I have an FX process $X_t = X_0 \exp((r_d-r_f)t+ \sigma W_t)$. Now clearly $E[X_t] = F_{0,t}^X$. i.e. a forward contract of the process $X$ starting at time 0 and maturing at time $t$.

What if I want to look at a forward contract at a later date. i.e. $F_{a,b}^X$. Where $0 < a<b<t$. How would I be able to rewrite this in terms of current time? Surely it's not the value $F_{0,b}^X - F_{0,a}^X$. Put it another way, how do I determine $E[F_{a,b}^X]$

  • 1
    $\begingroup$ I don't understand the question. (1) as you define it $E[X_t \vert \mathcal{F}_0] \ne F^X(0,t)$, at least not if $F^X(0,t)=X_0\exp((r_d-r_f)t)$, you miss the $\sigma^2/2t$ term which should compensate for the exponential of $\sigma W_t$ (i.e. making it a martingale with constant mean 1); (2) do you simply want to compute $F^X(t_1,t_2) = E[X_{t_2} \vert \mathcal{F}_{t_1}]$ ? $\endgroup$ – Quantuple May 4 '16 at 12:23
  • $\begingroup$ (2) Yes I do want this term. (1) Why is this incorrect. Isn't the expected asset price equal to its forward? $\endgroup$ – Jim May 4 '16 at 12:36

[Question 1]

Let us define \begin{align} X_t &= X_0 \exp((r_d-r_f-\frac{1}{2}\sigma^2)t + \sigma W_t) \\ &= X_0 \exp((r_d-r_f)t) \mathcal{E}(\sigma W_t) \end{align} then, in that case $$ E(X_t \vert \mathcal{F}_0) = X_0 \exp((r_d-r_f)t) = F^X(0,t) $$ only because $$ \mathcal{E}(\sigma W_t) $$ is a stochastic exponential (strictly positive martingale with mean 1).

Yet $E(X_t \vert \mathcal{F}_0) \ne F^X(0,t)$ for $ X_t = X_0 \exp((r_d-r_f)t + \sigma W_t) $ as you define it in your question.

[Question 2]

Under the domestic risk-neutral measure, the dynamics of the FOR/DOM (1 unit of foreign currency expressed in domestic currency) exchange rate $X_t$ should write (to preclude arbitrage opportunities) $$ \frac{dX_t}{X_t} = (r_d - r_f)dt + \sigma W_t $$ Applying Itô, to the function $f(t,X_t)=\ln(X_t)$ gives $$ d\ln(X_t) = (r_d - r_f - \frac{1}{2}\sigma^2)dt + \sigma W_t $$ which one can easily integrate e.g. from $t_1$ to $t_2$ (assuming $0 < t_1 < t_2 < T$) to obtain $$ \ln(X_{t_2}) - \ln(X_{t_1}) = (r_d - r_f - \frac{1}{2}\sigma^2)(t_2-t_1) + \sigma (W_{t_2}-W_{t_1}) $$ or equivalently $$ X_{t_2} = X_{t_1} \exp \left( (r_d - r_f - \frac{1}{2}\sigma^2)(t_2-t_1) + \sigma (W_{t_2}-W_{t_1}) \right) $$

From the above the forward FOR/DOM exchange rate at $t_1$ with maturity $t_2$ computes as \begin{align} F^X(t_1,t_2) &= E\left[ X_{t_2} \vert \mathcal{F}_{t_1} \right] \\ &= X_{t_1} \exp \left( (r_d - r_f)(t_2 - t_1) \right) \end{align}

[Edit] \begin{align} E_0 \left[ F^X(t_1,t_2) \right] &= E_0 \left[ X_{t_1} \exp \left( (r_d - r_f)(t_2 - t_1) \right) \right] \\ &= E_0 \left[ X_{t_1} \right] \exp \left( (r_d - r_f)(t_2 - t_1) \right) \\ &= F(0,t_1) \exp \left( (r_d - r_f)(t_2 - t_1) \right) \\ &= X_0 \exp \left( (r_d - r_f) t_1 \right) \exp \left( (r_d - r_f)(t_2 - t_1) \right) \\ &= F(0,t_2) \end{align}

this is only normal since $$ E [ X_{t_2} \vert \mathcal{F}_0 ] = F(0,t_2) = E[ E[ X_{t_2} \vert \mathcal{F}_{t_1}] \vert \mathcal{F}_0] $$ by the tower integral property.

  • $\begingroup$ Again, great response. So does that mean there is no way to express $F^X (t_1,t_2)$ in terms of something in the form $F^X(0, ...)$ $\endgroup$ – Jim May 4 '16 at 13:40
  • $\begingroup$ You're welcome, don't hesitate to mark these answers as accepted instead of just upvoting (so that the topic can be closed). No offense but I think you're not sure about what you want to compute exactly. $F^X(t_1,t_2)$ is the forward rate at $t_1$ for delivery at $t_2$... how could it possiblity depend on $t_0$ since this is in the past and we are dealing with Markovian models. $\endgroup$ – Quantuple May 4 '16 at 13:47
  • $\begingroup$ Maybe it is not $F^X(t_1,t_2)$ that you need but something else. $\endgroup$ – Quantuple May 4 '16 at 13:48
  • $\begingroup$ Well the main thing I want to compute is $E(F^X(t_1,t_2))$ $\endgroup$ – Jim May 4 '16 at 14:20
  • $\begingroup$ Well this is simply $F^X(0,t_2)$ the forward value as seen of today... did you expect something else? I edited my answer to show this. $\endgroup$ – Quantuple May 4 '16 at 14:29

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.