8.2 The Arbitrage-Free Valuation Framework & Binomial Interest Rate Trees

8.2 The Arbitrage-Free Valuation Framework & Binomial Interest Rate Trees

Core Insight: Binomial interest rate trees provide the computational foundation for valuing bonds with embedded options. Unlike equity binomial models where the underlying stock price moves up or down, in a fixed income binomial tree, the fundamental stochastic variable is the one-period short rate. An arbitrage-free tree is calibrated so that it reproduces the exact market prices of benchmark option-free bonds.

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1. Principles of Arbitrage-Free Valuation

An arbitrage opportunity is an investment strategy that generates riskless profit without net capital investment. Fixed income arbitrage analysis distinguishes between two formal definitions:

• Type I Arbitrage (Value Additivity / Free Lunch): Occurs when two portfolios producing identical future cash flows under all states of the world trade at different market prices today, generating immediate positive cash flow with zero future liability.
• Type II Arbitrage (Dominance / Free Lottery): Occurs when a costless investment produces non-negative cash flows in all future states with a strictly positive probability of a positive cash flow in at least one state.

The No-Arbitrage Condition and Strip Replication

An arbitrage-free framework ensures that any coupon bond can be replicated perfectly by a portfolio of zero-coupon bonds (strips). If a coupon bond trades below the sum of its stripped component values ($P_{\text{bond}} < \sum CF_t \cdot P(t)$), an arbitrageur buys the bond, strips the coupons and principal, sells the individual strips, and locks in a Type I arbitrage profit.

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2. Binomial Interest Rate Tree Construction

Lognormal Interest Rate Dynamics

A binomial interest rate tree models the evolution of the one-period forward rate. The period length ($\Delta t$) is typically 1 year or 6 months.

Time 0                     Time 1                                Time 2
                                                             i_2,UU = i_2,LL * e^(4*sigma)
                                    i_1,U = i_1,L * e^(2*sigma)    /
          i_0                      /                        --- i_2,UL = i_2,LL * e^(2*sigma)
(Current 1-Period Rate) --------->                         \   
                                   \                        i_2,LL
                                    i_1,L

Key Structural Rules of the Lognormal Tree:

1. Equal Risk-Neutral Probabilities: At every node, the probability of an upward move is exactly $p = 0.50$, and the probability of a downward move is $1 - p = 0.50$.
2. Lognormal Spread Condition: Rates at adjacent vertical nodes at time $t$ are linked by the volatility parameter $\sigma$: $i_{t,U} = i_{t,L} e^{2\sigma} \iff \ln(i_{t,U}) - \ln(i_{t,L}) = 2\sigma$ $i_{t,UU} = i_{t,LL} e^{4\sigma}, \quad i_{t,UL} = i_{t,LL} e^{2\sigma}, \quad i_{t,LL}$
3. Non-Negativity: Because $i_{t} = i_{t,LL} e^{2k\sigma} > 0$, the lognormal model ensures that short-term interest rates cannot drop below zero.
4. Proportional Volatility: The standard deviation of the rate change is proportional to the level of rates ($\sigma \times i_t$), matching empirical bond market behavior.

Iterative Par Curve Calibration

The tree is calibrated sequentially from $t = 0$ forward using benchmark on-the-run Treasury par bonds:

• $t = 0$: $i_0$ is set directly to the current 1-year spot rate ($z_1$).
• $t = 1$: Given $\sigma$, find the single base rate $i_{1,L}$ such that a 2-year par bond discounted through nodes $i_{1,U} = i_{1,L}e^{2\sigma}$ and $i_{1,L}$ yields exactly par value ($100.00$) at $t = 0$.
• $t = 2$: Given $\sigma$, solve for $i_{2,LL}$ such that a 3-year par bond discounted through nodes $i_{2,UU}, i_{2,UL}, i_{2,LL}$ yields exactly par ($100.00$) at $t = 0$.

3. Backward Induction Valuation of Option-Free Bonds

Backward Induction Formula

Valuation starts at bond maturity ($t = T$) where the bond value is known with certainty ($V_T = 100$), and works backward through the tree step-by-step to $t = 0$.

At any given node at time $t$, the node value $V$ is the expected discounted value of the next period's bond values plus coupon payment $C$:

$V = \frac{1}{2} \left[ \frac{V_u + C}{1 + i_t} + \frac{V_d + C}{1 + i_t} \right] = \frac{0.5(V_u + C) + 0.5(V_d + C)}{1 + i_t}$

Where:

• $V_u$ = Bond value at the upper adjacent node at time $t + 1$.
• $V_d$ = Bond value at the lower adjacent node at time $t + 1$.
• $C$ = Annual coupon payment received at time $t + 1$.
• $i_t$ = One-period forward rate at the current node at time $t$.

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4. Full Worked 3-Period Backward Induction Valuation Model

Calibrated Tree Parameters

Consider a calibrated 2-year binomial interest rate tree with annual volatility $\sigma = 15%$:

• Time $t = 0$: $i_0 = 3.0000%$
• Time $t = 1$: $i_{1,U} = 5.3994%$, $i_{1,L} = 4.0000%$ (Note: $4.0000 \times e^{2 \times 0.15} = 4.0000 \times 1.349859 = 5.3994%$)
• Time $t = 2$: $i_{2,UU} = 8.0991%$, $i_{2,UL} = 6.0000%$, $i_{2,LL} = 4.4449%$

We value a 3-year, 5.50% annual coupon bond with face value $FV = 100$.

t = 0                      t = 1                             t = 2                         t = 3
                                                                                        (Maturity: $100)
                                                      i_2,UU = 8.0991%
                                                      V_2,UU = (100+5.50)/1.080991
                                                      V_2,UU = $97.5956 ---------------> $100 + $5.50
                              i_1,U = 5.3994%        /
                              V_1,U = $99.5693 -----   i_2,UL = 6.0000%
                             /                       \ V_2,UL = (100+5.50)/1.060000
i_0 = 3.0000%               /                         V_2,UL = $99.5283 ---------------> $100 + $5.50
V_0 = $101.4429 ----------->                         /
                            \                         i_2,LL = 4.4449%
                             \                        V_2,LL = (100+5.50)/1.044449
                              i_1,L = 4.0000% -----   V_2,LL = $101.0098 --------------> $100 + $5.50
                              V_1,L = $101.4116      \
                                                                                        (Maturity: $100)

Step 1: Calculate Node Values at Time $t = 2$

At $t = 3$, all terminal payoffs are $100 + 5.50 = 105.50$.

• Node 2,UU: $V_{2,UU} = \frac{105.50}{1 + 0.080991} = \frac{105.50}{1.080991} = \mathbf{97.5956}$
• Node 2,UL: $V_{2,UL} = \frac{105.50}{1 + 0.060000} = \frac{105.50}{1.060000} = \mathbf{99.5283}$
• Node 2,LL: $V_{2,LL} = \frac{105.50}{1 + 0.044449} = \frac{105.50}{1.044449} = \mathbf{101.0098}$

Step 2: Calculate Node Values at Time $t = 1$

• Node 1,U: $V_{1,U} = \frac{0.5 \times (97.5956 + 5.50) + 0.5 \times (99.5283 + 5.50)}{1 + 0.053994} = \frac{0.5 \times 103.0956 + 0.5 \times 105.0283}{1.053994} = \frac{104.0620}{1.053994} = \mathbf{98.7311}$
• Node 1,L: $V_{1,L} = \frac{0.5 \times (99.5283 + 5.50) + 0.5 \times (101.0098 + 5.50)}{1 + 0.040000} = \frac{0.5 \times 105.0283 + 0.5 \times 106.5098}{1.040000} = \frac{105.7691}{1.040000} = \mathbf{101.7011}$

Step 3: Calculate Present Value at Time $t = 0$

$V_0 = \frac{0.5 \times (98.7311 + 5.50) + 0.5 \times (101.7011 + 5.50)}{1 + 0.030000} = \frac{0.5 \times 104.2311 + 0.5 \times 107.2011}{1.030000} = \frac{105.7161}{1.030000} = \mathbf{102.6370}$

The Fundamental Tree Invariance Rule: For any option-free bond, pricing the bond via backward induction on a calibrated binomial tree produces the exact same price as discounting its cash flows across the spot rate curve ($V_0 = \sum \frac{C}{(1+z_t)^t} + \frac{FV}{(1+z_T)^T}$). The binomial tree adds value specifically when evaluating pathways with embedded options where cash flows become contingent on future interest rates.

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5. Pathwise Valuation vs. Monte Carlo Simulation

Pathwise Valuation Mechanics

An alternative, mathematically identical valuation method is pathwise valuation. For an $N$-period binomial tree, there exist exactly $2^{N-1}$ unique interest rate paths.

To value a bond using pathwise valuation:

1. Identify all $2^{N-1}$ possible interest rate sequences from $t = 0$ to $t = T-1$.
2. Discount the bond's cash flows along each individual path $k$ to compute path present value $PV_k$: $PV_k = \sum_{t=1}^T \frac{CF_t}{\prod_{j=0}^{t-1} (1 + i_{j,k})}$
3. Calculate the simple arithmetic mean across all $2^{N-1}$ paths: $V_0 = \frac{1}{2^{N-1}} \sum_{k=1}^{2^{N-1}} PV_k$.

Why Monte Carlo Simulation is Required for Path-Dependent Securities

MBS Prepayment Burnout Insight: Homeowners' decisions to refinance mortgage debt depend not just on the current interest rate, but on whether mortgage rates were even lower in previous months. Borrowers who had the financial capacity to refinance already did so during prior rate dips (burnout effect). Because a standard recombining binomial tree cannot record whether a node was reached via a prior low-rate path, Monte Carlo simulation is mandatory for pricing mortgage-backed securities.