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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): $d\,X_t=\mu X_t dt + ... 8 1. This integral is not Ito's Integral. Indeed$Y_t$is a random time change with time change rate$\frac{W_t}{1+W_t^2}.(Oksendal, Sixth edition,page 147) 2. Sometimes this trick is useful.Indeed we assume that we are going to solve Riemann integral !. Let $$f''(x)=\frac{-2x}{(1+x^2)^2}$$ then $$f'(x)=\left(\frac{1}{1+x^2}\right)+c_1$$ and $$f(x)=\... 7 Stochastics are usually applied in the field of derivatives pricing. In this setting the task is to price a derivative such that it fits into the landscape of tradable instruments (no-arbitrage). We work using the risk-neutral measure - usually denoted by Q. The measure is derived from other traded instruments. In risk analysis (e.g. calculate the VaR, ES ... 7 I will assume a white noise is a process (\varepsilon_t) with zero mean, no autocorrelation and constant variance \sigma^2 > 0 while a random walk is a process (x_t) defined by$$ x_{t+1} = x_t + \varepsilon_{t+1} $$where \varepsilon is a white noise. 1) No since Var(x_{t+1}) = Var(x_t) + Var(\varepsilon_{t+1}) is stricly increasing while ... 7 X_t being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function f of t and X_t. Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming X_t is here a continuous, square integrable stochastic process)$$ df(t,X_t) = \frac{... 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 maturityT$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 $$\Pi_0 = e^{-rT} E^{\mathbb{Q}} \left[ \phi(... 6 Let$$Y_t = \int_0^t N_u duwhere (N_t)_{t \geq 0} figures a Poisson process with intensity \lambda. Using the stochastic Fubini theorem we have that: \begin{align} Y_T &= \int_0^T N_t dt \\ &= \int_0^T \int_0^t dN_u dt \\ &\color{lightgray}{= \int_0^T \int_0^T \mathbf{1}\{u \in [0,t]\} dN_u\ dt} \\ &\color{lightgray}{= \int_0^T \... 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 ... 5 IMHO the problem isn't stated correctly indeed, in the sense that the Radon-Nikodym derivative provided as the "solution" is not the unique way to define a measure \mathbb{Q} equivalent to \mathbb{P} and under which X_t is a martingale. Just take\frac {d\mathbb{Q}}{d\mathbb{P}} =\mathcal{E}\left(-\int_0^t \cos(s) dW_s + a\right)$$for any a \in \... 4 This is wrong! Notice that dX_t=\mu(t,X_t)dt + \sigma(t,X_t)dW is a shorthand for$$\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 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 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 The first process $$B_{t+dt} = B_t + Z$$ whereZ$is independent of$(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$r_t, which is the forward rate over an infinitesimal interval. That is, \begin{align*} r_t = \lim_{\Delta \rightarrow 0}\frac{1}{\Delta}\left(\frac{1}{P(t, t+\Delta)}-1 \right), \end{align*} whereP(t, u)$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 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 $$f_{\lambda}(t,r)=E^{(t,r)}\left[e^{-\lambda r_{T}}\right]$$ whereE^{(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] $$is a martingale. Applying Itô's lemma to M_t = f_{\lambda}(t,r_t) as you did is a good starting point. But doing this, leaves you with an SDE, not a PDE.... 3 Your problem probably comes from the notations used. Let the Moment Generating Function (MGF) of a random variable X be defined as$$ 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]$can be ... 3 We assume that \begin{align*} \frac{dX_t}{X_t} &= (r+\pi Y_t)dt + \pi\sigma dW_t,\tag{1}\\ dY_t &= -\lambda Y_t + dB_t.\tag{2} \end{align*} From(2), \begin{align*} Y_t = Y_0 e^{-\lambda t}+ e^{-\lambda t}\int_0^t e^{\lambda u} dB_u. \end{align*} Moreover, from(1), \begin{align*} \ln X_T &= \ln X_0 + (r-\frac{1}{2}\pi^2\sigma^2)T + \pi \... 3  My "answer" below is not a really an answer for I have completely misinterpreted your original question. I thought you asked about the covariance of 2 processes over a given time horizon (i.e. for a fixed\omega$) and not the covariance of two random variables (fixed$t$). Also note that$\text{cov}(x,y)=0$does not mean that$x$and$y\$ are ...