A message from our CEO about the future of Stack Overflow and Stack Exchange. Read now.

New answers tagged

2

Note merely that $B_t=B_s+(B_t-B_s)$ which is the sum of independent normally distributed random variables. In particular, $B_s$ is $\mathbb{F}_s$-measurable and $B_{t-s}$ is independent of $\mathbb{F}_s$. Thus, \begin{align*} \mathbb{E}_s[Z_t] &= \mathbb{E}_s\left[\exp\left(-\frac{1}{2}\theta^2t+\theta B_t\right)\right] \\ &= \mathbb{E}_s\left[\exp\...


3

Here are the things you need to correct in your code: Although you are setting a seed, you are generating the random numbers twice, and therefore they are not identical. Try this: rand = np.random.randn(N) y = np.exp(-volvol**2.0/2 * time_step + volvol * s * rand) y1 = rand You also need to multiply the sigma by the square root of the timestep in the SDE ...


1

You should replace the differential of the correlated process dBt with its value in the volatility equation, then replace dW~t in the same equation with: dW~t=dWt+ 2*theta*square_root(Vt)*dt you will get an formula with Vt,W1t and W2t. You can then simulate the volatility.


1

Let $Y = \log X$, then: $$\begin{align} Y &= Y_0 + (\mu-\frac{\sigma^2}{2})t + \sigma W_t \\ EY_t &=Y_0 + (\mu-\frac{\sigma^2}{2})t \\ EY_tEY_s &= Y_0^2 + Y_0 (\mu-\frac{\sigma^2}{2}) (t+s) + (\mu-\frac{\sigma^2}{2})^2 t s \\ E(Y_tY_s) &= E\left((Y_0 + (\mu-\frac{\sigma^2}{2})t + \sigma W_t) (Y_0 + (\mu-\frac{\sigma^2}{2})s + \sigma W_s)\...


2

Step 1: Know your distribution Since $\int_0^t W_s\mathrm{d}s\sim N\left(0,\frac{1}{3}t^3\right)$, we have \begin{align*} S_t &= S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \int_0^t W_s\mathrm{d}s \right) \\ &\overset{d}{=} S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \sqrt{\frac{1}{3}t^3} Z \right) \\ &\overset{d}{=} S_0 \exp\left( ...


5

As @Canardini pointed out, \begin{align*} E\big(I_t^2\big) &= E\left(\int_0^t f(s) W_s ds\int_0^t f(u) W_u du\right)\\ &= \int_0^t\!\int_0^t f(s)f(u)\min(s,u)dsdu\\ &= \int_0^t\left(\int_0^u f(s)f(u) s ds + \int_u^t f(s)f(u) u ds \right)du\\ &= \int_0^t \int_0^u sf(s) f(u)ds du + \int_0^t \int_u^t uf(u)f(s) ds du\\ &=2\int_0^t uf(u) \...


5

Using Fubini's argument, assuming that $f$ is deterministic $$E(I_t^2) = E\left(\int_0^t f(s) W_s ds\int_0^t f(u) W_u du\right)=\int_0^t\int_0^t{f(s)f(u)min(s,u)duds}$$ If $f$ is continuous(even piece wise) you can prove that $I_t$ is normally distributed.


1

There seems to be a consensus that the distribution of daily stock returns lies somewhere between the Gaussian Distribution and the Cauchy Distribution[1][2]. While the former will fail to encompass high volatility events, the latter typically exaggerates their occurence (tails are too fat). I recently implemented a script for calculating european option ...


Top 50 recent answers are included