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15

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

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

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 ...


6

$X_t$ 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) = ...


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 ...


4

For a martingale $dX=a(X,t)\,dt+b(X,t) dW(t)$ where $a$ and $b$ are not constant, your tree will not recombine in general [edit]. This is the main issue. See for instance: Florescu, I. and F. G. Viens (2008, March). Stochastic volatility: Option pricing using a multinomial recombining tree. Applied Mathematical Finance 15 (2), 151-181. It deals with the case ...


4

For a Brownian motion, if you wait $dt$, the variance will grow linearly with (proportionally to) $dt$. For a fractional Brownian motion, it will grow with a power law of $dt$, in fact in $dt^{H}$, where $H$ is the Hurst exponent. See wikipedia for more details. It means the fBM will somehow keep memory of the past. When $H$ is lower than 1/2, it will mean ...


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

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 ...


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 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 ...


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 ...


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

There is a shortcut around the Forward Equation when you are looking for the stationary distribution. Let me write $$ dX = \mu(X)dt +\sigma(X)dW $$ for $$ \mu(x)=b(1-x)-ax\ \text{ and }\ \sigma^2(x)=x(1-x) $$ The Forward Equation indeed states that the stationary distribution $p(x)$ will be satisfied for $\partial p/\partial t = 0$, therefore $$ ...


3

The more phenomenological definitions in his books are probably more helpful. Whether one uses the fractal dimension, Hurst coefficient, or exponential coefficient alpha, there is a value that corresponds to pure Brownian motion, a regime relative to this value that corresponds to persistence of motion, and the opposite regime that corresponds to ...


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 ...


2

I think that you are a bit confused: the support of the Black-Scholes model is $(0,+\infty)$, that is to say the underlying asset price is non-negative, like a stock. The Vasicek model has an OU process whose support is $(-\infty,+\infty)$, that is to say the underlying can be negative. Therefore all equivalent measures (of which the martingale is one) ...


2

This is a special case of the question of why $$ \int_0^T f(t) dW_t $$ is normally distributed for a continuous function $f(t).$ This Ito integral can be approximated by a sum $$ \sum_{i=0}^{N-1} f(i T/N) (W_{(i+1)T/N} - W_{i T/N}) .$$ The Brownian increments $(W_{(i+1)T/N} - W_{i T/N})$ are independent normally distributed random variables. The key point ...



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