# Tag Info

7

Firstly, $m_T=\min\limits_{t\in[0,T]} B_t = -\max\limits_{t\in[0,T]} -B_t \overset{Law}{=} -\max\limits_{t\in[0,T]} B_t = -M_T$. So, you can either consider the running maximum or minimum. Let $\tau$ be a stopping time and $(B_t)$ a Brownian motion. Then, \begin{align*} W_t =\begin{cases} B_t & t\leq \tau, \\ 2B_\tau - B_t & t\geq \tau, \end{cases} ...

7

Let $(W_t)$ be a standard Brownian motion and $a>0$. We define $X_t=e^{aW_t-\frac{1}{2}a^2t}$. Then, the process $(X_t)$ is adapted and integrable which are the first two conditions of being a martingale. Finally, for any $s<t$, \begin{align*} \mathbb{E}\left[ X_t\mid\mathcal{F}_s\right] &= \mathbb{E}\left[ e^{aW_t-\frac{1}{2}a^2t}\mid\mathcal{F}...

5

Set $Z_t := f(W_t,t)$ where $f(x,t) = e^{ax -\frac{1}{2}a^2t}.$ Then applying Ito's lemma we get $$dZ_t = aZ_tdW_t.$$ This means that $Z_t$ is a local martingale. For $Z_t$ to be a martingale you have to prove that $E \int_0^t|a e^{aW_t -\frac{1}{2}a^2u}|^2 du <\infty.$ For that you can use Fubini's theorem. \begin{align*} E \int_0^t|a e^{a W_u - \frac{1}{...

3

Your solution is correct. Generally speaking, for any $\alpha,\beta\in\mathbb{R}$, the drift $\mu_{X^{\alpha\beta}}$ of the process: $$X_t^{\alpha\beta}:=e^{\alpha t+\beta W_t}$$ will be equal to: $$\mu_{X^{\alpha\beta}}=\left(\alpha+\frac{\beta^2}{2}\right)X_t^{\alpha\beta}$$

2

Note that, for $t>s>0$, \begin{align*} X_t-X_s &= \frac{1}{t}\int_0^t udW_u - \frac{1}{s}\int_0^s udW_u\\ &=\frac{1}{t}\bigg(\int_s^t u dW_u + \int_0^s udW_u \bigg)- \frac{1}{s}\int_0^s udW_u\\ &=\frac{1}{t} \int_s^t u dW_u + \Big(\frac{1}{t} -\frac{1}{s}\Big)\int_0^s udW_u\\ &=\frac{1}{t} \int_s^t u dW_u - \frac{t-s}{t} X_s. \end{align*...

2

Using only words and no equations: Knowing the Variance (or standard deviation) of a Brownian Motion we can calculate the uncertainty in the future position of a particle. Knowing $\sigma^2$ and assuming the particle starts at $S_0$ we can say that $S_T$ will be in $[S_0-1.96 \sigma, S_0+1.96 \sigma]$ 95% of the time. In other words 95% of the trajectories ...

1

You state $X(t_j) - X(t_{j-1}) \backsim \mathcal{N}(0, \frac{t}{n})$. Thus: $$X(t_j) - X(t_{j-1}) = \sqrt{\frac{t}{n}} Z ,$$ where $Z \backsim \mathcal{N}(0, 1)$. Note that: \begin{align} & \mathbb{E} \sqrt{\frac{t}{n}} Z = 0 \\ & \mathbb{E} \left( \sqrt{\frac{t}{n}} Z \right)^2 = \frac{t}{n} \mathbb{E} Z^2 = \frac{t}{n} \\...

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