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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 $E[F_{a,b}^X]$

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    $\begingroup$ 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
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[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.

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  • $\begingroup$ 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

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