# 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 ...

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$... 3 Consider an (arithmetic) Ornstein-Uhlenbeck process as a model of the asset price$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}$$, which holds only for a Brownian motion W (and M_t martingale). So one cannot trivially replace W_t and W_t+\int\Theta(u)du=\widetilde{W}_t in 5.3.2 aswell by ... 2 I think you are on the right track here. You made a sign error in the first line, unfortunately:$$E[W_p W_q W_r] = E[W_r W_p^2 + W_pW_q^2 - W_qW_p^2]=\\ E[(W_r-W_q)W_p^2]+E[W_pW_q^2]= E[W_pW_q^2]  The first term is $0$ by independence (as $p<\text{min}(r,q)$ and the square does not affect independence). To take care of the second term we do the ...

2

\begin{align*} E\Big(W_t^3-3tW_t \mid \mathcal{F}_s\Big) &= E\Big((W_t-W_s+W_s)^3-3t(W_t-W_s+W_s) \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3+W_s^3+3(W_t-W_s)^2W_s + 3 (W_t-W_s)W_s^2\\ &\qquad \qquad -3t(W_t-W_s)-3tW_s \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3\Big) + W_s^3+3W_sE\Big((W_t-W_s)^2\Big)\\ &\qquad \qquad + 3W_s^2 ...

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