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## New answers tagged stochastic-calculus

1

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 $\pi = \max\{ t_i-t_{i-1}, i=1,\ldots, n\}$. ...

1

There must be many ways to answer the first question but I would like to point out the following. When the limit $\Delta t \to 0$ is not taken then $\Delta W$ is a normal distribution $\mathcal N(0,\sqrt{\Delta})$. This implies $\Delta W^2 - \Delta t$ is $\chi^2$ distribution with one degree of freedom with mean $0$ and standard deviation $\sqrt{2} \Delta ... 1 Royden's "Real Analysis" is a standard textbook for a first year grad course in Real Analysis. It covers all the integration topics nicely. A more stochastically oriented book would be "Probability with Martingales" by Williams, which covers integration as well. 3 We assume that$\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*} 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*} ... 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. ... 0 To solve this equation, let \begin{align*} M_t = e^{(\theta + \frac{1}{2}\sigma^2 ) t - \sigma W_t}. \end{align*} Then \begin{align*} dM_t = M_t\Big[\big(\theta +\sigma^2\big) dt - \sigma dW_t\Big]. \end{align*} Moreover, \begin{align*} d(M_t X_t) &= M_t dX_t + X_t dM_t + d\langle M, X \rangle_t\\ &=\theta\,\mu\, M_t dt. \end{align*} Then, ... 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\tauon both sides, we obtain that \begin{align*} ... 3 Your problem probably comes from the notations used. Let the Moment Generating Function (MGF) of a random variableXbe 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 ... 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 &= ... 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 toM_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 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) $$is a stochastic exponential (strictly positive martingale ... 2 For starters, the short rate model you mention in equation (1) is Cox-Ingersoll-Ross while the bond price in equations (2)-(4) correspond to the Vacisek model. So there is a problem somewhere, I would go for a typo in (1). Second, what you wrote seems fine to me, so there must definitely be yet another typo in your solution manual. Note that if there is no ... 3 Assume deterministic and constant interest rates. For an investor in the foreign economy i.e. a market participant that can only trade assets delivering a payout in the foreign currency, let us define$$ \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 ... 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 thatW_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 ... 0 I found the problem ,the partial derivatives were incorrectly derived. $$dM_t = \frac{1}{Y_t} dX_t - \frac{-X_t}{Y_t^2} dY_t + \frac{-1}{Y_t^2} dX_t dY_t + \frac{X_t}{Y_t^2} dY_t \quad / : \frac{Y_t}{X_t} \quad \quad (6)$$ $$\frac{dM_t}{M_t} = \frac{dX_t}{X_t} - \frac{dY_t}{Y_t} - \frac{dX_t dY_t}{X_t Y_t} + \frac{(dY_t)^2}{(Y_t)^2} \quad \quad \quad ... 2 What is written in attached slides is correct. However, what you have written is not correct. Setting M_t=\frac{X_t}{Y_t}, and applying Ito formula will lead to :$$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 - ... 2 Just to add an intuitive argument to @MJ73550's already very nice answer: When holding an American option - or any option callable by the holder for that matter -, the question you ask yourself before exercising it is whether the proceeds from early exercise (i.e. exercise now to get the option's intrinsic value) are greater than what you could expect to ... 3 I would not say there is no link to what you say but here would be my view. Intuitive explanation If you wait for a delay$h$before exercising, you lose your exercise right between$t$and$t+h$, this leads to a loss in value. Supermartingale property proof (to apply it in your case :$\phi_t=e^{-rt}(L-S_t)^+$) If we denote$\phi$the obstacle, and ... 0 Ok, so I have been thinking about it, and may have found the solution, but please correct me if I'm wrong. I guess the discounted process goes down, because when the holder of the option doesn't exercise it, as long as the price$S(t)$is less than the optimal exercise price$L^*$he's loosing cash from not investing into money market? 1 Let$Y_t := 2 S_t^1 S_t^2 $. Applying (multivariate) Itô to the function$f(t,S_t^1,S_t^2)=2 S_t^1 S_t^2$yields a stochastic differential equation for$Y_t$$$\frac{dY_t}{Y_t} = \frac{dS_t^1}{S_t^1} + \frac{dS_t^2}{S_t^2} + \rho \sigma_1 \sigma_2 dt$$ Re-applying Itô's lemma to the function$f(t,Y_t) = \ln(Y_t)\$ then yields $$d\ln Y_t = (\mu_1 + \mu_2 ... 1 Applying Itô's lemma to$$ Y_t := e^{\int_0^t b(v) dv} r_t You get \begin{align} dY_t &= b(t) e^{\int_0^t b(v) dv} r_t dt + e^{\int_0^t b(v) dv} dr_t + 0\\ &= e^{\int_0^t b(v) dv} (b(t) r_t dt + dr_t) \\ &= e^{\int_0^t b(v) dv} (a(t) dt + \sigma(t) dW_t) \end{align} where the last line is obtained by using the fact that dr_t = ...

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What can be shown is that the above expressions are equal in probability. First check the distribution. As any linear combination of a Gaussian is Gaussian the right hand side is Gaussian - the left hand side too. Then we need the 2 moments: The expected values - it is zero ... easy to see. Next what you did not specify is that the correlation between ...

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