# Tag Info

1

Given the well-known stylised facts of equity markets, I would go for a generic stochastic volatility model where log-asset prices, hence geometric returns, are driven by a standard Brownian motion (although this would explain the lack of returns' auto-correlation, it would also boil down to assuming their independence, which is a stronger assumption). ...

6

By definition, the payoff of a log-contract of maturity $T$ writes $$\phi(S_T) = \ln\left(\frac{S_T}{S_0}\right)$$ Let $\Pi_t$ denote the $t$-value of such a contingent claim. We are interested in the price at $t=0$, best known as the option premium. Theory tells us that the latter premium can be computed as \Pi_0 = e^{-rT} E^{\mathbb{Q}} \left[ ... 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*} ... 0 I presume you talk about local time. I hope it can help you : http://ebooks.cambridge.org/chapter.jsf?bid=CBO9780511662980&cid=CBO9780511662980A011 1 We assume that \begin{align*} \frac{dX_t}{X_t} &= (r+\pi Y_t)dt + \pi\sigma dW_t,\tag{1}\\ dY_t &= -\lambda Y_t + dB_t.\tag{2} \end{align*} From (2), \begin{align*} Y_t = Y_0 e^{-\lambda t}+ e^{-\lambda t}\int_0^t e^{\lambda u} dB_u. \end{align*} Moreover, from (1), \begin{align*} \ln X_T &= \ln X_0 + (r-\frac{1}{2}\pi^2\sigma^2)T + \pi ... 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 \tau on both sides, we obtain that \begin{align*} ... 0 The derivation in Appendix A of the paper Valuation of Equity-Indexed Annuities under Stochastic Interest Rates that you mentioned is Wrong: the Girsanov transformation is applied to an n-dimensional Brownian motion, where the components are independent. However, for the case here with n=2, the Brownian motions are dependent, we can not naively combine ... 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

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

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

I think you are right. The SDE does not attempt to describe the dynamics of the spot exchange rate with respect to random changes in interest rates. Rather, it describes the evolution of the FX rate as a drift term proportional to the rate differential, plus a random term. Specifically, it says that if domestic rates go up, the rate at which the foreign ...

1

The dynamics for the exchange rate $Q$ that converts one unit foreign currency to units of domestic currency is given by \begin{align*} dQ(t) = Q(t)\big[(r_d-r_f)dt + \sigma dW_t \big], \end{align*} where $r_d$ and $r_f$ are, respectively, the domestic and foreign interest rates. In your example, the exchange rate EUR/USD is to convert on unit EUR to ...

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. $$dM_t = \frac{1}{Y_t} dX_t - \frac{-X_t}{Y_t^2} dY_t + \frac{-1}{Y_t^2} dX_t dY_t + \frac{X_t}{Y_t^2} dY_t \quad / : \frac{Y_t}{X_t} \quad \quad (6)$$ $$\frac{dM_t}{M_t} = \frac{dX_t}{X_t} - \frac{dY_t}{Y_t} - \frac{dX_t dY_t}{X_t Y_t} + \frac{(dY_t)^2}{(Y_t)^2} \quad \quad \quad ... 0 A few years ago I asked a similar question on MO: http://mathoverflow.net/questions/22828/big-picture-concerning-ito-integral-stratonovich-integral-and-standard-results My take today is that you really don't need this heavy mathematical machinery for standard BS but as soon as you move on to more sophisticated (and realistic) models you surely do, so it is ... 2 What is written in attached slides is correct. However, what you have written is not correct. Setting M_t=\frac{X_t}{Y_t}, and applying Ito formula will lead to :$$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 - ...

Top 50 recent answers are included