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## Hot answers tagged stochastic-calculus

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 $$... 7 We know that (\tilde{W}_t) := (-W_t) is also a Wiener process so$$ 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 \, ,$$dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u is independent from the integrand S_u. So, \mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0, since ... 5$$ \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 ...

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

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 ...

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),$... 3 It is true that the self-financing property of the replicating portfolio seems not explicitly presumed nor shown in Shreve's derivation of the Black-Scholes formula. One may note that a replicating portfolio is by definition a self-financing portfolio which replicates the payoff. The problem as I see is that Shreve is just suggesting some portfolio and ... 3 About the integration problem: Your integrand is highly oscillatory, and the adaptive quadrature of Matlab doesn't handle such integrands very well. In general, I would recommend Mathematica when Matlab's standard procedures don't perform well. In this case, a Levin-type method would perform much better. The reason that quadv produces NaN values is because ... 3 A portfolio$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$0\leq t < T. Note that \begin{align*} S_T &= S_t \exp\bigg(\int_{t}^T \Big(r_s-\frac{\sigma_s^2}{2}\Big)ds + \int_t^T\sigma_s dW_s \bigg). \end{align*} Let \begin{align*} P(t, T) = \exp\bigg(-\int_t^T r_s ds ... 3 I am not sure if I understood your question correctly but I will try to answer it anyway. If you have a standard normal random vectorz \sim N(\mathbb{0},I_n)$(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 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. 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 ... 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$W$: ... 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

I am rather a fan of mathematical/statistical software for doing numerical finance (R/Matlab). But returning to your question: The commercial software UNRISK is based on mathematica, a computer algebra system. Usually you can use the Unrisk functions right in mathematica and price financial derivatives there. There also exists Jave interfaces if you want ...

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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