New answers tagged brownian-motion
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 ...
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