8

@Ivan's comment regarding the covariances is the key. Consider an equally spaced partition $\Pi_n = \left\{ t_0 = 0, t_1 = \Delta_n, \ldots, t_n = t \right\}$ of the interval $[0, t]$, where $t_i = i \Delta_n$ and $\Delta_n = t / n$ so that \begin{equation} X_t = \lim_{n \rightarrow \infty} X_n, \qquad X_n = \sum_{i = 1}^n W_{t_i} \left( t_i - t_{i - 1} \...


8

I can clarify 100% that $(dw)^2$= $dt$ and recommend you to accept it as a fact. Like any other differential, this differential is defined in terms of its integral: $$ \int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}[W(t_{k+1})-W(t_{k})]^{2} $$ Where $t_{k}=t_{0}+k(t_{1}-t_{0})/n$. Since $$ W(t_{k+1})-W(t_{k})=\sqrt{t_{k+1}-t_{k}}\...


8

We know that $(\tilde{W}_t) := (-W_t)$ is also a Wiener process so $$ E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r] $$ and that implies that $E[W_pW_qW_r] = 0$.


7

The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows: $$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$ where you'll need to find $C$ such that the equality holds (i.e. $C=W_s^3 + ...


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}...


7

What a great question! I've had a go at it below, I'd say I'm about 75% sure of the result I've got to but I'd love feedback from others. I'm going to use the definition of the Ito integral, \begin{align} \int^t_0 \sigma_s dW_s = \lim_{n \to \infty} \sum_{i=1}^n \sigma_{t_{i-1}} \bigl( W_{t_i} - W_{t_{i-1}} \bigr) \end{align} where $t_n = t$. Then, using the ...


6

yes, by definition. The integral is by definition the limit $$ \int_0^T f_s dW_s = lim_{ n \to \infty } \sum_{ s_i \in \mathcal{P}_n } f_{s_i} ( W_{ s_{i+1} } - W_{s_i} ), $$ where $\mathcal{P}_n$ is a partition of $n$ points of the interval $[0,T]$. In your case, for any partition, the sum above is telescopic and always evaluates to $W_T$. so the limit ...


4

If you examine a standard definition of a Wiener process, $W_t - W_0$ follows the normal distribution with mean zero and variance $t$. If you want the variance to be something else, you have to scale it.


4

In this particular case, the simplest way to compute the expected value is to write $\cos(x) = \Re(e^{ix})$ and use the formula for the characteristic function of a Gaussian variable: if $Z \sim \mathcal{N}(\mu,\sigma^2)$, $E[e^{iuZ}] = e^{iu\mu - \frac{1}{2}u^2 \sigma^2 }$ (simply write the expected value as an integral $\int_{\mathbb{R}} e^{iuz} \frac{1}{\...


4

This is wrong! Notice that $dX_t=\mu(t,X_t)dt + \sigma(t,X_t)dW$ is a shorthand for $$\int_0^tdX_s = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s$$ Integrating: $$X_t-X_0 = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s \text{ (eq.1)} $$ If we take expectations, remembering that $\mathbb{E}[\int_0^t\sigma(s,X_s)dW_s]=0$, we have $$\mathbb{E}[...


4

Just wanted to add to @StackG's great answer using a different approach. Please, double-check my solution as well because I'm not 100% sure. Let $\sigma_t$ be sufficiently regular such that $\dot{\sigma}_t \stackrel{def}{=}\frac{d \sigma}{dt}$ is well defined. Then, Ito's lemma: $$ d(\sigma_t W_t) = \dot{\sigma}_t W_t dt + \sigma_t dW_t $$ which in integral ...


3

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*...


3

I am curious about the target of your question. It is rarely put as $$ \mathbb{E}\left[\int_0^{\tau}X_t{\rm d}t\Bigg|X_{\tau}=a\right], $$ because, as per your statement, $\tau$ is defined as $$ \tau=\inf\left\{t>0:X_t=a\right\}. $$ Following this definition, it is a must that $X_{\tau}=a$. Hence it is unnecessary to be conditioned. As far as I know, ...


3

\begin{align*} E\Big(W_t^3-3tW_t \mid \mathcal{F}_s\Big) &= E\Big((W_t-W_s+W_s)^3-3t(W_t-W_s+W_s) \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3+W_s^3+3(W_t-W_s)^2W_s + 3 (W_t-W_s)W_s^2\\ &\qquad \qquad -3t(W_t-W_s)-3tW_s \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3\Big) + W_s^3+3W_sE\Big((W_t-W_s)^2\Big)\\ &\qquad \qquad + 3W_s^2 E(...


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 ...


3

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 ...


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

I believe that the process you postulate has a Beta conditional distribution. If my memory serves me well, I have encountered it in the book by Liptser and Shiryayev "Statistics of Random Processes" as the evolution of the conditional probability in a HMM. This was 10+ years ago, therefore I might be well off. In that case you should be sampling from Beta ...


2

The following is not a proof but some reasoning: If you consider the L2-limits then you see something along the lines: $$ dW^2 = \lim_{n \rightarrow \infty}\sum_{j=1}^n (W_{t_{j+1}} - W_{t_{j}})^2 \rightarrow t $$ For $W_1, W_2$ with correlation $\rho$ this transaltes to $$ dW^1 dW^2 = \lim_{n \rightarrow \infty}\sum_{j=1}^n (W_{t_{j+1}}^1 - W_{t_{j}}^1)(W_{...


2

$ \sigma S $ is in units of dollars per square root of a unit of time. $ \sigma $ is usually quoted as an annual or daily percentage. $ dX ^2 $ is in units of time, as $ E[(dX)^2] = dt $. Here is an online tutorial which you may find helpful. EDIT by kotozna: $\sigma$ has dimensions 1/(square root of time) and $dX$ has dimensions square root of time. ...


2

The proof uses the martingale property of the Ito integral. For an adapted stochastic process $X_t$ such that $$\mathbb{E}\int_0^{t}|X_s|^2ds <\infty$$ we have $$\mathbb{E}\int_0^{t}X_sdW_s =0$$ Now your result follows by setting $$X_t=\frac{1}{W_t^2+1}.$$ To see that the square integrability condition is satisfied note $$\mathbb{E}\int_0^{t}\frac{1}{(W_s^...


1

You state $X(t_j) - X(t_{j-1}) \backsim \mathcal{N}(0, \frac{t}{n})$. Thus: \begin{equation} X(t_j) - X(t_{j-1}) = \sqrt{\frac{t}{n}} Z , \end{equation} 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} \\...


1

The key is on the left hand side. Recall that the differential of log of x is: $d \ln x =\frac{1}{x}dx$ So you get: $\ln x_t-\ln x_0=at$ Which you will need to exponentiate to get rid of the log: $\frac{x_t}{x_0}=e^{at}$


1

The Bachelier model which assumes that price follows a normal distribution is a correct approximation for the Black-Scholes one for short times t. When time is short it's fine to ignore drift because the movements will be more driven by volatility For longer time intervals you cannot ignore drift any more. But the actual distribution of the stock price at ...


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