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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: $$\int_0^t f(t, \omega) d W(t, \omega) = \lim_{\| \Delta_n\| \to 0 } \sum_{i=1}^{n} f(\tau_i,\omega) \left ( W(t_i, \omega) - W(t_{i-1}, \omega) \right ) $$ , ...


10

These are all examples on Ito Formula in its general form (with quadratic variations):


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 $$ ...


8

Feynman–Kac Theorem: Assume that $F$ is a solution to the boundary value problem \begin{align} &F_t+\mu(t,x)F_x+\frac{1}{2}\sigma^2(t,x)F_{xx}-rF=0\\ &F(T,x)=\Phi(x), \end{align} Assume furthermore that the process $e^{-r_s}\sigma(s,X_s)F_s$ is in $\mathcal L^2$ where \begin{align} dX_s=\mu(s,x)ds+\sigma(s,x)dW_s, \end{align} then $F$ has the ...


7

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

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 ...


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 + ...


6

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): ...


6

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 ...


6

It's a lemma! Ito's Lemma gives the change of coordinates rule for stochastic calculus. The multiplication rule is a shorthand way of expressing it.


6

You know that $E\left[\int_{0}^{s}W_udu\right]=E\left[\int_{0}^{t}W_vdv\right]=0$. By definition \begin{align} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=E\left[\int_{0}^{s}W_u\,du\int_{0}^{t}W_v\,dv\right]-0 \end{align} then \begin{align} & ...


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

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

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 ...


5

The problem is equivalent to given to 2 independent standard normals $W$ and $Z$ the probability of $$ W > 0, \text{ and } W+Z<0. $$ or $$ W > 0, \text{ and } Z<-W. $$ Plotting this set we see it is the bottom half of the lower right quadrant. The probability of being in the lower right quadrant is clearly $0.25$ by symmetry. The probability ...


5

Note that, for $0 \leq s < t$, \begin{align*} W_t^3 &= (W_t-W_s+W_s)^3\\ &= (W_t-W_s)^3 + 3(W_t-W_s)^2 W_s + 3 (W_t-W_s) W_s^2 + W_s^3. \end{align*} Moreover, \begin{align*} E\big( (W_t-W_s)^3 \mid \mathcal{F}_s\big) &= E\big( (W_t-W_s)^3\big)\\ &= 0,\\ E\big((W_t-W_s)^2 W_s \mid \mathcal{F}_s\big) &= W_s E\big( (W_t-W_s)^2\big)\\ ...


4

I thought this was an interesting example to add. It concerns a "ratio model" of habit (as opposed to a "difference" model of habit). See, for example, Abel (1990, American Economic Review). Let $$ x_t = \lambda \int_{-\infty}^t e^{-\lambda(t-s)} c_s ds. $$ (For context, $x_t$ is a log habit index that is given by a geometric average of past consumption, ...


4

B1~N(0,1) and B2=B1+Z, for Z~N(0,1). From that E(B1*B1)=E(B1*B2)=1, E(B2*B2)=2. Therefore they are bivariate Gaussian with covariance matrix (1,1;1,2) therefore probability is around 12%, which is the volume over the bottom-right quadrant.


4

Maybe I'm missing something? Given $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$, you can write $f = (f_1,\ldots,f_m)$, where each $f_i:\mathbb{R}^n \rightarrow \mathbb{R}$. Apply Ito to each $f_i$ separately.


4

In this case it is just the notion that your payoff function should not explode at some point - made mathematically rigorous. Have a look at the following picture from wikipedia: Intuitively the Lipschitz condition (or Lipschitz continuity) ensures that your payoff function always remains entirely outside the white cone, so it cannot e.g. become ...


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 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 ...


3

What you need is to identify the distribution of the asset price $S_T$, conditional on the information set $\mathcal{F}_{t}$ at time $t$, for $0\leq t < T$. Note that \begin{align*} S_T &= S_t \exp\bigg(\int_{t}^T \Big(r_s-\frac{\sigma_s^2}{2}\Big)ds + \int_t^T\sigma_s dW_s \bigg). \end{align*} Let \begin{align*} P(t, T) = \exp\bigg(-\int_t^T r_s ds ...


3

I am not sure if I understood your question correctly but I will try to answer it anyway. If you have a standard normal random vector $z \sim N(\mathbb{0},I_n)$ (where $z,0 \in \mathbb{R}^{n\times1}$ and $I_n \in \mathbb{R}^{n\times n}$ is the identity matrix) and you want to transform it into a multivariate normal $x \sim N(\mu,\Sigma)$ you do it the ...


3

Formally, this is a shorthand for the quadratic variation. For a more rudimentary definition, $\langle W, W\rangle$ is a process such that $W^2-\langle W, W\rangle$ is a martingale. Moreover, $\langle W, W\rangle_t$ is a limit, in probability, of the variation \begin{align*} \sum_{i=1}^n|W_{t_{i}}-W_{t_{i-1}}|^2, \end{align*} over the partition ...


2

The logic from Bob Jansen is correct. The problem is abuse of ideas and notation the integral symbol from the deterministic world gets sloppily applied to random variables. Unlike normal $dt$, which is always positive, $dW_t$ can go 'backwards'. Thus increments of terms like $W_t dW_t$ have a first element that goes up and down with the second element ...



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