Deriving a Small-Volatility Convexity Adjustment for Bond Yields

Summary

The document examines an approximation used in deriving a convexity adjustment for a bond forward. It expands the bond price as a function of yield around today’s forward yield, then asks why the expected squared yield change can be approximated by the initial yield squared times yield volatility squared and time.

The answer supplies a model-based explanation: if yield follows geometric Brownian motion, its expected value remains at its initial level and its variance is the initial yield squared times the exponential of volatility squared times time, less one. The expected squared deviation from the initial yield equals that variance. When volatility squared times time is small, a first-order expansion of the exponential gives the stated approximation. This is a small-time, small-volatility result under the assumed process; other yield dynamics can produce a different expression. The document does not fully resolve the preceding Taylor expansion or clarify its expectation notation.

Key ideas

• A Taylor expansion of bond price links convexity to the curvature of price as a function of yield.
• Under geometric Brownian motion, the yield’s expected squared deviation from its initial value equals its variance.
• The variance is approximated by initial yield squared times volatility squared times time when that product is small.
• The approximation depends on the assumed yield process and may change under other models.

Tags

# Proof of the convexity adjustment formula


# Proof of the convexity adjustment formula












Let $y_0$ be the forward bond yield observed today for a forward contract with maturity $T$, $y_T$ be the bond yield at time $T$, $B_T$ be the price of the bond at time $T$ and let $\sigma_y$ be the volatility of the forward bond yield.

Suppose that $B_T = g(y_T)$ then expanding using a taylor series yields,

$$B_T = G(y_0)+(y_T-y_0)G'(y_0)+0.5G''(y_0)(y_T-y_0)^2$$

and then taking expectations we get

$$E_T(B_T) = G(y_0)+E_T(y_T-y_0)G'(y_0)+0.5G''(y_0)E_T(y_T-y_0)^2$$

as we are working in the risk neutral world, $E_T(B_T)=G(y_0)$,

and so

$$E_T(y_T-y_0)G'(y_0)+0.5G''(y_0)E_T(y_T-y_0)^2$$

Now apparently $E_T[(y_T-y_0)^2]$ is approximately equal to $\sigma_y^2y_0^2T$, but cannot see why this approximation is true.

## Answer by user30150 (score 5, accepted)

https://quant.stackexchange.com/a/36599

Well, you need to know what is the stochashtic model you are using for $y_T$, if you assume it's a geometric brownian motion you have this process :

$y_T = y_0 e^{\sigma W_T - \frac{1}{2} \sigma^2T} $

If you compute the expectation and variance you get

$ \mathbb{E}(y_T) = y_0$

and

$Var(y_T) = {y_0}^2( e^{\sigma^2 T }-1)$

As $y_0 $ is constant you have $\mathbb{E}((y_T-y_0)^2) = Var(y_T)+(\mathbb{E}(y_T)-y_0)^2$ (using the variance formula)

which gives $\mathbb{E}((y_T-y_0)^2) = y_0^2 (e^{\sigma^2 T}-1)$ if you have $\sigma^2 T$ small you can perfom a taylor's expansion and you'll have the result shown

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