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Here is a short list (to be edited and improved - community wiki) : Standard brownian motion (also called Wiener process) for which: $d\, W_t \sim \mathcal N(0, \sqrt{d t})$ Geometric brownian motion, used in the Black-Scholes model (1973): $d\,X_t = \mu X_t\,dt + \sigma X_t\,dW_t$ Constant elasticity of variance ("CEV") model (1975): $d\,X_t=\mu X_t dt + \... 12 I think to understand the martingale/local martingale distinction, it helps to bring in a third class of processes, the uniformly integrable martingale. I would argue that the local martingale and the non-uniformly integrable (true) martingale are actually fairly similar. The key property that a uniformly integrable martingale has is the so-called closure ... 9 Note that the Ito integral of a deterministic integrand$f: \mathbb{R}_+ \rightarrow \mathbb{R}$is normally distributed \begin{equation} \int_0^t f(u) \mathrm{d}W_u \sim \mathcal{N} \left( 0, \int_0^t f^2(u) \mathrm{d}u \right). \end{equation} In your case, we have$f(t) = e^{-\lambda t}$and thus \begin{equation} \int_0^t f^2(u) \mathrm{d}u = \... 7 If by 'solve' you mean how do we know that$\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 $$dY_t = [f_t(X_t,t) + f_x(X_t,t)\mu(X_t,t) + \... 6 In stochastic calculus, only stochastic integrals are defined. The differential form is just a notation. That is,$$dF=g(t)dW_t$$is just another expression for the integral$$F=\int_0^t g(s) dW_s.$$See, for example, in this book or this book, all Ito's lemmas are expressed in integral forms. For your question, note that F is not a function of t and ... 6 Here are two approaches that you could take to compute the variance of X_t. I am not making the conditioning explicit as it just complicates the notation but doesn't really add any additional insights. Compute \mathbb{E} \left[ X_t \right] and \mathbb{E} \left[ X_t^2 \right]. You can then you use that \begin{equation} \text{Var} \left( X_t \right)... 5 if Y=1 the stock price doesn't change since it's a percentage change not an absolute, so we have to subtract one when drift compensating. See my book Concepts etc for my discussion. 5 Apply the Ito product rule, noting the cov of a deterministic and stochastic term is zero:$$\begin{align} d\left(e^{at}r_t\right)&=e^{at} dr_t+r_t de^{at} \\[6pt] &=e^{at} dr_t+r_t e^{at} d(at) \\[6pt] &=e^{at} dr_t+r_t e^{at} a dt \end{align}5 It appears that you need to read some books such as Stochastic Differential Equations. For such type of equations, you need to use something called integrating factor such as the function e^{-\mu t} here. Note that \begin{align*} d\big(e^{-\mu t} X_t \big) &= X_t d\big(e^{-\mu t}\big) + e^{-\mu t} dX_t\\ &=-\mu e^{-\mu t} X_t dt + e^{-\mu t} (\mu ... 5 The issue is that you do not plot one sample path but for each time point t, you simply plot one possible realisation of the random variable S_t(\omega). Thus, you don't get a connected path. (Just as a minor, you would need brackets in the exponential in your for loop, i.e. X_analytic[i] <- X_analytic*exp((mu - 0.5*sigma^2)*time[i] + sigma*Z[i-1]... 4 It is a notation for quadratic variation of a stochastic process. 4 There is somewhere summary of methods that can be used to estimate parameters of SDE? If you'd like a brief survey, consider the following packages as well as the accompanying papers (note: you may want to follow the citations listed therein): https://cran.r-project.org/web/packages/Sim.DiffProc/ "Estimation of Stochastic Differential Equations with Sim.... 4 this is a well-known problem. One solution is to make volatility zero when rates exceed a certain high level. It's less problematic than it looks because any cash-flows generated will be divided by a rolling money market account which has huge value and so the deflated cash-flows are very small. 4 Under the stock numeraire measure, \frac{B_t}{S_t} is a Martingale. We can computed\frac{B_t}{S_t}= \frac{1}{S_t}dB_t -\frac{1}{S_t^2}B_tdS_t+\frac{1}{S_t^3}B_t\sigma^2S_t^2dt\\=\frac{B_t}{S_t}\left(rdt -\mu dt -\sigma dW_t +\sigma^2dt\right)$$so the growth rate \mu that makes this a Martingale is$$ \mu = r+\sigma^2.$$So the growth rate of the ... 4 In stochastic calculus, expressions of the type:$$ dX_t = a(t, X_t)dt + b(t, X_t) dW_t $$are called stochastic differential equations. What the one above means for example is that X_t has the following expression:$$ X_t = X_0 + \int_0^t a(u, X_u)du + \int_0^t b(u, X_u) dW_u $$The first integral is a regular one, and the second is called a stochastic ... 4 Using Itô's Lemma, notice that:$$d(tS_t)=tdS_t+S_tdt=dX_t+S_tdt$$Hence:$$X_t=tS_t-\int S_udu$$Using independence of Brownian increments, E(S_udW_u)=E(S_u)E(dW_u)=0, and the chain rule for the 4th step:$$\begin{align} E(X_t)&=E\left(\int dX_u\right) \\ &=\int uE(dS_u) \\ &=\int u\mu E(S_u)du \\ &=S_0\int u\mu e^{\mu u}du \\ &=S_0\... 4 That is not the SDE for the Heston model - it violates the affine property in the drift term. In other words, the paper has a typo. The correct SDE is: $$d v_t = \kappa (m-v_t) dt + \nu \sqrt{v_t} dw_t$$ wherev_t := \sigma_t^2$is the variance. Let$\xi_t^T := \mathbb{E}_t [ v_T]denote the forward variance and see that \begin{align} \xi_{t}^{T} &... 4 I am not providing a full proof but a reference for you to read up the details. The key step is mentioned below. Most models used in finance are Markovian which is kind of in line with the efficient market hypothesis. The key step of of seeing that the Heston process is Markovian is the following theorem. Let f be a bounded Borel function from \mathbb{... 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 In general SDE's are defined on a probability space which consists of a triplet (\Omega, P, B): the space \Omega, a probability measure P, and a sigma algebra B. In short, the sigma algebra consist on the set of all events that we can assign probability to. For SDE's driven by Brownian Motion this probability space is the so called Wiener space, ... 3 This is a corollary of Feynman-Kac theorem. For self-containedness, I re-produce the proof as follows. Assume that there exists a C^{1,2}-function F=F(t,x) defined on [0,T]\times\mathbb{R} that satisfies the PDE on the interior F_{t}+\beta F_{x}+\frac{1}{2}\sigma^{2}F_{xx}=0, $$and the boundary condition: F(T,x)=g(x). Consider the process \left(... 3 \require{cancel} Consider the following SDE$$ \frac{dF(t,T)}{F(t,T)} = \sigma(t,T) dW_t + (\exp(e^{-a(T-t)}dN_t)-1) + \mu_J(t,T)dt $$where N_t figures a standard Poisson process, supposedly independent of the standard Brownian motion W_t. This SDE should be interpreted by looking at N_t as what it is, namely a random counting process with, ... 3 Let us define the auxiliary process \Lambda_t=e^{\kappa t}\lambda_t. Note that:$$ \Lambda_t = \kappa e^{\kappa t} \int_0^t(\rho_s-\lambda_s)ds+\delta e^{\kappa t}\int_0^tdN_t$$Hence after a jump occurs at t:$$ \Lambda_t=\Lambda_{t-}+\delta e^{\kappa t}$$Therefore by Ito's lemma for jump-diffusion processes:$$ \begin{align} d\Lambda_t & = \... 3 We assume that, under the risk-neutral measureQ, \begin{align*} dP(t, T) = P(t, T)(r_t + \sigma(t, T)dW_t), \end{align*} where\{W_t, \, t \ge 0\}is a standard Brownian motion. Then \begin{align*} dL(t) &= \frac{1}{T-S}\bigg(\frac{dP(t, S)}{P(t, T)} -\frac{dP(t, S)}{P^2(t, T)}dP(t, T) \\ &\qquad + \frac{dP(t, S)}{P^3(t, T)} \langle dP(t, T), \,... 3 This has been indirectly discussed in this question. We assume that\{\mathcal{F}_t, \, t\ge 0\}$is the natural filtration generated by the Brownian motion$\{W_t,\, t \ge 0\}$. Moreover, let$B_t= e^{\int_0^t r_sds}$be the money market account value at time$t\ge0$. Let$V_T$be the option payoff at maturity$T\$. Then \begin{align*} V_t = B_tE_Q\left(\...
Assuming you are talking about unconditional expectation, in general you have $$\mathbb{E}[X_t] = \mathbb{E}[e^{W_t}] = e^{\mathbb{E}[W_t] + \frac{1}{2}\text{Var}(W_t) }$$ which yields $$\mathbb{E}[X_t]= e^{\frac{1}{2} t}$$ Hence, $$\mathbb{E}[X_2]= e$$