Hot answers tagged

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 "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 ...


2

Measure change is still the most natural approach for such problems. We assume that, under the measure $P$, \begin{align*} dX_t &= \mu X_t dt + \sigma X_t dW_t^1,\\ dY_t &= \mu Y_t dt + \sigma Y_t \left(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2 \right), \end{align*} based on the Cholesky decomposition, where $\{W_t^1, t \ge 0\}$ and $\{W_t^2, t \ge 0\}$ ...


2

I don't know if this is enough. But here is my understanding. Let's imagine a simple process like a Poisson process. It is naturally cadlag, because at the time you jump, you jump. Just before, you have not jumped. Mathematically, if the first jump occurs at $t$, $\forall s<t, N_s=0$ and $N_t=1$. It means that the jump occuring at time $t$ is $t$-...


1

Generally, the PCA approach proceeds as follows. Consider historical observation times $t_0 < t_1 < \cdots < t_K \le0$. For bucketing times $\delta_1 < \cdots \delta_n$, let $X_j(t_k) = \ln F(t_k, t_k+\delta_j)$. Moreover, let $\eta_j$, for $j=1, \ldots, n$, be a normal random variable with a sample set $\big\{X_j(t_k)-X_j(t_{k-1}\}_{k=1}^K\big\}...


1

Let's look at the formula for an ARMA(p, q) model $$ X_t = c + \sigma z_t + \sum_{i=1}^p \varphi_i X_{t-i} + \sigma\sum_{i=1}^{q}\theta_i z_{t-i} $$ where $z_t \in \mathcal{N}(0,1)$ for all $t$. Transform this into the continuous time counterpart below. $$ X_t = c + \sigma W_t + \int_0^{T_p}\varphi(v)X_{t-v} dv + \sigma\int_0^{T_q}\theta(v)W_{t-v}dv $$ ...


1

You find the connection between an AR(1) and an Ornstein-Uhlenbeck process here in QSE if you search. Taken the description form here (which is just the way to do it): $$ X_{n+1} = c + a X_n + b \varepsilon_k $$ and by setting $c=\theta \mu \Delta t, a=−\theta \Delta t$ and $b =\sigma \sqrt{\Delta t}$ you will get the discrete time approximation $$ X_{...


1

Hint: By Integration, we have $$x_t=x_{0}+\int_{0}^{t} \alpha_1(x_s,s)ds+\int_{0}^{t} \beta_1(x_s,s)dW_1(s)$$ $$y_t=y_{0}+\int_{0}^{t} \alpha_2(y_s,s)ds+\int_{0}^{t} \beta_2(y_s,s)dW_2(s)$$ then $$E[x_t]=x_0+E\left[\int_{0}^{t} \alpha_1(x_s,s)ds\right]$$ $$E[y_t]=y_0+E\left[\int_{0}^{t} \alpha_2(y_s,s)ds\right]$$ Now we apply Ito's lemma $$d(x_ty_t)=...


1

Note that, under measure $Q$, the dynamics is of the form \begin{align*} dS_t = S_t \big[(r+ \sigma \theta_t) dt + \sigma dW_t^Q \big]. \end{align*} Then, for $\Delta>0$ sufficiently small, \begin{align*} S_{t+\Delta} &= S_te^{\left(r-\frac{1}{2}\sigma^2\right)\Delta + \sigma \int_t^{t+\Delta} \theta_s ds + \sigma \left(W_{t+\Delta}^Q-W_t^Q\right)}\\ &...


1

Another Solution We should look for a solution of the form $$X(t)=U(t)V(t)$$ where $$dU_t=-\theta\,U_tdt+\sigma\,U_t\,dW_t$$ and $$dV_t=\alpha(t)dt+\beta(t)dW_t$$ $U$ is a geometric Brownian motion, therefore $$U(t)=U(0)\,e^{-(\theta+\frac{1}{2}\sigma^2)t+\sigma W_t}$$ let $U(0)=1$, this yields $V(0)=X(0)$. Now we should find $\alpha(t)$ and $\beta(t)$. $...


1

Relatively quick Solution If $U$ and $V$ be normally distributed with means $\mu_u\,,\,\mu_v$, variances $\sigma^2_u\,,\,\sigma^2_v$ and correlation $\rho$ then we can show ( by definition of expectation and apply joint density function ) $$\mathbb{E}\left[\left(e^U-e^V\right)^+\right]={\large{e^{\mu_u+\frac{1}{2}\sigma_u^2}}}\Phi\left(d_1\right)-{\large{e^...


1

First, let us formulate the problem mathematically: A symmetric random walk starts at 0 and moves up or down one unit (with equal probability) every 1 second. The are two absorbing barriers located at H and -L, with $H,L>0$. Given infinite time, what is the probability $p_H$ that H will be hit before -L is hit and what is the probability $p_L$ that -L ...



Only top voted, non community-wiki answers of a minimum length are eligible