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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 $E[W_pW_qW_r] = 0$.


6

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 + ...


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

Shreve's theorem also called "Girsanov II" indeed represents a special case of the general "Girsanov I" from Wiki above, with $$Y_t:=W_t,$$$$X_t:=-\int_0^t\Theta_udW_u$$ We can show: $$[Y,X]=-\int_0^t\Theta_udu$$ by using general Stochastic Calculus rules (e.g. p.37, 6.6 here): ...


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 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

For any $s \geq t$, note that \begin{align*} r_s = r_t + \sigma\int_t^s dW_u + \int_t^s \theta_u du. \end{align*} Then, \begin{align*} \int_t^T r_s ds &= (T-t)r_t + \sigma\int_t^T\int_t^s dW_u ds + \int_t^T \int_t^s\theta_u du ds\\ &=(T-t)r_t + \sigma\int_t^T\int_u^T ds\, dW_u +\int_t^T\int_u^T\theta_u ds du\\ &=(T-t)r_t + \sigma\int_t^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 ...


4

In the integral $$\int_0^t S_u dW^{*}_u \, ,$$ $dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u$ is independent from the integrand $S_u$. So, $\mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0$, since ...


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

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 ...


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 portfolio $V_t(\alpha_t,\beta_t)$ (for stock $S_t$ and zerobond $B_t$) is self-financing iff: $$V_t=\alpha_tS_t+\beta_t B_t$$ It further implies $$dV_t=\alpha_tdS_t+\beta_tdB_t$$ To replicate a derivative $C(S_t,t)$ by a self-financing portfolio of stock and bond, set: $$dV_t=dC_t$$ The dynamics of $dC$ can be specified using Ito's Lemma on ...


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

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 ItĂ´'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

First of all, a filtration $( \mathscr{F}_t )_{t \geq 0 }$ is a "set" of sigma algebras indexed usually by time t that are increasing. That is, for every $t>0$, $\mathscr{F}_t$ is a sigma algebra and $\mathscr{F}_t \subseteq \mathscr{F}_T$ for all $0\leq t \leq T$. The canonical example, is the filtration generated by a process, say Brownian Motion $W$: ...



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