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## Hot answers tagged wienerprocess

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

@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} \...

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. $C=W_s^3 + ... 7 I can clarify 100% that$(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 $$W(t_{k+1})-W(t_{k})=\sqrt{t_{k+1}-t_{k}}\... 6 yes, by definition. The integral is by definition the limit$$ \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} ), $$where \mathcal{P}_n is a partition of n points of the interval [0,T]. In your case, for any partition, the sum above is telescopic and always evaluates to W_T. so the limit ... 4 This is wrong! Notice that dX_t=\mu(t,X_t)dt + \sigma(t,X_t)dW is a shorthand for$$\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 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}{\...

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

 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 ...

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

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, ...

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. ... 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 ... 1 Based on the form of your equation, we can consider the SDE \begin{align*} dX_t = \sigma dW_t, \end{align*} where W is a standard Brownian motion. Since, for 0 \leq t \leq T, \begin{align*} X_T = X_t + \sigma (W_T-W_t), \end{align*} based on Feynman–Kac formula, the solution is given by \begin{align*} F(t, x) &= E\left(X_T^2 \mid X_t = x\right)\\ &... 1 The first part has already been answer by @Uditg_ucla, so I am only providing answer of your 2nd part. Rewriting your SDE in more sophisticated way:dS=k(b-S)dt+\sigma S dz$$You want SDE for S^2. Using Taylor series, it can be written as:$$df(S)=f'(S)dS + \frac{1}{2!}f''(S)(dS)^2+\cdotsdf(S)=2SdS+(dS)^2df(S)=2S[k(b-S)dt+\sigma S dz]+\...

1

Not sure I fully understand your question. However, I'd suggest using the Ito's lemma (second equation on wikipedia page https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma) to solve for dY. In both cases, dY will have both a drift term and a stochastic term. The coefficient of the stochastic term will indicate what sort of probability process Y follows. e.g., ...

1

You can use that $f(t,W_t)\in C^2$ is Martingale iff:$$\partial_t f+\frac{1}{2}\partial_{WW}f= 0$$ We get:$$\partial_t f=-3W_t$$$$\partial_{WW}f=6W_t$$ Finally: $$-3W_t+3W_t= 0$$ q.e.d. The proof of theorem follows by writing out $f(t,W_t)$ via Ito formula. Proof of theorem:

1

Note that $\{W_t \mid t \geq 0\}$ is a martingale. Then, for $0<p<q<r$, \begin{align*} E(W_pW_qW_r) &= E\Big( E(W_pW_qW_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q E(W_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q^2\Big)\\ &=E\Big(W_p(W_q-W_p+W_p)^2\Big)\\ &=E\Big(W_p(W_q-W_p)^2+W_p^3+2W_p^2(W_q-W_p) \Big)\\ &=E(W_p)E\Big((W_q-...

1

I think you are on the right track here. You made a sign error in the first line, unfortunately: $$E[W_p W_q W_r] = E[W_r W_p^2 + W_pW_q^2 - W_qW_p^2]=\\ E[(W_r-W_q)W_p^2]+E[W_pW_q^2]= E[W_pW_q^2]$$ The first term is $0$ by independence (as $p<\text{min}(r,q)$ and the square does not affect independence). To take care of the second term we do the ...

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