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, 2016 at 13:01
  • 3
    $\begingroup$ 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, 2016 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, 2018 at 13:27

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.