Exam FM — Common Traps Flashcards

Using the annuity-immediate factor and forgetting the timing shift. Due payments come at the **start** of each period, one period earlier, so they're worth $(1+i)$ more: $\ddot{a}_{\overline{n|}}=(1+i)\,a_{\overline{n|}}$. Likewise $\ddot{s}_{\overline{n|}}=(1+i)\,s_{\overline{n|}}$. Trap: applying $a_{\overline{10|}}$ when payments are at times $0,1,\dots,9$.

Same number of payments, different valuation date. $a_{\overline{n|}}$ values at one period before the **first** payment; $\ddot{a}_{\overline{n|}}$ values **at** the first payment. So $\ddot{a}_{\overline{n|}}=a_{\overline{n|}}(1+i)=1+a_{\overline{n-1|}}$ and $\ddot{s}_{\overline{n|}}=s_{\overline{n+1|}}-1$. Memorize one identity each direction so you can convert under pressure.

Don't assume! If the cash flow timing isn't stated, **annuity-immediate** is the default on FM, but always re-read. The phrase "payments at the beginning of each period," "annuity-due," or "first payment today" signals due — multiply the immediate factor by $(1+i)$. Drawing a timeline of $t=0,1,2,\dots$ before plugging in prevents the whole class of timing errors.

$8\%$ is a **nominal** annual rate $i^{(4)}=0.08$, not effective. The per-quarter rate is $\frac{i^{(m)}}{m}=\frac{0.08}{4}=0.02$. Either work in quarters at $2\%$, or convert to an annual effective rate $i=\left(1+\frac{i^{(m)}}{m}\right)^{m}-1=1.02^{4}-1\approx 8.24\%$. Never discount an annual nominal rate as if it were effective.

They use *different* per-period operations. Effective: $\left(1+\frac{i^{(m)}}{m}\right)^{m}=\left(1-\frac{d^{(p)}}{p}\right)^{-p}=1+i=v^{-1}$. Trap: treating $d^{(m)}/m$ like an interest rate. The per-period **discount** factor is $\left(1-\frac{d^{(m)}}{m}\right)$, applied as a *deduction in advance*, not $\frac{1}{1+\,\cdot}$.

Match the rate to the **payment** period. The annuity factor needs the *annual effective* rate over the payment interval: $i=\left(1+\frac{0.12}{12}\right)^{12}-1\approx 12.6825\%$. Trap: plugging $1\%$ (the monthly rate) or $12\%$ (nominal) directly into a yearly-payment annuity. First convert the conversion-period rate to the payment-period effective rate.

Convert the annual effective rate **down** to the monthly effective rate before valuing: $j=(1+i)^{1/12}-1=1.06^{1/12}-1\approx 0.4868\%$. Then use $n=12\times(\text{years})$. Trap: dividing $6\%$ by $12$ — that's only valid for a *nominal* rate, and an annual effective rate is not nominal.

Immediate (first payment at $t=1$): PV $=\dfrac{1}{i}$. Due (first payment at $t=0$): PV $=\dfrac{1}{d}=\dfrac{1+i}{i}=\dfrac{1}{i}+1$. Trap: using $\frac{1}{i}$ for a perpetuity starting **today**. Since $d=\frac{i}{1+i}$, the due perpetuity is exactly $(1+i)$ times the immediate one.

Wrong constant. The standard increasing perpetuity-immediate is $(Ia)_{\overline{\infty|}}=\dfrac{1}{i}+\dfrac{1}{i^{2}}=\dfrac{1+i}{i^{2}}$, **not** $\frac{1}{i^2}$ alone. The increasing perpetuity-**due** is $(I\ddot a)_{\overline{\infty|}}=\dfrac{1}{d^{2}}$. Memorize the due form as $\frac{1}{d^2}$ and the immediate as $\frac{1+i}{i^2}$ to avoid mixing them.

It fails when redemption $C\neq$ face $F$. The correct test is the sign of $Fr-Ci$ (coupon minus yield-on-redemption): **premium** if $Fr>Ci$, **discount** if $Fr<Ci$. Price form: $P=C+(Fr-Ci)\,a_{\overline{n|}}$. Comparing $r$ to $i$ directly only works when $C=F$.

A **premium** ($P>C$, $Fr>Ci$) writes the book value **down** to $C$. Each coupon $Fr$ splits into interest $i\cdot B_{t-1}$ plus a *write-down* of principal — the "premium amortization," which is $(Fr-Ci)v^{n-t+1}$ and **grows** over time. Trap: treating the whole coupon as interest income, or moving book value the wrong direction (discount accretes **up**).

Interest in payment $t$ is $I_t=i\cdot B_{t-1}$ — the rate times the **prior** (outstanding) balance, *not* the original loan. Principal repaid is $P_t=R-I_t$. For a level payment $R$, principal grows **geometrically**: $P_{t+1}=P_t(1+i)$, so $P_t=R\,v^{n-t+1}$. Trap: using $i\times$ original balance, or assuming a level principal split.

Both must give the same answer. **Prospective:** $B_k=R\,a_{\overline{n-k|}}$ (PV of remaining payments). **Retrospective:** $B_k=L(1+i)^{k}-R\,s_{\overline{k|}}$ (accumulated loan minus accumulated payments). Trap: using $a_{\overline{n|}}$ or $s_{\overline{n|}}$ instead of the *remaining* term $n-k$ / *elapsed* term $k$.

When the sinking-fund rate $j$ equals the loan rate $i$, total outlay equals the amortization payment exactly. The sinking-fund outlay is $L\cdot i + \dfrac{L}{s_{\overline{n|}j}}$ (interest to lender plus deposit). Trap: forgetting the lender still gets $L\cdot i$ **every** period — the deposit alone is not the total payment. If $j<i$, sinking fund costs **more**.

Under the sinking fund method the **stated loan balance stays at $L$** the whole term (interest-only to the lender); only the *net* balance $L-\text{(fund value)}$ declines. Contrast with amortization, where the outstanding balance itself drops each period. Trap: amortizing the loan when the problem specifies a sinking fund — the fund accumulates at rate $j$, separate from the loan at rate $i$.

**Dollar-weighted** (money-weighted) is an IRR: it solves the equation of value, so large deposits before strong sub-periods inflate it — it *is* affected by timing and amount of external cash flows. **Time-weighted** chains sub-period returns $\prod(1+r_k)-1$ and **removes** the effect of deposits/withdrawals, measuring the manager's performance. Trap: using the wrong one for the question asked.

$i\approx\dfrac{I}{A+\sum_t C_t(1-t)}$, where $I=B-A-\sum C_t$ is interest earned, $A$ is the start balance, and each contribution $C_t$ is weighted by the fraction of the year it's invested, $(1-t)$. Trap: weighting deposits by $t$ instead of $(1-t)$, or forgetting that a withdrawal is a **negative** $C_t$.

**Multiply**, don't add: $1+i_{TW}=\prod_{k}(1+r_k)$ where each $r_k=\dfrac{\text{value before next flow}}{\text{value after previous flow}}-1$. Compute each sub-period return on the balance **just after** the cash flow that started it. Trap: averaging the sub-returns, or letting a deposit leak into the return numerator (it must reset the base, not count as gain).

The realized (reinvestment) yield, not the bond's stated yield. Accumulate the coupon stream at the reinvestment rate: total at maturity $=Fr\,s_{\overline{n|}j}+C$, then solve $P(1+i')^{n}=Fr\,s_{\overline{n|}j}+C$ for $i'$. Trap: assuming coupons compound at the bond's yield $i$ — that's only true when $j=i$.

Two rates, two roles. Level payments of $1$ for $n$ years, each reinvested to time $n$ at rate $j$, accumulate to $s_{\overline{n|}j}$ — this uses the **reinvestment** rate, not $i$. If only the interest is reinvested at $j$, the value is $n+i\,(Is)_{\overline{n-1|}j}$ style builds. Trap: blending the earning rate $i$ and reinvestment rate $j$ into a single $s_{\overline{n|}}$.

For a **premium** bond (priced above redemption), the issuer calls **early** to your disadvantage, so you must price to the **earliest** call date — that gives the lowest price (yield-to-worst). For a **discount** bond, the worst case is the **latest** date (maturity). Rule: premium → assume earliest call; discount → assume latest. Price at the worst-case date and quote that price.

There's no formula shortcut: compute the price (or yield) at **every** possible redemption date and take the **minimum price** the investor would accept (equivalently the lowest yield). The premium/discount rule narrows it but you should still check each candidate when the call price varies by date. Trap: stopping at the first call date without confirming it's truly the worst.

Divide by $(1+i)$: $D_{mod}=\dfrac{D_{mac}}{1+i}$. Macaulay is the PV-weighted average time $\dfrac{\sum t\,v^{t}CF_t}{\sum v^{t}CF_t}$; modified is $-\frac{1}{P}\frac{dP}{di}$. Trap: using $D_{mac}$ directly in the price-change estimate $\Delta P\approx -P\,D_{mod}\,\Delta i$. Under a continuous force, $D_{mod}=D_{mac}$ — but only then.

$\Delta P\approx -P\cdot D_{mod}\cdot\Delta i$ — note the **minus** (price falls as yield rises) and that it's **modified** duration. With convexity: $\Delta P\approx -P\,D_{mod}\,\Delta i+\tfrac{1}{2}P\,C\,(\Delta i)^2$. Trap: dropping the negative sign, or pairing the estimate with Macaulay duration instead of modified.

Only by coincidence. Portfolio duration is the **present-value-weighted** average: $D_P=\dfrac{\sum_k P_k D_k}{\sum_k P_k}$, weighting each bond's duration by its **price** (market value), not a simple mean. Trap: $(D_1+D_2)/2$ when the holdings have different values. Dollar durations *do* simply add: $\sum_k P_k D_{mod,k}$.

You need asset **convexity $\geq$** liability convexity: $C_A\geq C_L$ (assets more spread out around the duration point). The three conditions: (1) $PV_A=PV_L$, (2) $D_A=D_L$ (durations match), (3) $C_A\geq C_L$. Trap: writing $C_A\leq C_L$, or matching dollar duration but forgetting the present values must be equal first.

**Redington** protects against *small* yield changes (local: matched PV and duration, asset convexity $\geq$ liability). **Full immunization** protects against shifts of *any* size, but typically requires a single liability bracketed by asset cash flows on both sides (one before, one after the liability date), with matched PV and duration. Trap: claiming Redington covers large shifts — it only guarantees the local minimum.

$\delta=\ln(1+i)$, equivalently $1+i=e^{\delta}$ and $v=e^{-\delta}$. Accumulation under constant force: $a(t)=e^{\delta t}$. Trap: treating $\delta$ as if it were $i$ (it's the *continuously-compounded* rate, always slightly **less** than $i$), or forgetting $\delta=i^{(\infty)}=\lim_{m\to\infty}i^{(m)}$.

Accumulation from time $a$ to $b$ is $\exp\!\left(\displaystyle\int_{a}^{b}\delta_t\,dt\right)$, and the discount factor is its reciprocal $\exp\!\left(-\int_{a}^{b}\delta_t\,dt\right)$. Trap: plugging $\delta_t$ in as $e^{-\delta_t t}$ without integrating, or dropping/flipping the sign of the exponent. Always integrate the force over the interval first.

Picking a **comparison date** but accumulating/discounting some terms to a *different* date. Every cash flow must be valued at the **same** point in time: $\sum (\text{inflows})\,v^{t}=\sum(\text{outflows})\,v^{t}$ at one chosen date. Trap: mixing dates, which silently changes the equation. The chosen date doesn't affect the answer — but it must be consistent across all terms.

Keep one consistent sign rule: inflows positive, outflows negative (or vice versa), then set the net PV to $0$: $\sum_t C_t\,v^{t}=0$. Trap: writing deposits and withdrawals with the same sign, or moving a term across the equals sign without negating it. For IRR, you're solving $NPV(i)=0$ with the signs intact.

Use the **BA II Plus cash-flow/TVM worksheet** or linear interpolation between two trial rates; there's rarely a closed form. Trap: accepting an interpolated $i$ without plugging it back to check $NPV\approx 0$, and forgetting non-level cash flows can give **multiple** IRRs (sign changes in the flow stream). Verify the root and confirm the cash-flow signs change only as expected.

Payments that **grow** need the increasing factor: $(Ia)_{\overline{n|}}=\dfrac{\ddot{a}_{\overline{n|}}-n\,v^{n}}{i}$, not $a_{\overline{n|}}$. For a level perpetuity it's $\frac{1}{i}$; for the increasing perpetuity $\frac{1+i}{i^2}$. Trap: treating an arithmetic-increasing stream as level, or confusing $(Ia)$ with $(I\ddot a)$ (due version, divide by $d$).

Payments $1,(1+g),(1+g)^2,\dots$ valued at rate $i$ collapse to a level annuity at the **rate** $i^{*}=\dfrac{1+i}{1+g}-1=\dfrac{i-g}{1+g}$: PV $=\dfrac{1}{i-g}\left(1-\left(\frac{1+g}{1+i}\right)^{n}\right)=\dfrac{a_{\overline{n|}i^{*}}}{1+g}$. Trap: applying the arithmetic $(Ia)$ formula to a **geometric** grower, or fumbling the $i^{*}$ substitution.

When you're given the **spot-rate curve** $s_1,s_2,\dots$: each cash flow at time $t$ must be discounted at *its own* spot rate, $CF_t(1+s_t)^{-t}$, **not** a single YTM. The YTM is the one rate that reproduces that price. Trap: using a flat rate when a term structure is provided. Forward rates link spots via $(1+s_t)^{t}=(1+s_{t-1})^{t-1}(1+f_{t-1,t})$.

To replace payments $C_t$ at times $t$ with a single equal payment at time $\bar t$, the **exact** date solves $\sum C_t v^{t}=\left(\sum C_t\right)v^{\bar t}$. The *approximate* method of equated time uses the dollar-weighted average time $\bar t=\dfrac{\sum t\,C_t}{\sum C_t}$ — which slightly **overstates** $\bar t$ (true value is a bit earlier). Trap: quoting the approximation as exact.