\,v^{t+2}\,CF_{t}}{P}$. It is the **second-order** term capturing the curvature of the price-yield relationship. Because the price curve is convex, $C>0$ for ordinary bonds, so the duration-only estimate always understates the true price for both up and down moves.

$\dfrac{\Delta P}{P}\approx -D_{mod}\,\Delta i+\tfrac{1}{2}\,C\,(\Delta i)^{2}$, so $P(i+\Delta i)\approx P\left[1-D_{mod}\,\Delta i+\tfrac{1}{2}C(\Delta i)^{2}\right]$. The convexity term is always **positive** (it adds back the curvature the linear term misses) and matters most for large $\Delta i$.

$\Delta i=0.005$. Duration term: $-D_{mod}\,\Delta i=-2.631\times0.005=-0.013155$. Convexity term: $\tfrac{1}{2}C(\Delta i)^{2}=0.5\times9.681\times0.005^{2}=0.000121$. $\dfrac{\Delta P}{P}\approx -0.013034\Rightarrow P\approx 1053.46(1-0.013034)\approx 1039.73$ — matching the exact reprice to the cent.

For a single payment at time $n$, the modified convexity is $\dfrac{n(n+1)}{(1+i)^{2}}$. At $n=4$, $i=0.05$: $\dfrac{4\times5}{1.05^{2}}=\dfrac{20}{1.1025}\approx 18.14$. (Longer-dated zeros have rapidly growing convexity because the $n(n+1)$ factor dominates.)

The price-yield curve is **convex** (bows above its tangent line). Duration is the slope of the tangent at the current yield, so the tangent line lies **below** the actual curve on both sides. For a yield rise the true price falls less than predicted; for a yield drop it rises more — in both cases the actual price exceeds the linear estimate. The positive convexity correction repairs this.

With equal duration, the two react identically to small parallel moves, but the higher-convexity bond outperforms for **large** moves in either direction: it falls less when rates rise and gains more when rates fall. This asymmetric benefit means convexity is a desirable property — investors will generally pay (accept a slightly lower yield) for it.

At the current yield: (1) **PV match** — $PV_{A}=PV_{L}$. (2) **Duration match** — $D_{mod,A}=D_{mod,L}$ (equivalently $\dfrac{dP_{A}}{di}=\dfrac{dP_{L}}{di}$). (3) **Convexity** — $C_{A}>C_{L}$ (asset cash flows more spread out than liability cash flows). Together these make surplus $PV_{A}-PV_{L}$ a local minimum at the current rate, so any small shift leaves surplus $\geq 0$.

PV-and-duration matching makes the surplus function's first derivative zero at the current rate (a stationary point), but that point could be a maximum or a saddle. The convexity condition $C_{A}>C_{L}$ forces the second derivative of surplus to be **positive**, guaranteeing a local **minimum** of zero surplus — so small moves in either direction can only increase surplus. Omitting condition (3) is the classic Redington mistake.

$PV_{L}=\dfrac{1000}{1.10^{2}}=\dfrac{1000}{1.21}\approx 826.45$. Being a single cash flow at $t=2$, its Macaulay duration is exactly $2$ years. Any immunizing asset portfolio must match $PV\approx826.45$ and duration $2$.

Let the PVs of the two assets be $x_{1}$ (at $t=1$) and $x_{3}$ (at $t=3$). PV match: $x_{1}+x_{3}=826.45$. Duration match: $\dfrac{1\cdot x_{1}+3\cdot x_{3}}{826.45}=2\Rightarrow x_{1}+3x_{3}=1652.89$. Subtract: $2x_{3}=826.45\Rightarrow x_{3}=413.22=x_{1}$. Faces: $A_{1}=413.22\times1.10\approx 454.55$; $A_{3}=413.22\times1.10^{3}\approx 550.00$.

Redington protects only against **small** rate shifts (it is a local result). Full immunization protects against a single parallel shift of **any size**. It is typically achieved by funding each liability with two asset cash flows that **straddle** the liability's date (one before, one after) while matching PV and duration. The earlier/later asset flows guarantee surplus $\geq 0$ for every shift, not just infinitesimal ones.

Place asset cash flows at times $t_{1}<t_{L}<t_{2}$ (they straddle the liability) and require: (1) $PV_{A}=PV_{L}$ at the current rate, and (2) duration match, $D_{A}=D_{L}=t_{L}$ — equivalently the PV-weighted average asset time equals $t_{L}$. With a single liability, these two conditions plus the straddle guarantee surplus $\geq 0$ for any size parallel shift.

The two asset cash flows **straddle** the liability date ($1<2<3$), PV and duration are matched, and there is a single liability. Those are exactly the full-immunization conditions, so surplus $\geq 0$ holds for a parallel shift of any magnitude. You can confirm the asset convexity exceeds the liability's: spreading flows to $t=1$ and $t=3$ around a point mass at $t=2$ raises convexity, satisfying Redington's third condition automatically.

You select assets whose cash flows **exactly equal** the liability cash flows at every date, so each liability is paid by an asset flow arriving the same day. There is **no** reinvestment risk and **no** interest-rate risk of any kind — protection is absolute, not just for parallel shifts. The trade-off is cost and rigidity: a perfectly matching asset set may be expensive or unavailable, whereas immunization needs only PV/duration matching.

Start at $t=2$ (only the bond pays then): $1.05\,F_{2}=1000\Rightarrow F_{2}=\dfrac{1000}{1.05}\approx 952.38$. At $t=1$: bond coupon $0.05F_{2}\approx47.62$ plus the zero $F_{1}$ must total $1000$, so $F_{1}=1000-47.62\approx 952.38$. Holding $\approx952.38$ face of each instrument exactly funds both liabilities with no rate risk.

The final liability date is typically funded by a single instrument (e.g. the longest bond's redemption), so its face is pinned first. Working backward, each earlier date's requirement is reduced by the coupons that earlier-maturing instruments and the already-chosen longer bonds pay on that date. Solving forward leaves multiple unknowns entangled at the late dates; solving backward isolates one unknown at a time.

Yes — all three Redington conditions hold (PV match, duration match, asset convexity exceeds liability convexity), so the surplus is protected against **small** parallel rate movements. If, additionally, the asset cash flows straddle the single liability date, the position is **fully** immunized and protected against a parallel shift of any size.

Redington's third condition fails, so the surplus is at a local **maximum** rather than a minimum. A small move of the yield in **either** direction can push the surplus **negative** — the firm is exposed despite matching PV and duration. The fix is to lengthen/spread the asset cash flows (raise $C_{A}$) so asset convexity exceeds liability convexity.

Duration and convexity summarize sensitivity to a single rate $i$ moving uniformly. The matching conditions are derived assuming the **whole** curve shifts by the same $\Delta i$. If the curve **twists** (short and long rates move by different amounts — a non-parallel shift), a matched portfolio can still lose surplus. This residual is **reinvestment / yield-curve-shape risk**, which only exact cash-flow matching fully eliminates.

**Cash-flow matching** is strongest (no rate risk at all, any curve move) but most costly/restrictive. **Full immunization** protects against any-size **parallel** shift using straddling assets with PV + duration match. **Redington** is weakest — only **small** parallel shifts — but needs just PV match, duration match, and $C_{A}>C_{L}$. Protection rises and flexibility falls as you move from Redington to full to exact matching.

$D_{eff}=\dfrac{P_{-}-P_{+}}{2\,P_{0}\,\Delta i}$, using a $\pm50$ bp shock ($\Delta i=0.005$). $D_{eff}=\dfrac{1067.45-1039.73}{2\times1053.46\times0.005}=\dfrac{27.72}{10.5346}\approx 2.631$ years. This numerical estimate matches the analytic modified duration of $2.631$ — handy when a closed-form $D_{mod}$ is awkward.