forward - Convexity Adjustment for Futures

Let $B_t$ be the cash account numeraire. The future and forward prices at time t are expressed as:

$$Fut = E_t^Q\left[S_T\right],$$ $$Fwd = \frac{E_t^Q[S_T/B_T]}{E_t^Q[1/B_T]}.$$

Where

$$\frac{dS(t)}{S(t)} = \mu dt + \sigma dW_s^Q(t),$$ $$dr(t) = -Kr(t)dt+ \alpha dW_r^Q(t),$$ $$ = \rho dt.$$

Where $K$ is the mean reversion of the short interest rate $r$.

How is the convexity adjustment calculated in order to express the forward price in terms of the future price?

Answer

We assume that, under the probability measure $Q$,

$$\begin{align*} dS_t &= S_t\big(r_t dt + \sigma dW_s(t)\big),\\ dr_t &= -k\, r_t dt + \alpha dW_r(t),\tag{1} \end{align*}$$ where $d\langle W_s(t), W_r(t)\rangle_t = \rho dt$. From $(1)$, for $s\ge t$, $$\begin{align*} r_s = e^{-k(s-t)}r_t + \alpha\int_t^s e^{-k(s-u)} dW_r(u). \end{align*}$$ Then, for $T\ge t$, $$\begin{align*} \int_t^T r_s ds &=\frac{r_t}{k}\left(1-e^{-k(T-t)} \right)+\alpha \int_t^T\!\!\!\int_t^s e^{-k(s-u)} dW_r(u) ds\\ &=\frac{r_t}{k}\left(1-e^{-k(T-t)} \right)+\alpha \int_t^T\!\!\!\int_u^T e^{-k(s-u)} ds dW_r(u) \\ &=\frac{r_t}{k}\left(1-e^{-k(T-t)} \right)+\alpha \int_t^T\frac{1}{k}\left(1-e^{-k(T-u)} \right) dW_r(u)\\ &=r_t\beta(t, T)+\alpha \int_t^T \beta(u, T) dW_r(u), \end{align*}$$ where $$\beta(t, T)=\frac{1}{k}\left(1-e^{-k(T-t)} \right).$$ Therefore, $$\begin{align*} E^Q\left(\frac{1}{B_T} \mid \mathcal{F}_t\right) &=\frac{1}{B_t}E^Q\left(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t \right)\\ &=\frac{1}{B_t} e^{-r_t\beta(t, T) + \frac{\alpha^2}{2} \int_t^T \beta^2(u, T) du}. \end{align*}$$ Moreover, $$\begin{align*} E^Q\left(S_T \mid \mathcal{F}_t\right) &= S_t E^Q\left(e^{\int_t^T r_s ds - \frac{\sigma^2}{2} (T-t) + \sigma \int_t^T dW_s(u)} \right)\\ &=S_t E^Q\left(e^{r_t\beta(t, T)+\alpha \int_t^T \beta(u, T) dW_r(u) - \frac{\sigma^2}{2} (T-t) + \sigma \int_t^T dW_s(u)} \right)\\ &=S_te^{r_t\beta(t, T)+ \frac{\alpha^2}{2} \int_t^T \beta^2(u, T) du +\alpha \sigma \rho \int_t^T\beta(u, T) du}. \end{align*}$$ Consequently, $$\begin{align*} C(t, T) &= \frac{Fut}{Fwd}\\ &=\frac{E^Q\left(S_T \mid \mathcal{F}_t\right)}{E\left(\frac{S_T}{B_T} \mid \mathcal{F}_t\right)/E^Q\left(\frac{1}{B_T} \mid \mathcal{F}_t\right)}\\ &=\frac{S_te^{r_t\beta(t, T)+ \frac{\alpha^2}{2} \int_t^T \beta^2(u, T) du +\alpha \sigma \rho \int_t^T\beta(u, T) du}}{\frac{S_t}{B_t} B_t e^{r_t\beta(t, T) - \frac{\alpha^2}{2} \int_t^T \beta^2(u, T) du}}\\ &=e^{\alpha^2\int_t^T \beta^2(u, T) du +\alpha \sigma \rho \int_t^T\beta(u, T) du}. \end{align*}$$

Don't forget the 1/2 in normal variable's characteristic function.