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8

From $(2)$, \begin{align*} \ln S_t &=\ln F_{t, t} \\ &= \ln F_{0, t}-\frac{1}{2}\int_0^t\sigma^2 e^{-2\lambda (t-s)}ds+\int_0^t \sigma e^{-\lambda(t-s)} dB_s\\ &=\ln F_{0, t}-\frac{\sigma^2}{4\lambda} \left(1-e^{-2\lambda t}\right)+e^{-\lambda t}\int_0^t \sigma e^{\lambda s} dB_s. \end{align*} Then, \begin{align*} \lambda e^{-\lambda t}\int_0^t \...


3

For $0 < T_0\le T$, consider the option with payoff, at the option maturity $T_0$, of the form \begin{align*} \max(F_{T_0, T}-K, \, 0).\tag{1} \end{align*} Note that \begin{align*} F_{T_0, T} &= F_{0, T}\exp\left(-\frac{\sigma^2}{2}\int_0^{T_0} e^{-2\lambda (T-t)} dt+\sigma \int_0^{T_0}e^{-\lambda (T-t)} dB_t\right). \end{align*} Let \begin{align*} \...


3

[Edit] My "answer" below is not a really an answer for I have completely misinterpreted your original question. I thought you asked about the covariance of 2 processes over a given time horizon (i.e. for a fixed $\omega$) and not the covariance of two random variables (fixed $t$). Also note that $\text{cov}(x,y)=0$ does not mean that $x$ and $y$ are ...


3

You can proceed similarly to this question. For $0 < T_0\le T$, consider the option with payoff, at the option maturity $T_0$, of the form \begin{align*} \max(F_{T_0, T}-K, \, 0).\tag{1} \end{align*} Note that \begin{align*} F_{T_0, T} &= F_{0, T}\exp\Bigg(-\frac{1}{2}\int_0^{T_0} \left[\left(h_1e^{-\lambda (T-t)}+h_0\right)^2 + h_2^2e^{-2\lambda (T-...


2

The formula $F^X(t,T) = E_t^d\left(X_T \right)$, under the domestic risk-neutral measure, is problematic. Note that, at time $t$, the forward exchange rate $F^X(t,T)$, for maturity $T$, is the exchange rate such that the payoff $X_T-F^X(t,T)$ has a zero value at $t$. That is, \begin{align*} B_t^d E_d\left(\frac{X_T-F^X(t,T)}{B_T^d} \mid \mathcal{F}_t\right)=...


2

I think you should look at it the other way around. Let $X_t$ denote the FOR/DOM spot exchange rate, i.e. 1 unit of foreign currency = $X_t$ units of domestic currency at time $t$. The FX forward rate $F^X(t,T)$ is defined as $$ F^X(t,T) = X_t \frac{B_f(t,T)}{B_d(t,T)} $$ by basence of arbitrage opportunity. To understand this, consider the following ...


1

Generally, the PCA approach proceeds as follows. Consider historical observation times $t_0 < t_1 < \cdots < t_K \le0$. For bucketing times $\delta_1 < \cdots \delta_n$, let $X_j(t_k) = \ln F(t_k, t_k+\delta_j)$. Moreover, let $\eta_j$, for $j=1, \ldots, n$, be a normal random variable with a sample set $\big\{X_j(t_k)-X_j(t_{k-1}\}_{k=1}^K\big\}...


1

Hint: By Integration, we have $$x_t=x_{0}+\int_{0}^{t} \alpha_1(x_s,s)ds+\int_{0}^{t} \beta_1(x_s,s)dW_1(s)$$ $$y_t=y_{0}+\int_{0}^{t} \alpha_2(y_s,s)ds+\int_{0}^{t} \beta_2(y_s,s)dW_2(s)$$ then $$E[x_t]=x_0+E\left[\int_{0}^{t} \alpha_1(x_s,s)ds\right]$$ $$E[y_t]=y_0+E\left[\int_{0}^{t} \alpha_2(y_s,s)ds\right]$$ Now we apply Ito's lemma $$d(x_ty_t)=...


1

Your approach is correct. But unfortunately it is not applicable. In the case i.e. $\frac{1}{2}<H<1$, Dai and Heyde have defined a stochastic integral as limit of Riemann sums.Their approach does not satisfy the property $E[\int_{0}^{t}f(s)dB_{H}s]=0$ (Why?!!). For this reason, Duncan have introduced a new stochastic integral with zero mean which is ...


1

Another Solution We should look for a solution of the form $$X(t)=U(t)V(t)$$ where $$dU_t=-\theta\,U_tdt+\sigma\,U_t\,dW_t$$ and $$dV_t=\alpha(t)dt+\beta(t)dW_t$$ $U$ is a geometric Brownian motion, therefore $$U(t)=U(0)\,e^{-(\theta+\frac{1}{2}\sigma^2)t+\sigma W_t}$$ let $U(0)=1$, this yields $V(0)=X(0)$. Now we should find $\alpha(t)$ and $\beta(t)$. $...



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