LIBOR in Arrears¶

In a standard LIBOR-based derivative, the rate $L(T_i)$ is fixed at $T_i$ and paid at the end of the accrual period $T_{i+1}$. In a LIBOR-in-arrears structure, the rate is fixed and paid on the same date $T_i$. Because the natural measure for $L_i$ is $\mathbb{Q}^{T_{i+1}}$, evaluating the expectation of $L_i(T_i)$ under $\mathbb{Q}^{T_i}$ (the measure appropriate for payment at $T_i$) introduces a convexity correction. This section derives the correction formula, explains its financial intuition, and provides a worked numerical example.

Standard vs. Arrears Payment¶

Standard (Natural) Payment¶

A standard floating payment based on LIBOR:

Fixing date: $T_i$ (observe $L_i(T_i)$)
Payment date: $T_{i+1} = T_i + \delta_i$
Cashflow: $\delta_i \, L_i(T_i)$ paid at $T_{i+1}$

Under the $T_{i+1}$-forward measure, $L_i(t)$ is a martingale, so

\[ \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)] = L_i(0) \]

No adjustment is needed for standard payment.

Arrears Payment¶

A LIBOR-in-arrears payment:

Fixing date: $T_i$ (observe $L_i(T_i)$)
Payment date: $T_i$ (same date!)
Cashflow: $\delta_i \, L_i(T_i)$ paid at $T_i$

Pricing requires evaluating $\mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)]$. Since $L_i$ is a martingale under $\mathbb{Q}^{T_{i+1}}$ but not under $\mathbb{Q}^{T_i}$, this expectation is not equal to $L_i(0)$.

Derivation of the Convexity Correction¶

Step 1: Pricing Under the Natural Measure¶

The present value of the arrears payment is

\[ V_0 = P(0, T_i) \, \mathbb{E}^{\mathbb{Q}^{T_i}}\!\left[\delta_i \, L_i(T_i)\right] \]

We need to compute $\mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)]$.

Step 2: Change of Measure¶

Using the Radon--Nikodym derivative from $\mathbb{Q}^{T_{i+1}}$ to $\mathbb{Q}^{T_i}$:

\[ \frac{d\mathbb{Q}^{T_i}}{d\mathbb{Q}^{T_{i+1}}}\bigg|_{\mathcal{F}_{T_i}} = \frac{P(T_i, T_i) / P(0, T_i)}{P(T_i, T_{i+1}) / P(0, T_{i+1})} = \frac{1 / P(0, T_i)}{P(T_i, T_{i+1}) / P(0, T_{i+1})} \]

Since $P(T_i, T_{i+1}) = 1/(1 + \delta_i L_i(T_i))$:

\[ \frac{d\mathbb{Q}^{T_i}}{d\mathbb{Q}^{T_{i+1}}}\bigg|_{\mathcal{F}_{T_i}} = \frac{P(0, T_{i+1})}{P(0, T_i)} \cdot (1 + \delta_i L_i(T_i)) \]

Step 3: Compute the Adjusted Expectation¶

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] = \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}\!\left[L_i(T_i) \cdot \frac{P(0, T_{i+1})}{P(0, T_i)} \cdot (1 + \delta_i L_i(T_i))\right] \]

Since $P(0, T_{i+1}) / P(0, T_i) = 1/(1 + \delta_i L_i(0))$ (to first order), this simplifies to

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] = \frac{\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}\!\left[L_i(T_i)(1 + \delta_i L_i(T_i))\right]}{1 + \delta_i L_i(0)} \]

Step 4: Expand the Numerator¶

\[ \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}\!\left[L_i(T_i)(1 + \delta_i L_i(T_i))\right] = \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)] + \delta_i \, \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^2] \]

Using the martingale property, $\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)] = L_i(0)$.

For the second moment under lognormal dynamics:

\[ \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^2] = L_i(0)^2 \, e^{\sigma_i^2 T_i} \]

where $\sigma_i$ is the Black implied volatility.

Step 5: The Convexity Correction Formula¶

Combining:

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] = \frac{L_i(0) + \delta_i \, L_i(0)^2 \, e^{\sigma_i^2 T_i}}{1 + \delta_i L_i(0)} \]

For moderate $\sigma_i^2 T_i$, using $e^{\sigma_i^2 T_i} \approx 1 + \sigma_i^2 T_i$:

\[ \boxed{\mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] \approx L_i(0) + \frac{\delta_i \, L_i(0)^2 \, \sigma_i^2 \, T_i}{1 + \delta_i L_i(0)}} \]

The convexity correction is

\[ \text{Correction} = \frac{\delta_i \, L_i(0)^2 \, \sigma_i^2 \, T_i}{1 + \delta_i L_i(0)} \]

Financial Intuition¶

Why the Correction Is Positive¶

The correction is always positive: the arrears expectation exceeds the forward rate. The intuition is:

When $L_i(T_i)$ is high, the bond $P(T_i, T_{i+1}) = 1/(1 + \delta_i L_i(T_i))$ is low (discount factor is low), so the present value of the arrears payment is higher than the standard payment
When $L_i(T_i)$ is low, $P(T_i, T_{i+1})$ is high, but the rate itself is low
The asymmetry (rates appear both in the payment and in the discounting) creates a positive bias

This is a manifestation of Jensen's inequality: the function $L \mapsto L(1 + \delta L)$ is convex in $L$.

Size of the Correction¶

The correction scales as:

$L_i(0)^2$ --- larger for higher rate levels
$\sigma_i^2$ --- larger for higher volatility
$T_i$ --- larger for longer fixing periods
$\delta_i$ --- larger for longer accrual periods

Worked Example¶

LIBOR-in-Arrears Convexity Correction

Parameters:

Forward rate: $L_i(0) = 5.0\% = 0.05$
Volatility: $\sigma_i = 20\%$
Fixing time: $T_i = 5.0$ years
Accrual fraction: $\delta_i = 0.5$ (semiannual)
Discount factor: $P(0, T_i) = 0.78$

Step 1: Convexity correction

\[ \text{Correction} = \frac{0.5 \times 0.05^2 \times 0.04 \times 5.0}{1 + 0.5 \times 0.05} = \frac{0.5 \times 0.0025 \times 0.2}{1.025} = \frac{0.000250}{1.025} = 0.000244 = 2.44 \text{ bp} \]

Step 2: Adjusted expectation

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] = 5.00\% + 2.44 \text{ bp} = 5.024\% \]

Step 3: Present value of arrears payment

\[ V_0 = 0.78 \times 0.5 \times 0.05024 = 0.01959 \]

Comparison: Without the correction, $V_0 = 0.78 \times 0.5 \times 0.05 = 0.01950$. The convexity adjustment adds $0.93 per $10,000 notional per period.

Exact Formula (Without Approximation)¶

The exact result (no Taylor expansion) is

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] = \frac{L_i(0)(1 + \delta_i L_i(0) \, e^{\sigma_i^2 T_i})}{1 + \delta_i L_i(0)} \]

For the parameters in the example above:

\[ = \frac{0.05 \times (1 + 0.5 \times 0.05 \times e^{0.04 \times 5})}{1.025} = \frac{0.05 \times (1 + 0.025 \times 1.2214)}{1.025} = \frac{0.05 \times 1.03054}{1.025} = 0.05027 \]

The exact value (5.027%) is close to the approximation (5.024%), confirming the accuracy of the first-order expansion.

Arrears Caplet¶

Payoff¶

An arrears caplet pays $\delta_i \max(L_i(T_i) - K, 0)$ at time $T_i$ (instead of $T_{i+1}$).

Pricing¶

\[ \text{Arrears Caplet} = P(0, T_i) \, \mathbb{E}^{\mathbb{Q}^{T_i}}\!\left[\delta_i \max(L_i(T_i) - K, 0)\right] \]

Under the change of measure, this can be expressed as

\[ \text{Arrears Caplet} = P(0, T_{i+1}) \, \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}\!\left[\delta_i \max(L_i(T_i) - K, 0) \cdot (1 + \delta_i L_i(T_i))\right] \]

Expanding the product and using properties of the lognormal distribution, the arrears caplet can be decomposed into a standard caplet plus a correction involving the second moment of $L_i$.

Key Takeaways¶

LIBOR-in-arrears pays the rate on the same date it is fixed, rather than at the end of the accrual period
The convexity correction is $\delta_i L_i(0)^2 \sigma_i^2 T_i / (1 + \delta_i L_i(0))$, always positive
The correction arises from the measure change between the $T_{i+1}$-forward and $T_i$-forward measures
Financial intuition: convexity of the payoff in the rate creates a Jensen's inequality effect
The correction is most significant for long-dated, high-volatility, high-rate environments
The exact formula uses $e^{\sigma_i^2 T_i}$ rather than the linear approximation $1 + \sigma_i^2 T_i$

Exercises¶

Exercise 1. A LIBOR-in-arrears payment fixes and pays at $T_i = 3$ years. The forward LIBOR rate is $L_i(0) = 4.2\%$, the Black implied volatility is $\sigma_i = 18\%$, and the accrual fraction is $\delta_i = 0.25$ (quarterly). Compute both the approximate and exact convexity corrections and the corresponding adjusted forward rates. What is the percentage error of the approximation?

Solution to Exercise 1

Given: $L_i(0) = 0.042$, $\sigma_i = 0.18$, $T_i = 3$, $\delta_i = 0.25$.

Approximate convexity correction:

\[ \text{Correction}_{\text{approx}} = \frac{\delta_i \, L_i(0)^2 \, \sigma_i^2 \, T_i}{1 + \delta_i L_i(0)} \]

Substituting:

\[ = \frac{0.25 \times 0.042^2 \times 0.18^2 \times 3}{1 + 0.25 \times 0.042} = \frac{0.25 \times 0.001764 \times 0.0324 \times 3}{1.0105} \]

Numerator: $0.25 \times 0.001764 \times 0.0324 \times 3 = 0.25 \times 0.001764 \times 0.0972 = 0.000042866$.

\[ \text{Correction}_{\text{approx}} = \frac{0.000042866}{1.0105} = 0.00004242 = 0.4242 \text{ bp} \]

Approximate adjusted forward rate: $L_i(0) + 0.4242 \text{ bp} = 4.2000\% + 0.0424\% = 4.2424\%$.

Exact convexity correction:

Using the exact formula:

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] = \frac{L_i(0)\bigl(1 + \delta_i L_i(0) \, e^{\sigma_i^2 T_i}\bigr)}{1 + \delta_i L_i(0)} \]

We compute $e^{\sigma_i^2 T_i} = e^{0.0324 \times 3} = e^{0.0972} = 1.10208$.

\[ = \frac{0.042 \times (1 + 0.25 \times 0.042 \times 1.10208)}{1.0105} = \frac{0.042 \times (1 + 0.011571)}{1.0105} = \frac{0.042 \times 1.011571}{1.0105} \]

\[ = \frac{0.042486}{1.0105} = 0.042044 \]

Exact correction: $0.042044 - 0.042000 = 0.000044 = 0.44$ bp.

Exact adjusted forward rate: $4.2044\%$.

Percentage error of approximation:

\[ \text{Error} = \frac{|0.4242 - 0.44|}{0.44} \times 100\% \approx 3.6\% \]

The approximation underestimates the exact correction by about 3.6%, which is small because $\sigma_i^2 T_i = 0.0972$ is modest, so the linear Taylor expansion $e^x \approx 1 + x$ is accurate.

Exercise 2. Starting from the Radon--Nikodym derivative

\[ \frac{d\mathbb{Q}^{T_i}}{d\mathbb{Q}^{T_{i+1}}}\bigg|_{\mathcal{F}_{T_i}} = \frac{P(0, T_{i+1})}{P(0, T_i)} \cdot (1 + \delta_i L_i(T_i)), \]

derive the exact formula

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] = \frac{L_i(0)\bigl(1 + \delta_i L_i(0)\,e^{\sigma_i^2 T_i}\bigr)}{1 + \delta_i L_i(0)} \]

by computing $\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^2]$ explicitly under lognormal dynamics. State all assumptions used.

Solution to Exercise 2

Assumptions:

Under $\mathbb{Q}^{T_{i+1}}$, the forward LIBOR rate $L_i(t)$ follows lognormal (Black) dynamics: $dL_i/L_i = \sigma_i \, dW^{T_{i+1}}$.
$\sigma_i$ is constant.
The relation $P(T_i, T_{i+1}) = 1/(1 + \delta_i L_i(T_i))$ holds (simple compounding).
$P(0, T_{i+1})/P(0, T_i) = 1/(1 + \delta_i L_i(0))$ (consistent with the forward rate definition).

Derivation:

Starting from the Radon--Nikodym derivative:

\[ \frac{d\mathbb{Q}^{T_i}}{d\mathbb{Q}^{T_{i+1}}}\bigg|_{\mathcal{F}_{T_i}} = \frac{P(0, T_{i+1})}{P(0, T_i)} \cdot (1 + \delta_i L_i(T_i)) \]

The change-of-measure formula gives:

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] = \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}\!\left[L_i(T_i) \cdot \frac{P(0, T_{i+1})}{P(0, T_i)} \cdot (1 + \delta_i L_i(T_i))\right] \]

Since $P(0, T_{i+1})/P(0, T_i) = 1/(1 + \delta_i L_i(0))$:

\[ = \frac{1}{1 + \delta_i L_i(0)} \, \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}\!\left[L_i(T_i)(1 + \delta_i L_i(T_i))\right] \]

Expanding:

\[ = \frac{1}{1 + \delta_i L_i(0)} \left(\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)] + \delta_i \, \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^2]\right) \]

Computing $\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^2]$:

Under lognormal dynamics, $L_i(T_i) = L_i(0) \exp\!\left(-\tfrac{1}{2}\sigma_i^2 T_i + \sigma_i W(T_i)\right)$ where $W(T_i) \sim \mathcal{N}(0, T_i)$.

For a lognormal variable $X = e^{\mu + \sigma Z}$ with $Z \sim \mathcal{N}(0,1)$:

\[ \mathbb{E}[X^2] = e^{2\mu + 2\sigma^2} \]

Here $\mu = \ln L_i(0) - \tfrac{1}{2}\sigma_i^2 T_i$ and $\sigma^2 = \sigma_i^2 T_i$, so:

\[ \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^2] = \exp\!\left(2\ln L_i(0) - \sigma_i^2 T_i + 2\sigma_i^2 T_i\right) = L_i(0)^2 \, e^{\sigma_i^2 T_i} \]

Combining: Using the martingale property $\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)] = L_i(0)$:

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] = \frac{L_i(0) + \delta_i \, L_i(0)^2 \, e^{\sigma_i^2 T_i}}{1 + \delta_i L_i(0)} = \frac{L_i(0)\bigl(1 + \delta_i L_i(0) \, e^{\sigma_i^2 T_i}\bigr)}{1 + \delta_i L_i(0)} \]

This is the exact formula. $\blacksquare$

Exercise 3. Explain why the map $L \mapsto L(1 + \delta L)$ is convex in $L$ for $\delta > 0$, and use Jensen's inequality to give a one-line proof that $\mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] > L_i(0)$ without performing any measure-change calculation.

Solution to Exercise 3

Define $f(L) = L(1 + \delta L) = L + \delta L^2$. We compute the second derivative:

\[ f'(L) = 1 + 2\delta L, \qquad f''(L) = 2\delta \]

Since $\delta > 0$, we have $f''(L) = 2\delta > 0$ for all $L$, so $f$ is strictly convex.

One-line proof via Jensen's inequality:

The present value of the arrears payment can be written as $P(0, T_{i+1}) \, \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[f(L_i(T_i))]/(1+\delta_i L_i(0))$, and since $f$ is convex and $L_i$ is a martingale under $\mathbb{Q}^{T_{i+1}}$, Jensen's inequality gives:

\[ \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[f(L_i(T_i))] > f\!\left(\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)]\right) = f(L_i(0)) = L_i(0)(1 + \delta_i L_i(0)) \]

(strict inequality holds whenever $L_i(T_i)$ is non-degenerate). Since $\mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)]$ equals $\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[f(L_i(T_i))]/(1+\delta_i L_i(0))$ (from the measure change), we get:

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] > \frac{L_i(0)(1+\delta_i L_i(0))}{1+\delta_i L_i(0)} = L_i(0) \]

No explicit measure-change calculation is needed; the convexity of $f$ and the martingale property suffice. $\blacksquare$

Exercise 4. Consider a swap that pays LIBOR-in-arrears quarterly for 5 years on a notional of $100 million. The forward LIBOR rates for each quarterly period are approximately $L_i(0) = 5\%$, the implied volatilities are $\sigma_i = 20\%$, and the accrual fractions are $\delta_i = 0.25$. Estimate the total convexity adjustment (summed over all 20 periods) in dollar terms. How does this compare to a single basis point on the fixed leg?

Solution to Exercise 4

Given: 20 quarterly periods, $L_i(0) = 0.05$, $\sigma_i = 0.20$, $\delta_i = 0.25$, notional $N = \$100{,}000{,}000$.

The convexity correction for period $i$ (fixing at $T_i = 0.25 i$) is:

\[ C_i = \frac{\delta_i \, L_i(0)^2 \, \sigma_i^2 \, T_i}{1 + \delta_i L_i(0)} = \frac{0.25 \times 0.0025 \times 0.04 \times T_i}{1 + 0.0125} = \frac{0.000025 \, T_i}{1.0125} \]

The total correction summed over all periods:

\[ \sum_{i=1}^{20} C_i = \frac{0.000025}{1.0125} \sum_{i=1}^{20} T_i = \frac{0.000025}{1.0125} \sum_{i=1}^{20} 0.25i \]

\[ \sum_{i=1}^{20} 0.25i = 0.25 \times \frac{20 \times 21}{2} = 0.25 \times 210 = 52.5 \]

\[ \sum_{i=1}^{20} C_i = \frac{0.000025 \times 52.5}{1.0125} = \frac{0.0013125}{1.0125} = 0.001296 \]

This is the sum of the rate adjustments (in absolute terms). The dollar value of each period's correction is $N \times \delta_i \times C_i$. The total dollar convexity adjustment (undiscounted) is:

\[ \text{Total} = N \times \delta_i \times \sum_{i=1}^{20} C_i = 100{,}000{,}000 \times 0.25 \times 0.001296 = \$32{,}407 \]

Comparison to one basis point on the fixed leg:

One basis point on the fixed leg over 5 years (20 quarterly periods) is:

\[ \text{1 bp} = N \times 0.0001 \times \sum_{i=1}^{20} \delta_i \times P(0, T_{i+1}) \approx N \times 0.0001 \times 20 \times 0.25 \times \bar{P} \]

Assuming an average discount factor $\bar{P} \approx 0.95$:

\[ \text{1 bp} \approx 100{,}000{,}000 \times 0.0001 \times 5 \times 0.95 = \$47{,}500 \]

The total convexity adjustment of approximately $32,400 is about 0.68 basis points on the fixed leg. This is material for a $100 million swap and would need to be incorporated in pricing.

Exercise 5. An arrears caplet pays $\delta_i \max(L_i(T_i) - K, 0)$ at time $T_i$ with $K = 5\%$, $L_i(0) = 5\%$, $\sigma_i = 22\%$, $T_i = 2$, and $\delta_i = 0.5$. The standard (non-arrears) caplet with the same parameters pays at $T_{i+1}$. Explain qualitatively why the arrears caplet is more expensive than the standard caplet. Write down the pricing formula for the arrears caplet in terms of an expectation under $\mathbb{Q}^{T_{i+1}}$ and identify the additional term compared to the standard Black caplet formula.

Solution to Exercise 5

Qualitative explanation of why the arrears caplet is more expensive:

The standard caplet pays $\delta_i \max(L_i(T_i) - K, 0)$ at $T_{i+1}$, while the arrears caplet pays the same amount at $T_i$ (earlier). Receiving a positive cashflow earlier is always more valuable due to time-value of money, but the effect here is subtler: the cashflow is large precisely when rates are high (i.e., when $L_i(T_i) > K$), and high rates mean that discounting from $T_{i+1}$ back to $T_i$ provides a larger benefit. This positive correlation between the payoff magnitude and the discount-factor advantage creates an additional convexity premium.

Pricing formula under $\mathbb{Q}^{T_{i+1}}$:

\[ \text{Arrears Caplet} = P(0, T_i) \, \mathbb{E}^{\mathbb{Q}^{T_i}}\!\left[\delta_i (L_i(T_i) - K)^+\right] \]

Using the change of measure:

\[ = P(0, T_{i+1}) \, \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}\!\left[\delta_i (L_i(T_i) - K)^+ \cdot (1 + \delta_i L_i(T_i))\right] \]

Expanding the product:

\[ = P(0, T_{i+1}) \, \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}\!\left[\delta_i (L_i - K)^+\right] + P(0, T_{i+1}) \, \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}\!\left[\delta_i^2 \, L_i \, (L_i - K)^+\right] \]

The first term is exactly the standard Black caplet price. The second term, $P(0, T_{i+1}) \, \delta_i^2 \, \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)(L_i(T_i) - K)^+]$, is the additional term that makes the arrears caplet more expensive. This term involves $\mathbb{E}[L \cdot (L-K)^+]$, which under lognormal dynamics can be computed in terms of the second moment of $L$ conditional on $L > K$. It is always positive, confirming that the arrears caplet price exceeds the standard caplet price.

Exercise 6. Suppose interest rates are modeled under a normal (Bachelier) framework instead of the lognormal (Black) framework, so that $L_i(T_i) \sim \mathcal{N}(L_i(0), \sigma_N^2 T_i)$ under $\mathbb{Q}^{T_{i+1}}$. Rederive the LIBOR-in-arrears convexity correction under normal dynamics and show that it takes the form

\[ \text{Correction}_{\text{normal}} = \frac{\delta_i \, \sigma_N^2 \, T_i}{1 + \delta_i L_i(0)} \]

Compare this to the lognormal correction and discuss when the two formulas give materially different results.

Solution to Exercise 6

Setup: Under normal dynamics, $L_i(T_i) = L_i(0) + \sigma_N \sqrt{T_i} \, Z$ where $Z \sim \mathcal{N}(0,1)$ under $\mathbb{Q}^{T_{i+1}}$.

Under $\mathbb{Q}^{T_{i+1}}$, $L_i$ is a martingale: $\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)] = L_i(0)$.

Compute the second moment:

\[ \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^2] = \text{Var}(L_i(T_i)) + \left(\mathbb{E}[L_i(T_i)]\right)^2 = \sigma_N^2 T_i + L_i(0)^2 \]

Apply the same measure change formula:

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)] = \frac{\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)(1 + \delta_i L_i(T_i))]}{1 + \delta_i L_i(0)} \]

Expanding:

\[ = \frac{L_i(0) + \delta_i(\sigma_N^2 T_i + L_i(0)^2)}{1 + \delta_i L_i(0)} = \frac{L_i(0)(1 + \delta_i L_i(0)) + \delta_i \sigma_N^2 T_i}{1 + \delta_i L_i(0)} \]

\[ = L_i(0) + \frac{\delta_i \, \sigma_N^2 \, T_i}{1 + \delta_i L_i(0)} \]

Therefore:

\[ \text{Correction}_{\text{normal}} = \frac{\delta_i \, \sigma_N^2 \, T_i}{1 + \delta_i L_i(0)} \]

Comparison with the lognormal correction:

The lognormal correction is $\delta_i L_i(0)^2 \sigma_i^2 T_i / (1 + \delta_i L_i(0))$, while the normal correction is $\delta_i \sigma_N^2 T_i / (1 + \delta_i L_i(0))$.

The two agree when $\sigma_N = L_i(0) \sigma_i$ (i.e., when the normal volatility equals the lognormal volatility times the rate level), which is the standard at-the-money volatility equivalence.

The formulas give materially different results when rates move far from their initial level. If rates double from $L_i(0)$ to $2L_i(0)$, the lognormal correction would scale as $(2L_i(0))^2 \sigma_i^2 = 4 L_i(0)^2 \sigma_i^2$ (quadratic in the rate level), while the normal correction remains $\sigma_N^2$ (independent of the rate level). In a low-rate environment (e.g., $L_i(0) = 0.5\%$), the lognormal correction is tiny ($\propto L_i(0)^2$) while the normal correction can still be significant if $\sigma_N$ is non-negligible. Conversely, in a high-rate environment, the lognormal correction dominates.

Exercise 7. A structured note pays $\delta_i \, L_i(T_i)^2$ at time $T_i$ (a "LIBOR-squared" in-arrears payment). Using the change of measure from $\mathbb{Q}^{T_{i+1}}$ to $\mathbb{Q}^{T_i}$, derive the convexity-adjusted expectation $\mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)^2]$ under lognormal dynamics. You will need the third moment of a lognormal random variable. Express your result in terms of $L_i(0)$, $\sigma_i$, $T_i$, and $\delta_i$.

Solution to Exercise 7

LIBOR-squared in arrears: We need $\mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)^2]$.

Using the change of measure:

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)^2] = \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}\!\left[L_i(T_i)^2 \cdot \frac{P(0, T_{i+1})}{P(0, T_i)} \cdot (1 + \delta_i L_i(T_i))\right] \]

\[ = \frac{1}{1 + \delta_i L_i(0)} \left(\mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^2] + \delta_i \, \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^3]\right) \]

Second moment (computed in Exercise 2):

\[ \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^2] = L_i(0)^2 \, e^{\sigma_i^2 T_i} \]

Third moment of a lognormal variable:

If $L_i(T_i) = L_i(0) \exp(-\frac{1}{2}\sigma_i^2 T_i + \sigma_i W(T_i))$, then for any integer $k$:

\[ \mathbb{E}[L_i(T_i)^k] = L_i(0)^k \exp\!\left(\frac{k(k-1)}{2}\sigma_i^2 T_i\right) \]

For $k = 3$:

\[ \mathbb{E}^{\mathbb{Q}^{T_{i+1}}}[L_i(T_i)^3] = L_i(0)^3 \, e^{3\sigma_i^2 T_i} \]

Combining:

\[ \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)^2] = \frac{L_i(0)^2 \, e^{\sigma_i^2 T_i} + \delta_i \, L_i(0)^3 \, e^{3\sigma_i^2 T_i}}{1 + \delta_i L_i(0)} \]

Factoring:

\[ = \frac{L_i(0)^2 \, e^{\sigma_i^2 T_i}\!\left(1 + \delta_i \, L_i(0) \, e^{2\sigma_i^2 T_i}\right)}{1 + \delta_i L_i(0)} \]

The convexity correction for the LIBOR-squared payment is the difference between this expression and the "natural" second moment:

\[ \text{Correction} = \mathbb{E}^{\mathbb{Q}^{T_i}}[L_i(T_i)^2] - L_i(0)^2 \, e^{\sigma_i^2 T_i} \]

\[ = L_i(0)^2 \, e^{\sigma_i^2 T_i} \left(\frac{1 + \delta_i L_i(0) \, e^{2\sigma_i^2 T_i}}{1 + \delta_i L_i(0)} - 1\right) = \frac{\delta_i \, L_i(0)^3 \, e^{\sigma_i^2 T_i}(e^{2\sigma_i^2 T_i} - 1)}{1 + \delta_i L_i(0)} \]

Note that this correction involves the third moment through the $e^{3\sigma_i^2 T_i}$ term, and it scales as $L_i(0)^3$ (compared to $L_i(0)^2$ for the standard LIBOR-in-arrears correction), making it more sensitive to the rate level.