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6

The dynamics \begin{align*} \frac{dS_t}{S_t} =\mu dt + \sigma dW_t. \end{align*} is under the real-world measure $\mathbb{P}$. Then, \begin{align*} d\ln S_t =\Big(\mu-\frac{1}{2}\sigma^2 \Big) dt + \sigma dW_t. \end{align*} Therefore, \begin{align*} \ln S_T = \ln S_t + \Big(\mu-\frac{1}{2}\sigma^2 \Big)(T-t) + \sigma \big(W_T-W_t\big).\tag{1} \end{align*} ...


4

In general $dX^2$ is an ad-hoc or heuristic form of $d\langle X, X\rangle_t$, where $\langle X, X\rangle_t$ is the quadratic variation, which is defined by \begin{align*} \langle X, X\rangle_t = \lim_{\pi\rightarrow 0} \sum_{i=1}^n (X_{t_i}-X_{t_{i-1}})^2. \end{align*} Here, $0=t_0 < \cdots < t_n = t$, and $\pi = \max\{ t_i-t_{i-1}, i=1,\ldots, n\}$. ...


3

We assume that $\gamma(s, t)$ is differentiable with respect to $t$. Then, \begin{align*} dx_t = \left(\int_0^t \frac{\partial\gamma(s, t)}{\partial t} dW_s \right)dt + \gamma(t, t) dW_t. \end{align*}


3

Assume deterministic and constant interest rates. For an investor in the foreign economy i.e. a market participant that can only trade assets delivering a payout in the foreign currency, let us define $$ \tilde{X}_t = \tilde{X}_0 \exp \left(\left(r_f-r_d-\frac{\sigma_\tilde{X}^2}{2}\right)+\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right) $$ $$ Y_t ...


3

Your problem probably comes from the notations used. Let the Moment Generating Function (MGF) of a random variable $X$ be defined as $$ M_X(u) := E[e^{uX}] $$ From this definition, it entails that $$ E(X^n) = M_X^{(n)}(u=0) = \frac{d^{n} M_X}{ d u^{n}}(u=0) $$ Knowing this, the function $$ f_{\lambda}(t,r)=E[e^{-\lambda {r_{T}}}|r_t=r] $$ can be ...


3

Here's my 2 cents: a) Conditional expectations can always be seen as martingales (this is a direct consequence of the tower property). Thus, we here have that $$ M_t := E^*[e^{-\lambda {r_{T}}}|r_t] $$ is a martingale. Applying Itô's lemma to $M_t = f_{\lambda}(t,r_t)$ as you did is a good starting point. But doing this, leaves you with an SDE, not a ...


3

Let $$ f_{\lambda}(t,r)=E^{(t,r)}\left[e^{-\lambda r_{T}}\right] $$ where $E^{(t,r)}$ denotes the expectation conditional on $r_{t}=r$. We assume $f$ is smooth for the remainder. Let $\theta=T\wedge\inf\left\{ s>t\colon\left|r_{s}-r\right|>1\right\} $. By the Markov property of $\{r_{t}\}$, $$ ...


2

I don't think just knowing $dW_t\,dW_t=dt$ is enough.We assume $$d{{X}_{t}}=\mu (t,{{X}_{t}})dt+\sigma (t,{{X}_{t}})d{{W}_{t}}\,\,\,\,\,(1)$$ The idea behind the Milstein scheme is that the accuracy of the discretization can be increased by expanding the coefficients$\mu =\mu (t,{{X}_{t}})$ and $\sigma =\sigma (t,{{X}_{t}})$ via Ito’s lemma. This is sensible ...


2

Based on Cholesky decomposition, \begin{align*} W_t^A &= W_t^1,\\ W_t^B &= \rho W_t^1 + \sqrt{1-\rho^2}W_t^2, \end{align*} where $(W_t^1, t \ge 0)$ and $(W_t^2, t \ge 0)$ are two independent standard Brownian motions. Then \begin{align*} A_t &= A_0\exp\Big(\big(a-\frac{1}{2}\sigma_A^2\big)t + \sigma_A W_t^1 \Big),\\ B_t &= ...


2

From Equation (6), $B(t,T)=-t+c(T)$ for some function $c(T)$. $1=P(t,t)=e^{-A(t,t)-(c(t)-t)r_t}$ or $A(t,t)+(c(t)-t)r_t=0,\,\forall (r_t,t)$. So $c(t)=t, A(t,t)=0,\forall t$. For Equation (8) you have missed the square on $\sigma$ and a factor of $\frac13$. Then you just need to substitute in the function for $b(s)$ and integrate the following to get the ...


2

[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 ...


2

For starters, the short rate model you mention in equation (1) is Cox-Ingersoll-Ross while the bond price in equations (2)-(4) correspond to the Vacisek model. So there is a problem somewhere, I would go for a typo in (1). Second, what you wrote seems fine to me, so there must definitely be yet another typo in your solution manual. Note that if there is no ...


2

Apply Ito's lemma to $\ln M_t$, we obtain that \begin{align*} d\ln M_t &= \frac{1}{M_t} dM_t -\frac{1}{2} \frac{1}{M_t^2} d\langle M, M\rangle_t\\ &=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} \frac{1}{M_t^2}\left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right)M_t^2dt\\ &=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} ...


2

You almost get there. However, you ca not conclude that $\rho^2$ is a constant based on $(10)$. Note that, from your $(7)$ and $(8)$, \begin{align*} \frac{\rho(z_t)^2}{\beta} e^{\beta \tau} (e^{\beta \tau} - 1) = -h'(\tau)+e^{\beta \tau}h'(0). \end{align*} Taking derivative with respect to $\tau$ on both sides, we obtain that \begin{align*} ...


2

What is written in attached slides is correct. However, what you have written is not correct. Setting $M_t=\frac{X_t}{Y_t}$, and applying Ito formula will lead to : $$dM_t=\frac{dX_t}{X_t} M_t -\frac{dY_t}{Y_t} M_t + M_t \frac{d<Y>_t}{Y^2_t}-\frac{d<X,Y>_t}{Y^2_t}$$ which gives you in your case : $$dM_t = (\mu_x dt+\sigma_x dZ^1_t)M_t - ...


1

Applying Itô's lemma to the Black-Scholes SDE and integrating from $t$ to $t+\Delta t$ gives: $$ S_{t+\Delta t} = S_t e^{(r-\frac{1}{2}\sigma^2)\Delta t + \sigma \sqrt{\Delta t}Z} $$ with $Z \sim N(0,1)$, showing that $S_{t+\Delta t}$ given $S_t$ is log-normally distributed. It is then straightforward to write, for any compact $\mathcal{A} = [a_1,a_2]$ ...


1

Note that $${{f}_{W(t)\left| W(s) \right.}}\left(x\left| y \right. \right)=\frac{{{f}_{ W(s),W(t)}}\left( x,y \right)}{{{f}_{ W(s)}}\left( y \right)}=\frac{1}{\sqrt{2\pi(t-s)}}\exp \left[-\frac{{{(x-y)}^{2}}}{2(t-s)} \right]$$ By application of Ito's lemma we have $$ln\,S_{t+\Delta t}=ln\,S_t+\left((\mu-\frac{1}{2}\sigma^2)\Delta t+\sigma(W_{t+\Delta ...


1

Royden's "Real Analysis" is a standard textbook for a first year grad course in Real Analysis. It covers all the integration topics nicely. A more stochastically oriented book would be "Probability with Martingales" by Williams, which covers integration as well.


1

I assume that the problem is $$\max_{\pi} E\left(\ln Z_T^{\Pi} \right).$$ Note that $\ln Z_t^{\Pi} = \ln X_t^{\Pi} -\ln X_t^{\rho}$. Moreover, \begin{align*} d\ln Z_t^{\Pi} &= d\ln X_t^{\Pi} -d\ln X_t^{\rho}\\ &=\Big[\big(\mu \pi - \frac{1}{2}\sigma^2 \pi^2\big) - \big(\mu \rho- \frac{1}{2}\sigma^2 \rho^2\big) \Big]dt + \sigma(\pi-\rho)dW_t. ...


1

Just use the fact that $$ \sigma_X W_t^1 + \sigma_Y W_t^2 = \sqrt{ \sigma_X^2 + \sigma_Y^2 + 2\rho\sigma_X\sigma_Y } W_t $$ holds in probability assuming that $W_t^1$ and $W_t^2$ are 2 correlated Brownian motions with $$ d\langle W_t^1, W_t^2 \rangle_t = \rho dt $$ and $W_t$ is a new standard Brownian motion defined over the same probability space. Simply ...


1

Equations (1) to (3) are correct. Your investment strategy is then, $\forall t > 0$ $$ X_t = \theta _ t S_t $$ Provided you use this strategy as part of self-financing portfolio you can write the P&L over an infinitesimal time interval as $$ dV_t = \theta_ t dS_t $$ assuming zero safe rate, i.e. that any cash required to finance your long stock ...



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