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

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?

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    $\begingroup$ @Richard The future definition is obtained from the expectation that makes the continuous margining zero. Hence, $E_{t-1}^Q\left[\frac{Fut_t-Fut_{t-1}}{B_t}\right]=0$. That is; $E_{t-1}^Q[Fut_t] = Fut_{t-1}$, and by the law of iterated expectations the formula above. The forward is derived from the assumption that $E_{t}^Q\left[\frac{S_t-Fwd}{B_T}\right]=0$. Which is basically the contract's fair price. $\endgroup$ – ZeroCool Mar 21 '17 at 20:31
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    $\begingroup$ @ZeroCool: I think the drift term for $S$ should be $r(t)$. Otherwise, the correlation is not needed. $\endgroup$ – Gordon Mar 21 '17 at 20:48
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    $\begingroup$ @Richard: The forward price paid at time $T$, while known at time $t$ is the amount $K$ such the payoff at time $T$ is $S_T-K$, while value at time $t$ is zero. That is, $E\left(\frac{S_T-K}{B_T} \mid \mathcal{F}_t\right) = 0$. Therefore $K=\frac{E\left(S_T/B_T\mid \mathcal{F}_t\right)}{E\left(1/B_T\mid \mathcal{F}_t\right)}$. $\endgroup$ – Gordon Mar 21 '17 at 20:57
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    $\begingroup$ +1 Gordon. @Richard This is equivalent to writing the Radon Nikodym derivative of the change of measure between the risk-neutral and forward measures. $\endgroup$ – Quantuple Mar 22 '17 at 8:45
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    $\begingroup$ Theoretical convx adj rely on vol and correlation. But mkt prices are subject to supply and demnd. Some cnvxt adj have even been negative (eur STIRS)! Reason being: some market makers trade swaps, FRAs or bonds with one clearhouse (CCP) or CP and then trade on a futures exchange in STIRs. Their net delta is zero across all products but their delta facing one CCP relative to another considerably changes. The execution charge is minimal and the saving on initial margin (due to improved exposure) at each CCP is considerably reduced. Thereby distorting the prices considerably. $\endgroup$ – Attack68 Feb 8 '18 at 7:31
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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.

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