stochastic calculus - "Expectation" of a FX Forward

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

Answer

[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.

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[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

$$\begin{align} F^X(t_1,t_2) &= E\left[ X_{t_2} \vert \mathcal{F}_{t_1} \right] \\ &= X_{t_1} \exp \left( (r_d - r_f)(t_2 - t_1) \right) \end{align}$$

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[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.