Tag Info

Hot answers tagged

3

The equation $$\text{cov} \left( \int_0^T a(t) \, dW_t, \int_0^T b(t) \, dW_t \right) = \mathbb{E} \left( \int_0^T a(t) b(t) \, dW_t \right)$$ does not hold. Instead it should read $$\text{cov} \left( \int_0^T a(t) \, dW_t, \int_0^T b(t) \, dW_t \right) = \mathbb{E} \left( \int_0^T a(t) b(t) \, \color{red}{dt} \right).$$ This follows from Itô's isometry ...


2

First of all, there are several typos in your calculations (e.g. it should read $\int_0^t W_s^2 \,ds$ instead of $\int_0^t W_t^2 \, ds$). Your calculation goes wrong when you write $$\mathbb{E} \left( \int_0^t W_s^3 \, dW_s \right) = \frac{\mathbb{E}(W_t^4)}{4} - \frac{3}{2} \int_0^t V(W_s) \, ds = \frac{\color{red}{3t^4}}{4} - \frac{3t^2}{4}.$$ (I don't ...


2

Every continuous local martingale is a continuous time change of a Brownian Motion in a possibly extended probability space. So, modulo some technicalities, all the properties obtained from integration w.r.t. Brownian Motion translates, one way or another, to properties when integrating w.r.t. to continuous time local martingales.


1

Let $f(x) := x^a$ for some fixed $a>0$. Then $$f'(x) = a x^{a-1} \qquad f''(x) = a (a-1) x^{a-2}.$$ Since by Itô's formula $$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s+ \frac{1}{2} \int_0^t f''(X_s) \, d\langle X \rangle_s$$ we get $$\begin{align*} Y_t - Y_0 &= f(X_t)-f(X_0) \\ &= a \int_0^t X_s^{a-1} \,d X_s + \frac{1}{2} a (a-1) \int_0^t ...


1

See here for the fact that $\int_0^t B_u du$ follows the normal distribution with zero mean and variance $\dfrac{t^3}{3}$. See here for moment generating function(MGF) of normal distribution


1

If $X \colon \mathbb{R} \longrightarrow \mathbb{R^{+}}\bigcup\left\{0\right\}$ is defined as $ X(x) = x^2$, then observe that, for $b\ge a\ge 0$, $$ \begin{eqnarray*} ...


1

Note that $$\begin{align*}B_T^3 &=\sum_{j=1}^n B_{s_j}^3 - B_{s_{j-1}}^3 = \sum_{j=1}^n (B_{s_{j-1}}+(B_{s_j}-B_{s_{j-1}}))^3-B_{s_{j-1}}^3 \notag \\ &= \underbrace{\sum_{j=1}^n (B_{s_j}-B_{s_{j-1}})^3}_{=: I_1}+ 3 \underbrace{\sum_{j=1}^n \cdot B_{s_{j-1}} \cdot (B_{s_j}-B_{s_{j-1}})^2}_{=:I_2} + 3 \underbrace{\sum_{j=1}^n B_{s_{j-1}}^2 \cdot ...


1

This is a consequence from the Clarke-Ocone Theorem, and uses Malliavin derivative. See also the Clarke-Ocone formula paragraphe here. If you want a technical reference, see this introductory course, mainly theorem 1 p. 18.


1

Hint: Case 1: $g_1$, $g_2$ and $K$ are simple functions. Then the claim follows from a straight-forward calculation (since the stochastic integrals can be calculated explicitly). Case 2: $K$ is a simple function and $g_1,g_2$ are deterministic measurable (square integrable) functions. Approximate $g_1$ and $g_2$ by simple functions and use step 1. Case 3: ...


1

Hint: The stochastic integral $$M_t := \int_0^t e^{-as} \, dB_s$$ is a martingale. Hence, $\mathbb{E}M_t = \mathbb{E}M_0=0$. In order to calculate the variance of $X_t$ use Itô's isometry. Alternative approach: Since stochastic integrals are martingales, we have $$\mathbb{E}X_t- \mathbb{E}X_0 = \int_0^t a \cdot \mathbb{E}X_s \, ds,$$ i.e. $m(t) := ...


1

$$ Y_t = \mathrm{e}^{2B_t-\alpha t} $$ using Ito $$ dY_t = -\alpha Y_t dt + 2Y_tdB_t + \frac{4}{2}Y_t dB_t^2 = -\alpha Y_t dt + 2Y_tdB_t + 2Y_t dt = (2-\alpha)Y_tdt + 2Y_tdB_t $$ thus martingale requires a driftless SDE thus $\alpha = 2$. Was this the way you performed it?


1

Ito's Lemma: For suitable stochastic process $X_t(t, B_t)$, $$ dX_t{}={}\left(\dfrac{\partial}{\partial t}X_t{}+{}\mu_b\dfrac{\partial}{\partial b}X_t{}+{}\dfrac{1}{2}\sigma_b^2\dfrac{\partial^2}{{\partial b}^2}X_t\right)dt{}+{}\sigma_b\dfrac{\partial}{\partial b}X_tdB_t\,. $$ where $\mu_b{}={}0$ and $\sigma_b{}={}1$ for brownian motion, $B_t$ . From ...



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