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Quuestion1.

I make a solution $r(t)$ used by Ito's lemma

$r(t)=e^{-a t}r(0)+\int _{0}^{t}e^{a (s-t)}\theta (s)ds+\sigma e^{-a t}\int _{0}^{t}e^{a u}\,dB^{1}(u)$

Is this right?

and I try to make solution of $P(t)$ and $X(t)$.But because of my lack of understand, I couldn't solve that.

Question2.

I cannot understand problem not at all. what is mean of $\frac{Z_T}{\beta(T)}|F_t$ and $\frac{W_T}{\beta(T)} | F_t$

Question 1.

Let $(Ω, F, P)$ be a complete probability space, and let $B (t) = (B^1 (t), B^2 (t))$, $t \in [0, T]$ be a two-dimensional standard Brownian motion. Let $(F_t)_{ t \in [0, T]}$ be a fit that satisfies the standard condition that $B (t)$ generates. A US dollar-denominated, default-free zero-coupon bond that pays a repayment of US \$ 1 at maturity $T$, but gives the price $P (t)$ at time $t \in [0, T]$ as the solution to the next stochastic differential equation It is assumed that

$dP(t)=(r(t)+\lambda b(t))P(t)dt - b(t)P(t)dB^1(t), $

$P(T)=1$

Where $b (t)$ is a deterministic positive value function of time $t, r (t)$ is a short rate of US dollars, and the following Hull-White model is used.

$dr(t)= (\theta(t)-a\cdot r(t))dt +\sigma \cdot dB^1(t),$

$ r(0)>0$,Constant

$\theta (t)$ is a deterministic function of time$ t$, and $a, \sigma$ are positive value constants. The spot exchange rate $X (t)$ for the US $ -Yen is assumed to be the solution of the following stochastic differential equation.

$dX (t)= \mu Χ(t)dt + \sigma_1 X(t)dB^1(t) + \sigma_2 Χ(t)dB^2(t),$

$ X (0 ) =Χ_0 > 0.$

Here, $\mu$, $\sigma_1$, and $\sigma_2$ are positive constants. Use the Ito formula and answer the following. Please also explain the calculation process.

(1) Suppose that a Japanese investor has invested in the top zero coupon bond. This investor recognizes the market value on a yen basis. Calculate $dS (t)$ for the value $S (t) = X (t) P (t)$ for this investor. (Please express the last equation by $dt, dB^1 (t), dB^2 (t)$.)

Question 2.

Let $(Ω, F, P)$ be a complete probability space, and let $B (t) = (B^1 (t), B^2 (t))$,$ t ∈ [0, T] $be a two-dimensional standard Brownian motion. The filtration generated by this Brownian motion and satisfying the standard condition is $(F_t)_{ t \in [0, T]}$. Suppose$ F = F_T$. Suppose that the prices $\beta(t), X (t) $ and $ Y (t)$ of each of the deposit, stock $X$ and stock $Y$ are given by the solution of the following stochastic differential equation.

$d\beta(t) =r \beta(t)dt,\beta(0)=1.$

$dX(t)=\mu_x X(t)dt+\sigma_x X(t)dB^1(t),X(0)=x>0.$

$dY(t)=\mu_y Y(t)dt+\sigma_y Y(t)dB^2(t),X(0)=y>0.$

The parameters are $r, x, \mu_X, \sigma_X, y, \mu_Y, \sigma_Y> 0$. Let $Q$ be the equivalent Martingale measure with $\beta (t)$ as the numeraire. Let $Q-(F_t)-$standard Brown's motion be $\hat{B} (t) = (\hat{B}^1 (t), \hat{B}^2 (t))$.

Answer the following questions. Please explain the calculation process.

(1)Payoff at time $T$

$W_T:=\sqrt{X(T)}$

The price at time t of the derivative that gives

$\Pi (t)=\beta (t) E^Q [\frac{W_T}{\beta(T)} | F_t]$

Find the ratio of $\sqrt{X(t)}$,$\Pi(t)/\sqrt{X(t)}$.

(2)Payoff at time $T$

$Z_T := X(T) \cdot Y(T)$

The price at time $t$ of the derivative that gives

$\Phi (t) =\beta(t)E^Q[\frac{Z_T}{\beta(T)}|F_t]$

Find the ratio of $Z(t)= X (t) \cdot Y(t)$ , $\Phi(t)/Z(t)$ .

Cautions 1. The settings for this problem generally do not have a ratio of 1.

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  • $\begingroup$ Do you really need to solve for $P_t$ and $X_t$ in the first question? It only asks you to compute $dS_t = d(X_t P_t)$ which is a simple application of Ito's lemma. $\endgroup$ – Freelunch Jun 14 at 7:56
  • $\begingroup$ you mean i don't need to know $S(t)$ to compudte $dS(t)$?? $\endgroup$ – ddss321 Jun 14 at 8:01
  • $\begingroup$ No, using Ito's lemma we have $dS_t = dX_t P_t + X_t dP_t + dX_t dP_t$. Assuming $B^1_t$ and $B^2_t$ are uncorrelated the $dB^1_tdB^2_t$ terms can be set to zero. I also don't see the need for an explicit expression for $r_t$. $\endgroup$ – Freelunch Jun 14 at 8:35
  • $\begingroup$ Thnks for your help. I think that the answer of problem 1. is that $dS(t)=(\mu+r(t)+\lambda b(t))S(t)dt +$$ (\sigma_{1}-b(t))S(t)dB^{1}(t)+\sigma_{2} S(t)dB^{2}(t)$. $\endgroup$ – ddss321 Jun 14 at 10:47
  • $\begingroup$ Looks like you forgot the cross term, $dX_tdP_t = -b(t) \sigma_1 S_t dt$. $\endgroup$ – Freelunch Jun 14 at 12:03

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