# Tag Info

11

My understanding is because the Ito's integration definition keeps the martingale property. With Brownian motion $W(t, \omega)$ defined, to define stochastic integration in a Riemann–Stieltjes style: $$\int_0^t f(t, \omega) d W(t, \omega) = \lim_{\| \Delta_n\| \to 0 } \sum_{i=1}^{n} f(\tau_i,\omega) \left ( W(t_i, \omega) - W(t_{i-1}, \omega) \right )$$ , ...

9

In fact Ito and Stratonovich calculus are both mathematically equivalent. In the following paper you can e.g. see that both derivations lead to the same result, i.e. the Black-Scholes equation: Black-Scholes option pricing within Ito and Stratonovich conventions by J. Perello, J. M. Porra, M. Montero and J. Masoliver From the abstract: Options ...

8

I can clarify 100% that $(dw)^2$= $dt$ and recommend you to accept it as a fact. Like any other differential, this differential is defined in terms of its integral: $$\int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}[W(t_{k+1})-W(t_{k})]^{2}$$ Where $t_{k}=t_{0}+k(t_{1}-t_{0})/n$. Since $$... 7 These are all examples on Ito Formula in its general form (with quadratic variations): 6 We know that (\tilde{W}_t) := (-W_t) is also a Wiener process so$$ E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r] $$and that implies that E[W_pW_qW_r] = 0. 5 If by 'solve' you mean how do we know that \ln S_t is the right change of variable, then you can go by the following (not rigorous) line of thought: Ito's fomula suggests that given an SDE$$dX_t = \mu(X_t,t)dt+\sigma(X_t,t)dW_t$$and a function f(x,t): the SDE for the process Y_t=f(X_t,t) will satisfy$$dY_t = [f_t(X_t,t) + f_x(X_t,t)\mu(X_t,t) + ...

5

The convexity of the exponential function of the stochastic variable $W$ makes its expectation greater than the exponentiation of the expectation of $W$. This is an example of Jensen's inequality, $E[e^{\sigma W}]> e^{\sigma E[W]}=1$. $\sigma$ can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ...

5

Suppose that there are multiple martingale measures $Q_1$ and $Q_2$ that attain the minimal variance. Then the convex combination $Q_* := \frac{1}{2}Q_1 + \frac{1}{2}Q_2$ is also a martingale measure. Due to the strict convexity of $f(x) = x^2$, it can be shown that $$E_P \left[\frac{dQ_*}{dP}^2 \right] < \frac{1}{2} E_P \left[ \frac{dQ_1}{dP}^2 ... 5$$ \textbf{Preface} $$I am assuming log normal asset but this is not clear from the question? Or rather I have misinterpreted the question! Well as I see it from a a purely mathematical exercise$$ d\left(\dfrac{S_t}{M_t}\right) =\frac{1}{M_t}dS_t - \frac{S_t}{M_t^2}dM_t +O(dt^2) $$using Ito's lemma. Then we can sub in the original processes yields ... 5 The second theorem called "Girsanov II" is indeed a special case of the general "Girsanov I" from above with$$Y_t=W_t,X_t=-\int_0^t\Theta_udW_u$$. We can show that$$[Y,X]=-\int_0^t\Theta_udu$$using general Stochastic Calculus rules (e.g. see p.37, 6.6 here):$$[Y,X]=[W_t,-\int_0^t\Theta_udW_u]=-\int_0^t\Theta_ud[W_u,W_u]=-\int_0^t\Theta_udu$$since ... 5 The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows:$$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]where you'll need to find C such that the equality holds (i.e. C=W_s^3 + ... 4 I would calculate it this way, \mathbb{E}[(W_s+W_t−2W_0)^2] = \mathbb{E}\left[\left((W_s-W_0)+(W_t-W_0)\right)^2\right]\\ \hspace{4cm}=\mathbb{E}[(W_s-W_0)^2]+\mathbb{E}[(W_t-W_0)^2]+2\mathbb{E}[(W_s-W_0)(W_t-W_0)] \\ \hspace{4cm}=s+t+2\mathbb{E}[W_sW_t]\\ \hspace{4cm}=s+t+2\min(s,t) 4 If the loss distribution is normal with mean \mu and variance \sigma^2, then the Value-at-Risk and Expexted Shortfall (or CVaR) at level \alpha \in (0, 1) are \begin{align*} \mbox{VaR}_\alpha & = \mu + \sigma \Phi^{-1}(\alpha) , \\ \mbox{ES}_\alpha & = \mu + \sigma \frac{\phi\{\Phi^{-1}(\alpha)\}}{1 - \alpha} , \end{align*} where \phi ... 4 You derivation here is flawed because you are deriving with respect to two processes and you do not take into account that the variable W_t is stochastic and hence S_t is as well. So, to derive S_t from dS_t, you have to apply Ito's Lemma, see this question for details. This is the "classic" way you see it. If you want to do it the other way ... 3 Consider an (arithmetic) Ornstein-Uhlenbeck process as a model of the asset price X_t:dX_t = \kappa(\mu-S_t)dt + \sigma dW_t$$where \mu is the mean-reversion level, \sigma is a volatility parameter, W_t is Brownian motion, and \kappa is the reversion speed. An Ornstein-Uhlenbeck process will revert to the mean infinitely often if \kappa ... 3 For Q1, the function a(t) is the instantaneous correlation. The form given by (2) is basically the Cholesky decomposition. Of course, you may directly show, uisng Levy's characterization, that$$ \widetilde{W}(t) = \int_0^t\bigg[\frac{1}{\sqrt{1-||a(t)||^2}} dZ(t) -\frac{a(t)^T}{\sqrt{1-||a(t)||^2}} dW^B(t) \bigg] $$is a standard scalar Brownian motion ... 3 Q1:$$(1)\rightarrow(2)$$(1): a(t) is the instantaneous correlation of \rho(Z_t,W_t) because:$$\rho(dZ_t,dW_t)=\dfrac{Cov(dZ_t,dW_t)}{\sigma_{dZ_t}\sigma_{dW_t}}=\dfrac{E(dZ_t\cdot dW_t)}{\sqrt{dt} \sqrt{dt}}=\dfrac{\langle dZ_t, dW_t\rangle}{t}=a(t)$$\Rightarrow (2) holds as following, in the 1-dim case: dZ_t\sim N(0,dt), ... 3 About the integration problem: Your integrand is highly oscillatory, and the adaptive quadrature of Matlab doesn't handle such integrands very well. In general, I would recommend Mathematica when Matlab's standard procedures don't perform well. In this case, a Levin-type method would perform much better. The reason that quadv produces NaN values is because ... 3 If you allow X_t to be two dimensional then a model with a stock price X_t^1 and its variance process X_t^2 (stochastic volatility) would fit your definition. In such cases to my knowledge we often don't have a closed form of the density of X_T^1 but in some cases we have a closed form of the Laplace transform. An example is the Heston model. 3 This is a good shorter reference: http://www.impan.pl/CZM/tankov.pdf. Cont and Tankov have also written a longer book about modelling with Levy processes that I think is really good. There's going to be a strong connection between the sequence of jump times and the Levy measure \nu. In a single unit of time,  \nu(dx) is a measure (not necessarily a ... 3 The initial condition for the backward Kolmogorov PDE is that$$ u(0,x) = g(x) $$for all x in the relevant domain and not just at a particular point. So if your functions f and g agree only at a single point the initial conditions are in fact different. 3 I will try to answer this a bit differently. The rigorous answer: because Ito calculus tells us that we need the second order term. Look at$$ S_t = S_0\exp(\mu t + \sigma B_t). $$Assume that S_0 is known and fixed and look at by Ito's formula$$ d(S_t/S_0) = \mu dt + \sigma B_t + \frac{\sigma^2}{2} dt. $$Then with some abuse of notation:$$ ...

3

A key property of Brownian motion is independent increments. So if $x-1 > y$, then $$\mathbb{E}[\Delta W_x \Delta W_y] = 0$$ because the time intervals [x-1,x] and [y-1,y] do not overlap. If they do overlap, i.e. $x-1 \leq y < x$, then \begin{align} \mathbb{E}[\Delta W_x \Delta W_y] =&\ \mathbb{E}[(W_x - W_{x-1}) (W_y-W_{y-1})] \\ =&\ ...

3

It is true that the self-financing property of the replicating portfolio seems not explicitly presumed nor shown in Shreve's derivation of the Black-Scholes formula. One may note that a replicating portfolio is by definition a self-financing portfolio which replicates the payoff. The problem as I see is that Shreve is just suggesting some portfolio and ...

2

This interesting question provides excellent links to Dynamic Nelson-Siegel Term Structure Models for interest rates for No Arbitrage and exposes key formulation in an interesting way. Appendix in p37 of ssrn link says $\lambda$ is market price of diffusion risk. However, in the DNS model the $\lambda$ is eigenvalues of $\kappa$, which then part of ...

2

This will be the inverse process $$\frac{1}{S_t}$$ Applying Itô's formula the dynamics are then given by $$d\frac{1}{S_t}=\frac{-1}{S_t^2}dS_t+\frac{1}{S_t^3}dS_tdS_t$$ some simple algebra then leads to $$d\frac{1}{S_t}=\frac{1}{S_t}(\sigma^2 -r)dt+\frac{1}{S_t}\sigma dW_t$$

2

I don't know what you did when you tried pulling out $1-\alpha$, the correct expression would be $\lim_{\alpha \to 1} \frac{\mu(1-\alpha) + \sigma {\phi^{-1}(\alpha)}}{(1-\alpha)(\mu + \sigma \phi^{-1}(\alpha))}$. Anyhow, you can try using the substitution $\Phi^{-1}(\alpha) = x$, $x \to \infty$ and $\alpha = \Phi(x)$. Then the expression becomes ...

2

$$S_t = S_0\exp((r-\frac{\sigma^2}{2})t+\sigma W_t)$$ is not yet a martingale for it is not dirftless. From a probabilistic point of vew the "drift adjustment" comes into play so that the expected value of $S_t$ will be $e^{rt}$ rathern than $e^{(r+0.5\sigma^2)t}$. For the expected value of a log-normaly distributed variable with mean $\mu$ and vol ...

2

So we have the identity $$g(S,\sigma, t, C,C_t,C_S,...)=g(S, t,\sigma, V,V_t,V_S,...)$$ where $S$, $\sigma$, and $t$ are independent variables and $V=V(S,\sigma,t)$, $C=C(S,\sigma,t)$ are some unknown functions. But we can also treat the above identity formally and assume that the functions $C,C_t,C_S,...,V,V_t,V_S,...$ are themselves independent ...

2

You have $$\widetilde{W}_t=W_t+\int\Theta(u)du$$ which is in general not a Brownian motion, because it has a drift component. But 5.3.1 states $$M_t=M_0+\int \Gamma(u)dW_u\tag{5.3.1}$$ , which holds only for a Brownian motion $W$ (and $M_t$ martingale). So one cannot trivially replace $W_t$ and $W_t+\int\Theta(u)du=\widetilde{W}_t$ in 5.3.2 aswell by ...

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