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


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.

  • $\begingroup$ Last expression = 0, right? How is the last sentence "Now apparently..." connected to the rest of the post? $\endgroup$ – Mats Lind Aug 16 '16 at 13:01
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    $\begingroup$ This is from John Hull's book. It is likely based on the assumption that $y_t$ is log-normal and that $\sigma^2_y T$ is small. $\endgroup$ – Gordon Aug 16 '16 at 13:39
  • $\begingroup$ Just to clarify something, saying E(y_T)=y_0 is a little confusing. The point is to calculate an expression for precisely the term E(y_T). Is it the case that the term (E(y_T)-y_0)^2 drops off not because E(y_T)=y_0 but because the square of the difference (i.e. the convexity) is negligible? $\endgroup$ – questioner Sep 29 '18 at 13:27

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$


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

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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