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9 votes

Calculate value of Integral of Wiener process $\int_{0}^t e^{\lambda u } dZ_u$

This is just a Wiener integral (stochastic integral with respect to a Brownian motion and deterministic integrand), hence a centered Gaussian random variable with variance $$ \int_0^t{e^{2\lambda u}\...
siou0107's user avatar
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8 votes
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Integral of Wiener process w.r.t. time

@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 ...
LocalVolatility's user avatar
7 votes
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Probability Density Function of a Wiener Process Minimum

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 $\tau$ ...
Kevin's user avatar
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7 votes
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conditional expectation of stochastic integral

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{...
StackG's user avatar
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7 votes
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Proof that $\exp(aW(t)-0.5a^2t)$ is a martingale

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 ...
Kevin's user avatar
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5 votes
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Differentiating Wiener process

Let $\text dX_t=\mu_t\text dt+\sigma_t\text dW_t$ be an Itô process. Itô's Lemma tells us $$\text df(t,X_t)=\left(f_t+\mu_tf_x+\frac{1}{2}\sigma_t^2f_{xx}\right)\text dt+\sigma_tf_x\text dW_t.$$ You'...
Kevin's user avatar
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4 votes

Integral of the OU (Ornstein Uhlenbeck) process conditioned on hitting a threshold value for the first time

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=\...
hypernova's user avatar
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4 votes

Geometric Brownian Motion: Why is the Wiener process multiplied by volatility?

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 ...
Matthew Gunn's user avatar
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4 votes

conditional expectation of stochastic integral

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{\...
Gabriele Pompa's user avatar
4 votes

Integral of brownian motion wrt. time over [t;T]

The last integral is correct as $$\int_t^T W_s ds = \int_t^T (T-s) dW_s \sim N\left(0, \int_t^T(T-s)^2ds\right) = N\left(0,\frac{1}{3}(T-t)^3\right).$$ Ref. Arbitrage Theory in Continuos Time (Björk, ...
Landscape's user avatar
  • 548
3 votes

Integrated Brownian motion

Your derivation is correct. Even if we fix your obvious typo the formula $$ \textstyle\int_t^TW_s\,ds=\int_t^T(T-s)\,dW_s $$ is wrong. There is no doubt that \begin{align} TW_T&=\textstyle\int_0^...
Kurt G.'s user avatar
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3 votes
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Independence of increments of the stochastic process $\frac{1}{t}\int_0^t u dW_u $

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 ...
Gordon's user avatar
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3 votes
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Calculation of a process's drift

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: $$\...
Daneel Olivaw's user avatar
3 votes
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What is the difference between standard deviation, volatility and quadratic variation?

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 ...
nbbo2's user avatar
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3 votes

Are two stochastic processes independent if the Wiener processes inside are uncorrelated

[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. ...
Quantuple's user avatar
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2 votes

Two Wiener process under same martingale measure Q

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 \...
Richi Wa's user avatar
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2 votes
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Expectation on a function of Wiener Process

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 $$\mathbb{E}\int_0^{t}X_sdW_s =0$$ ...
fes's user avatar
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2 votes
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Calculate value of Integral of Wiener process $\int_{0}^t e^{\lambda u } dZ_u$

I'll try to use Ito's Lemma to come up with a solution. Ito's lemma states that: $$F(Z_t,t)=\int_0^t\left(\frac{\partial F}{\partial u}+\frac{\partial F}{\partial Z}a+0.5\frac{\partial^2 F}{\partial Z^...
Jan Stuller's user avatar
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2 votes

Arbitrage portfolio example

You can choose an arbitrage for the classical 1-dimensional Black-Scholes model, and not use $S_2$ at all. Such an arbitrage is e.g. presented in Example 3.5 in "Fractional Processes As Models In ...
mortenmcfish's user avatar
1 vote

Sample Wiener process constrained to open (initial), high (max), low (min), close (final)

You can do this (within some small $\epsilon$) using the reflection principle. Let's take the simpler case, where you only want to constrain the maximum. Adding the minimum is a fairly simple ...
Brian B's user avatar
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1 vote
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Moments of the integral of the exponential of Brownian motion/Normal random variable

You can use the formulas of Baxter and Brummelhuis (2011). They provide moment formulas for the time integral of geometric brownian motion X_t, using divided differences. The full derivation is in the ...
Achrbot's user avatar
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1 vote
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Regression of stochastic integral on Wiener process

By definition, $$ {\mathbb Cov}(M_t,W_T) = {\mathbb E}[M_t W_T] - {\mathbb E}[M_t] {\mathbb E}[W_T] = {\mathbb E}[M_t W_T] $$ since ${\mathbb E}[M_t] = {\mathbb E}[W_T] = 0 $. We now consider the ...
Gabriele Pompa's user avatar
1 vote

Differentiability of solutions of a stochastic differential equation

Rather non rigorously, $\frac{W(t)}{t} \sim N(0,\frac{1}{t}) $ if $t \to 0$ , we can see the variance goes to infinity. Hence Ito process is not differentiable.
Preston Lui's user avatar
1 vote

How to Evaluate Expected Value powered 4 of a Wiener Process?

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:...
Colin T Bowers's user avatar
1 vote

Can anyone explain to how Hull get from stock returns to continuously compounded stock returns?

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: ...
Magic is in the chain's user avatar
1 vote
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Interpretation of IV and its use in stock movement prediction

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 ...
Ezy's user avatar
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