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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): $d\,X_t=\mu X_t dt + ... 7 Stochastics are usually applied in the field of derivatives pricing. In this setting the task is to price a derivative such that it fits into the landscape of tradable instruments (no-arbitrage). We work using the risk-neutral measure - usually denoted by$Q$. The measure is derived from other traded instruments. In risk analysis (e.g. calculate the VaR, ES ... 7 I will assume a white noise is a process$(\varepsilon_t)$with zero mean, no autocorrelation and constant variance$\sigma^2 > 0$while a random walk is a process$(x_t)$defined by $$x_{t+1} = x_t + \varepsilon_{t+1}$$ where$\varepsilon$is a white noise. 1) No since$Var(x_{t+1}) = Var(x_t) + Var(\varepsilon_{t+1})$is stricly increasing while ... 7$X_t$being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function$f$of$t$and$X_t$. Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming$X_tis here a continuous, square integrable stochastic process) df(t,X_t) = \frac{... 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} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=\int_{0}^{s}\int_{0}^{t}E\,... 6 The dynamics \begin{align*} \frac{dS_t}{S_t} =\mu dt + \sigma dW_t. \end{align*} is under the real-world measure \mathbb{P}. Then, \begin{align*} d\ln S_t =\Big(\mu-\frac{1}{2}\sigma^2 \Big) dt + \sigma dW_t. \end{align*} Therefore, \begin{align*} \ln S_T = \ln S_t + \Big(\mu-\frac{1}{2}\sigma^2 \Big)(T-t) + \sigma \big(W_T-W_t\big).\tag{1} \end{align*} ... 6 By definition, the payoff of a log-contract of maturity T writes \phi(S_T) = \ln\left(\frac{S_T}{S_0}\right) $$Let \Pi_t denote the t-value of such a contingent claim. We are interested in the price at t=0, best known as the option premium. Theory tells us that the latter premium can be computed as$$ \Pi_0 = e^{-rT} E^{\mathbb{Q}} \left[ \phi(... 5 Mean reversion speed\kappa$is better interpreted with the concept of half-life, which can be calculated from$\text{HL} = \ln(2) / \kappa$. For example, if the mean reversion coefficient is$\kappa = 1.5$, then the half-life of the process is$\ln(2) / 1.5 = 0.46209812$years, or about 6 months. Let's assume that the current interest rate is 1% and the ... 5 The first process is a BM. The second does not exist in continuous time. The variance goes down too slowly with dt and the process blows up at the limit. You can break the (0,1) interval into 1, 100, 1000, 1000000 steps and see that happening. Variance of a martingale has to scale with dt: if it is too fast then the process dies, if it is too slow then ... 5 What can be shown is that the above expressions are equal in probability. First check the distribution. As any linear combination of a Gaussian is Gaussian the right hand side is Gaussian - the left hand side too. Then we need the 2 moments: The expected values - it is zero ... easy to see. Next what you did not specify is that the correlation between$...
Write $X_t = A_t B_t$ with $A_t = e^{(\lambda - \eta)t}$ and $B_t = \left(\frac{\eta}{\lambda} \right)^{N_t}$. Then $dX_t = A_t dB_t + B_t dA_t$ by the product rule of calculus. There are no second order terms since both $A_t$ and $B_t$ are finite variation (i.e. $\langle A_t, B_t\rangle$= 0). Next, $dA_t = (\lambda - \eta)A_t dt$, and dB_t = B_t \cdot \... 4 Milstein Scheme This scheme is described in Glasserman (2003) and in Kloeden and Platen (1992) for general processes.Hence, for simplicity, we can assume that the Stochastic Process is driven by the SDE \begin{align} &dX_t=\Xi(t,X_t)dt+\Sigma(t,X_t)dW_t\\ \end{align} Milstein discretization is, \begin{align} dX_{t+\Delta t}=X_t+\Xi(t,X_t)dt+\Sigma(t,X_t)... 4 I think all they are doing is integrating and estimating $$P(|W_t| \leq 2) = \int_{-2}^{2} \frac{d}{dr} P(W_t \leq r) dr$$ so $$P(|W_t| \leq 2) \leq 4 \sup \limits_{r \in [-2,2]} \frac{d}{dr}P(W_t \leq r)$$ The normal density is maximal at zero and we are done. 4 If you want to address interesting problems that are interesting for financial mathematics, I do not believe you have the good list. Pricing. For instance, most of explicit formulas for pricing that are not available yet will never be. In this direction, you should have a look at simulation techniques. See for instance Nonlinear Option Pricing. Interesting ... 4 This is wrong! Notice thatdX_t=\mu(t,X_t)dt + \sigma(t,X_t)dW$is a shorthand for $$\int_0^tdX_s = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s$$ Integrating: $$X_t-X_0 = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s \text{ (eq.1)}$$ If we take expectations, remembering that$\mathbb{E}[\int_0^t\sigma(s,X_s)dW_s]=0, we have $$\mathbb{E}[... 4 The first process$$ B_{t+dt} = B_t + Z $$where Z is independent of (B_s)_{s \le t} and follows a Gaussian distribution with mean 0 and varince dt is a standard Brownian motion (thus the variance of B_t is t). For the second process let us recall the definition from your link:$$ E[B^H_t B^H_s] = \frac12 ( t^{2H} + s^{2H} - |t-s|^{2H}), thus ... 4 Geometric Brownian Motion has independent increments but Ornstein-Uhlenbeck doesn't have this property. For more details you can look here. 4 First thing, Geometric Brownian motion do not have independent increments. It is only Wiener process or Brownian motion that have independent increment. Under GBM, the increments of process (assume stock prices) show markovian property. It means that changes in the process depend on the current price level. In layman terms, the magnitude of change in stock ... 4 A few points can be noted. The CIR model is usually for a short, or instantaneous, spot rate r_t, which is the forward rate over an infinitesimal interval. That is, \begin{align*} r_t = \lim_{\Delta \rightarrow 0}\frac{1}{\Delta}\left(\frac{1}{P(t, t+\Delta)}-1 \right), \end{align*} where P(t, u) is the price at time t of a zero-coupon bond with ... 4 Typically when running a Monte Carlo simulation we might simulate an SDE similar to \dfrac{dS}{S} = \mu\:dt + \sigma \: dW(t) $$by some appropriate method (e.g. Euler-Maruyama, Milstein, etc). We notice by dimensional analysis that if t is in units of \textrm{years} then \mu \sim \textrm{years}^{-1} and \sigma \sim \textrm{years}^{-1/2}. ... 3 EDIT: My reasoning below seems to be wrong. The process as you write it tends to infinity if a is big enough and positive and if \lambda_0 is positive. I would not call this process non-meanreverting OU. It is just an Ito process of a simple form. If we remove the stochastic part then we get$$ d\lambda_t = a \lambda_t dt with the solution (if \... 3 Not sure about the correctness of the first approach, but second approach uses 1 /\sqrt k to scale the variance of the total sum by k. So the difference of two processes (say W_t and W_{t+\Delta t}) generated by the random walk would have a variation of \Delta t, which satisfies one of conditions needed to get a Wiener's process. 3 To solve for U_t, we can proceed as follows. First, note that \begin{align*} d\left(e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} U_t \right) &= e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} U_t \left((\theta+\xi^2) dt -\xi dW_t\right) \\ &\qquad+ e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} dU_t -\xi^2e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} U_t dt\\ &... 3 Your notations are really hard to follow as you define \mathbb{P} twice at the beginning. The notation \mathbb{P} = \mathbb{\hat{P}} and \mathbb{P} =\mathbb{\tilde{P}} is not meaningful as the probability measure \mathbb{P} is already fixed and used for the real world probability measure. I think that this is the reason why you are getting confused. ... 3 Given efficient markets, asset prices should be unpredictable in the sense that any upcoming returns are uncorrelated with current or past returns. Hence for traded assets the price should follow something more similar to a GBM than an O-U process. However, many financial metrics are not prices; for example interest rates or volatility. O-U processes may ... 3 Below I assume that you meant: \psi (T) = \max (S_t - S_T, 0) which constitutes the payout of a forward start rather than a lookback option. If not please clarify your question... If you are looking for the option price V_0, assuming a Black-Scholes diffusion (GBM + constant interest rates), you have \begin{align*} V_0 &= P(0,T) E[ \psi (T) \vert \... 3 Let f_{\lambda}(t,r)=E^{(t,r)}\left[e^{-\lambda r_{T}}\right] $$where E^{(t,r)} denotes the expectation conditional on r_{t}=r. We assume f is smooth for the remainder. Let \theta=T\wedge\inf\left\{ s>t\colon\left|r_{s}-r\right|>1\right\} . By the Markov property of \{r_{t}\},$$ f_{\lambda}(t,r)=E^{(t,r)}\left[f_{\lambda}(\left(t+h\right)... 3 Here's my 2 cents: a) Conditional expectations can always be seen as martingales (this is a direct consequence of the tower property). Thus, we here have that $$M_t := E^*[e^{-\lambda {r_{T}}}|r_t]$$ is a martingale. Applying Itô's lemma toM_t = f_{\lambda}(t,r_t)$as you did is a good starting point. But doing this, leaves you with an SDE, not a PDE.... 3 Your problem probably comes from the notations used. Let the Moment Generating Function (MGF) of a random variable$X\$ be defined as $$M_X(u) := E[e^{uX}]$$ From this definition, it entails that $$E(X^n) = M_X^{(n)}(u=0) = \frac{d^{n} M_X}{ d u^{n}}(u=0)$$ Knowing this, the function $$f_{\lambda}(t,r)=E[e^{-\lambda {r_{T}}}|r_t=r]$$ can be ...