# Tag Info

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] $$and that implies that E[W_pW_qW_r] = 0. 6 In the integral$$\int_0^t S_u dW^{*}_u \, ,$$dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u is independent from the integrand S_u. So, \mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0, since ... 6 You derivation here is flawed because you are deriving with respect to two processes and you do not take into account that the variable W_t is stochastic and hence S_t is as well. So, to derive S_t from dS_t, you have to apply Ito's Lemma, see this question for details. This is the "classic" way you see it. If you want to do it the other way ... 6 The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows:$$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$$by using general Stochastic Calculus rules (e.g. p.37, 6.6 here): ... 6 It's a lemma! Ito's Lemma gives the change of coordinates rule for stochastic calculus. The multiplication rule is a shorthand way of expressing it. 5$$ \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 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. $$Plotting this set we see it is the bottom half of the lower right quadrant. The probability of being in the lower right quadrant is clearly 0.25 by symmetry. The probability ... 4 I thought this was an interesting example to add. It concerns a "ratio model" of habit (as opposed to a "difference" model of habit). See, for example, Abel (1990, American Economic Review). Let$$ 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 ... 4 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} & ... 4 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 ... 4 let Y_t=(X_t)^\alpha,thendY_t=\alpha Y_tdt+dW_t^P$$we define Q measure by$$\frac{dQ}{dP}=exp\left(-\alpha\int_{0}^{T}Y_t\,dW_t^p-\frac{1}{2}\alpha^2\int_{0}^{T}Y_t^2 dt\right)$$this shows that$$W_t^Q=W_t^P+\alpha\,\int_{0}^{t}Y_s\,ds$$is standard wiener process under Q measure, thus we have$$dW_t^P=dW_t^Q-\alpha\,Y_t dt$$and$$dY_t=\alpha ...

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

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),$... 3 What you need is to identify the distribution of the asset price$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 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,$\langle W, W\rangle_tis a limit, in probability, of the variation \begin{align*} \sum_{i=1}^n|W_{t_{i}}-W_{t_{i-1}}|^2, \end{align*} over the partition ... 3 Note that, for0 \leq s < t, \begin{align*} W_t^3 &= (W_t-W_s+W_s)^3\\ &= (W_t-W_s)^3 + 3(W_t-W_s)^2 W_s + 3 (W_t-W_s) W_s^2 + W_s^3. \end{align*} Moreover, \begin{align*} E\big( (W_t-W_s)^3 \mid \mathcal{F}_s\big) &= E\big( (W_t-W_s)^3\big)\\ &= 0,\\ E\big((W_t-W_s)^2 W_s \mid \mathcal{F}_s\big) &= W_s E\big( (W_t-W_s)^2\big)\\ ... 2 First of all, a filtration( \mathscr{F}_t )_{t \geq 0 }$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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